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Generalized totients over certain classes of n-square matrices

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Generalized totients over certain classes of n-square matrices
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LeVan, Marijo O'Connor, 1936-
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Language:
English
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iii, 53 leaves : illus. ; 28 cm.

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Subjects / Keywords:
Matrices ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF

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Thesis:
Thesis - University of Florida.
Bibliography:
Bibliography: leaf 52.
General Note:
Manuscript copy.
General Note:
Vita.

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Full Text
GENERALIZED TOTIENTS OVER CERTAIN
CLASSES of n-SQUARE MATRICES
By
MARIJO O'CONNOR LeVAN
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA August, 1964




ACKNO14LEDGMENT
To the members of her committee, and especially to Dr. John E. Maxfield, chairman, the author wishes to express her gratitude for their assistance and friendship.
ii




TABLE OF CONTENTS
Page
ACKNOWEDGIKEN .. .. .. ....... . .. i
INTRODUCTION .......... .1. ...
Chapter
I. ELEKI TARY THEOREMS 3
ii. DIVISION AND FACTORLIZATION .'7 ...
III. THE FUNCTIONpu(h)............ 18
IV. TE FUNCTIONS n(A), qAA),I n*(A) AIND R*(A). 25
V. TE FUNCTIONS CJk(A), Ink(A), CPk *(A) AND
k *(A) .. .. .. .. ..... .. .. . ...3
VI. THE FUNCTIONS n (k)(A), 4~(k)(A), (k)(A)
AND fl'(k) (A) ................. 39
VII. THE FUNCTIONS )z a(A) AND q(A) . . . 4
BIBL1IOGR.APHY . .. .. .. .. ... ... .....52
BIOGRLAPHICAL SKETCH ....................53




INTRODUCTION
Euler first investigated the totient in 1760 in
connection with his generalization of Fermat's theorem.1 He defined P(m) as the number of integers not exceeding m and relatively prime to m Among the more elementary (and interesting) properties of the totient are
mm m-1
(1) f(pm) = p pm- where p is prime
(2) if (a,b) = 1, then /(ab) = /(a) 0(b)
(3) i q(d) =m
dim
(4) F1. ) =~m
dim
where a(m) is the M8bius function.
There have been many generalizations of the Eulerfunction1 most of which generalize one of these four properties. R. D. von Sterneck 12 considered the function Hk(m) = ( (nl) (n2) (nk) i.e., the summation of the
[n.]=m
products of the totients of all collections of k integers each less than or equal to m whose least common multiple
L. E. Dickson, History of the Theory of Numbers
(vol. 1; New York: Chelsea Publishing Co., 1952) pp. 113-155
2Eckford Cohen, "Some Totient Functions," Duke Mathematical Journal, XXIII (1956), 515-522.
1




2
is m C. Jordan1'2 considered the function Jk(m) of the number of distinct k-tuples from a complete residue set (mod m) whose greatest common divisor was prime to m. E. Cohen2 considered the function Yk(m) ,the number of elements in a complete resude set (mod mk) which have no
k 2
k% power divisor in common with mk He proved that these three functions are equal.
This paper is a result of seeking the number of
unordered sets of k integers whose greatest common divisor is prime to m It generalizes the totient to certain k-square matrices, preserving properties (1), (3) and (4). In Chapter V it further generalizes Cohen's (and hence von Sterneck's and Jordan's) function to matrices.
Chapter VII generalizes Cohen's paper "A Class of
Residue Systems (mod r) and Related Arithmetical Functions
I
1 L. E Dickson, History of the Theory of Numbers
(vol. 1; New York: Chelsea Publishing Co., 1952) pp. 13-155.
2Eckford Cohen, "Some Totient Functions," Duke Mathematical Journal, XXIII (1956), 515-522.




CHAPTER I
ELEMENTARY THEOREMS
Let p be an odd prime and 5(p) the field of integers (mod p) represented by the least non-negative residue set. For each natural number m define 6(m,p) as the set of all m-square matrices over 4(p) whose characteristic equations have m roots.
Following the usual notation, let Im be the
m-square identity matrix and 0m the m-square zero matrix.
Define Jm+l = I A matrix N in 6.(m,p) will
be a regular Jordan form if M = NM1 N2 '-Ms (where A (DB is the direct sum of A and B), each M. = a.Im + 11
s
J m. = m a a 2 as and whenever a. =
' i=l
ai+ mi mi+1 (m,p) is the set of all regular Jorden forms in &(m,p) .
Throughout, M will be a (fixed) element of 4(m,p) with the additional requirement that 1 4 a. p-1 for all i
1
Let J(M) be the subset of O(m,p) consisting of
all matrices T having a Jordan form T which will satisfy
3




the matric inequality I T P where
m
s
P = G [(p-1)Im + Jm.] ( refers to the direct sumi=l
mation) and X = (x) Y = (y) iff xi. y. for all
i and j
Def. 1: Let S,T EJ (M) with their required Jordan forms S = (s ij) and T = (t ) respectively. Then S T (mod M,p) iff s t (nod ai) for all i and j
Let = BE (H):Im B & M] such that B,C E
implies B / C (mod M,p). Then every element of o(M) is congruent (mod M,p) to exactly one element in ?2. Hence N is a least non-negative residue set (mod M,p). If we denote the number of elements in ?9 by n(M) and 7h' is any collection of n(M) incongruent (mod N,p) elements of J (N), then Y?' is a complete residue set (mod M,p). Theorem 1: Congruence (mod M,p) is an equivalence relation. Proof: Let X1,X2,X3 2 -(M) with their required Jordan forms S = (s ij) T = (t ij) and U = (ui) respectively. Then we have
(i) s. s (mod a.) for all i and j
(ii) if sij tij and t.. uij then s.ij ui..
(iii) if s t then t si.. j




5
Theorem 2: If B C and D E (mod M,p) then there
exist matrices S and T in & (M) such that
(i) S D T = E (mod M,p)
(ii) B + S C + T (mod M,p)
(iii) BS SB CT E TC (mod M,p)
Proof: Let X = (xij) Y = (yij) Z = (zij) and W = (wij) be the Jordan forms of B, C, D and E, respectively. Then there exist non-singular matrices P and Q such that B = PXP-1 and C = QYQ- Define S = PZP-1 and T = QWQ-1
By definition S and D have the same Jordan form, hence are congruent (mod M,p). Similarly, T E (mod M,p).
Hence applying the transitivity proved in theorem 1, S mD E T .
Since X B C Y we know x.. yij (mod ai) 10j da.)
for all i,j Further Z D r E W (mod M,p) so that z. w. (mod ai) for all i,j Hence
B + S = PXP-1 -1
B+S=PP + PZP
= P(X + Z)P-1
= P(xij zij)Pij 10
P(yij w.ij)P-1 (mod M,p)
ij 10j
= P(Y + w)P-l
Q(Y + W)Q-1 (mod M,p)




6
-1 -1
= QYQ + QWQ
=C+T
The proof of the third part of this theorem will be found in the proof of lemma 1.




CHAPTER II
DIVISION AND FACTORIZATION
Def. 2: Let B,C E(m,p) such that Im B 4 C (p-2)Im
+ Jm Then BIC (mod C,p) iff there exists DE, (m,p) such that BD a C (mod C,p) and BDi- IC I Lemma 1: Let M be as in the last section and B4e,(m,p),
a
Then BIN (mod M,p) iff B f3Bi where Bi = biIm +
i =1 i
i=l
Kmi b iai and Omi? Km* : Jm.
Proof: Let B satisfy the conditions of the lemma.
s
Define D = &Di where Di = i Im + Km such that
i=1i i
i=1
bid= a and K + Km= Jim (matric addition). If
i i I i .i
s
BD = F = (fj) then F = E Fi where fi = bidi = i=1
kiai for all i fii+1 = 0 if mii+l1 = 0 fii+l is bi of di if mii+l = 1 and for all other i,j f =ij 0 .
Since (bi,p) =- (di,p) = 1 for each i in each block, Fi, there exists at least one kxk (k < mi) minor of K Fi whose determinant is a non-zero scalar (namely the minor
7




8
formed by taking the first k elements of the 2nd through
m.
(k+l)st columns). Also, I Fil = (X-ka) Hence
m i 1 Hence
the elementary divisors of Imi- F. are k-1 ones and 21
m. s m. s m.
(X-kiai) But F = (ka) i and = a i
i=1 i=1
Therefore, k. = 1 for all i and the Jordan form of F
1 1
is NMi for all i .1 Therefore, BD = F M (mod M,p) and B IM (mod M,p).
Conversely, assume BIM (mod M,p). Then there exists D (m,p) such that Im 5 D 4 M and BD = F G M (mod M,
s
p) where F 2 D(aim + aJm.) Let F = (f
i=1 i 1
Then fl1 = all since f11 I a1 and f11 a al (mod al). Further, B = (bij)6 (m,p) and D = (d ij)e (m,p) imply that f11 = bldll Hence b1lI al Now if ml 1 ,
f12 / 0 since F1 and M1 must have the same Jordan form in 0(p) But fl2 = blldl2 + bl2d22 Hence either bl2 = 1 in which case b22 =- bl ,or dl2 = 1 in which case d22 = d11 or both. But al = f22 = f11 = d22b22
hence if either b22 = b11 or d22 = d11 the other equality must also hold. By repeating this argument mI 1 times we see that bii = ai for i = 1,2,...,ml and bii+l = 1
11
1F. R. Gantmacher, The Theory of Matrices (vol. 1; New York: Chelsea Publishing Co., 1959), pp. 159-151.




9
or 0 for i = 1,2,'*,ml-1 Moreover, since mm1ml++l = 0 we know that fmlml+1 = 0 and since 1 & b114 p 2 ,
1 t d11 C p 2 and 1 blld11 4 p 2 hence 1 bl +
d z p 1 since fmml = d mb + d b
11p ml1ml1+1 Mmim mlml+1 Mmiml+1 m 1ml1
= d 1bm m + bldm m 1 0 (mod p) we have bmm +1 =
1 ml1 1 m1+ 1l1
dm ml+ = 0 Hence we have described B1 = b1 ml + Kml where blia1 and Om & K Jm and B = B1 + B' .
1 ml 1
By repeating this argument on the remaining s 1
blocks we can similarly describe B' and the lemma follows.
Also, since the proof in no manner depends on the
location of the units on the super diagonal it is clear that BD 2 DB = M (mod M,p).
Returning to part iii of theorem 2 we have (in the notation of that theorem):
SB = (PzP-)(PxP-) = P(ZX)P- PXZP- = BS
SB = PZXP-1 QZXQ-1 QYWQ-l = CT and CT = (QYQ- )(QWQ-) QWYQ = TC .
Theorem 3: If B,C E (mp) BIC (mod C,p) and q is any odd prime such that' ii 2 for all i then BIC
(mod C,q). r
Proof: Since C & (m,p), C G= Ci where Ci =
i=1




10
c iIm + J and ci! p 2 for all i But, by hypothesis,
0 i q 2 for all i Therefore C EJ(m,q) Applying
r
lemma 1, if B/C (mod C,p) then B = Bi where Bi =
i=1
b.I + K where bi(a. and 0 m K 0 J Then
i ai. i i
b ci q 2 for all i and therefore B E-(m,q) Then again applying lemma 1, B,C t J(m,q) and B having
the form of the lemma show that BIC (mod C,q).
If ) (m) is defined as the set of all distinct Jordan forms over the natural numbers, theorem 3 allows the following restatements of definition 2 and lemma 1: Def. 2'. If Im 4. B 4 C where B,C E4(m) BIC iff there
exists D :A(m) such that BD F Z C (mod C,p) for any
odd prime p such that ci_ p 2 for all i and
r
F 6 I (CiIm + i Jm ) or B = PTB'D where B/C (P a
2.li i
i=1
permutation matrix).
s
Lemma 1. BIM iff B = (biI m + Km) where bi a
i=1 2
and Om ii Kmi Jmi for all i.
s
Def. 3: Let BIN B = (bi m + Km) such that
i=l
q q
mi = n where n = mi and n1 < n2< ...gn.
i=1 1=1




s
ll
Then M/B is defined to be the matrix D = G (dI + i=1
/ /
Ki) where bid. = a and K + K J .mi
m i 1 i m m m.
1 i i 1
It follows from lemma 1 that B M (mod Mp), but,
B (mod 14,p), but,
unless mi = 1 for all i there is at least one C such that BC M (mod M,p) and C X S (mod M,p). Def. 4: Let B,C e,-J(m) Then B and C are relatively prime [(B,C) = Im] if the only matrix which divides both B and C is the identity matrix.
Note that if (B,C) Im then one of B and C m-2
must be diagonal, or 12 I1 would divide both.
i=l
Convention: Whenever the determinant of a matrix is taken, it will be taken in the ring of integers. Def. 5: Let B Ex (m) Then
(1) B is a unit if IBI =1
(2) B is a prime if B is not a unit and DIB implies
D = I or D =B .
m
Theorem 4: Let P E J(m) Then P is prime iff P is diagonal and IP) = p,a prime. Proof: Assume P is prime. Then since P is not a unit, for any unit V I 'm., VP Hence P must be diagonal,




12
m
i.e., P xi Suppose for i j xi 1 and x 9 1
i=1
i-1 m
Define Q @ xk @ 1 ( P xk Then Q P since
k=l k=i+l
qii = 1 pi / 1 ; Q Im since qjj 1 but, by lemma 1, QP Hence there cannot be two distinct xi j 1 Let i-1 m
xi 1 x = ab Define R = 1 (G aQ ( 1 and
i-1 m k=l k=i+l
S = 1 ab 6 1 Then RS= P Hence one of
k=l k=i+1
R and S must be P and the other, Im Therefore x. = ab implies one of a or b is x. and the other 1, i.e., xi is a prime. Hence if P is prime, P is diagonal with one element a prime number and the others all 1, i.e., jPI = x a prime.
Conversely, let P be diagonal and (PL = p a prime.
Then P must have p as one entry and 1 as all other diagonal entries. Assume p is in the first row. Then,
m-1
by lemma 1, if QIP Q = x G 1 where xip But
i=1
then x = 1 or x = p Hence Q = Im or Q = P .
Def. 6: A non-unit D is a primitive divisor of M f(m)
if DIM and for all E E J(m) if EID and EIM then
E = D or E = Im




13
s
Theorem 5: Let M J(m) M = Z (aiIm + ) Then
i=1
M has a unique factorization into matrices V and D such that V is a unit and D is diagonal and VD M (mod M, p). Further, if, whenever i j (aa ) = 1 then D
n1 n2 nr
has a unique factorization D= P1 2 n. r
r n.
where each P. is a primitive divisor of M and V ( r Pi i=l1
M (mod M,p). s
Proof: Let M = where M i = a.I Define
i=l
s s
V= ) (Im. + and D= B a Im Then by lemma
i=l m i=l
1, V(M and DIM so this is a factorization of M ,
s
provided VD N (mod M,p). YD = ( (Imi + mi))
i=l
ai mi (aiIm + a Jm) (aIm + Jm M
i=1 i=l -m i=l
k
(mod M,p) since (ak ,p) = 1 for all i and for all k .
Suppose VE N (mod M,p) where V is a unit, V/N ,
and E is diagonal, E/M Then, by lemma 1, V = S S
& (Im. + Km.) where Om. ] Km ~ m and E = d i .
.2.l 2.1 1 1 m mi




s s
where dia E = e (dim + diKm) P 3
i=l i=1
(dilmI + Km) (mod M,p) since (dik,p) = 1 for all i and
1 1
k Hence if VE M (mod M,p), d.I + K = a.I + J r m m i m. im.
1 1 1 1
for all i; i.e., d. = a. and K = Jm. for all i .
1 1 m. m.
1 1
'-I
Therefore V = V and E = D and the factorization is unique.
Suppose (aia.) = 1 for all i = j Let a =
n.
n. n. ir
nil ni2 1iri
Pil Pi2 *. Pir Then if i = j Pik = Pil for all k and 1 For each j,k define Pjk =
j-1 s
f Im. &Pjk I m. &m Then Pjk M by
i=l 1i=j+l 1
lemma 1, and Pjk is primitive by the same lemma. Further,
n j-1 s s
since Pk a Im m if ri =
r i=l i=j+l i=1
S1.
r 7 Pjk = D ,we can renumber the Pjk to get the
k=l j=l
result of the theorem.
def
Def. 7: For m 1 define Z(m) =- ni : each n
i
is a natural number and a-= m i.e., o is a partition of m Define s(m) as the cardinality of the set Z(m) and f(O-) as the number of terms in the series .




15
ai xl Ci2 aiti
Theorem 6: Let M F (m,p) and ai = Pil Pi2 Piti
where (ai,a.) = 1 if i # j Then the number of divisors of M is
S
i (1 + ail)(l + Ui2) ** (1 + it)s(m )
1=1. 1
Proof: Applying lemma 1, BIM if and only if B =
S
S(biIm + Km ) where biai and 0m K m m i=l 1mi
By theorem 5, B may be uniquely factored into the product of a unit and a diagonal matrices, U = s s
, (Im. + Km.) and D = b bIm By applying lemma
i=l i=l
s
1, there are T s(m ) distinct U which may be chosen and,
i=l
ti
for each i there are (1 + d.ij.) choices for b. for
j=1
each U The theorem follows. Def. 8: If DIM and (D,M/D) = I then D is a
m
canonical divisor of M Theorem 7: If D is a canonical divisor of M and EIM there exist unique E1 E2 e (m,p) such that E1E2 E
(mod M,p), E ID and E21M/D .
Proof: Since (D,M/D) = Im one of them must be diagonal. We may assume this is D Then, applying lemma 1, we have




16
S 5
D f> diIm ; /f I + Lam O m. Lm J
Then by Theorem 5 M V1N, (mod M,p), N/l V2 2 (mod MY,p)
S S
where V1 = L'ft(Im + im. Nli G 2ZaiIm.
i=1 =
S S
V2 = fG(Im + Tm.) ;and N 2 = F(im
i=l 1 =1
Then DNM/ = D(N2V2)=(1 di im.)L(i ~im)
(L (Im + Laj) (7 dIm. ) i ~eim
i ii
we have iN2 =N and V=V Therefore Ta =J and
d a. for all i .Also, since Di is a canonical divisor
of M, (di, 'l 1
Laet Elm Then by lemma l and theorem 5, E UP
(modNM,p) where U=/Z (I~ +K ) Omi
1i 1 1 m im




1~7
5
and P= e iI~ i a and this factorization is
unique. Since e i ai d = a.i and (d.i,6j 1 1 there
exist unique bi,f3. such that bir3. = ei I bi\d and
5 5
*Define E= G bZ @bI miand E2
(f3.IM + K ) Then E 1 ID and E21M/D and E 1E2
E~b i I m (PiIm.i + K m bim+
b ) K (mod Njp) *The uniqueness of B1and E2
follows from the uniqueness of U and P and of b.i and




CHAPTER III
THE FUNCTION p.(M)
DefV. 9: Let M F/(m,P). Then
(2) if MN 12 ** PtP such that P.i
is prime for all i ; Pi P P. if i j ; if' ijj t t then IPJl # IP. ; if j >, t + 1, there is an i t such that jPij =1 Ii I; then )i(N) = (-l)
(3) for all other N )u(M) = 0
Proposition 1: If Di is a canonical divisor of' N then u(M) = i(D) ui(m/Dl)
Proof: If M is not diagonal, then either Di or N/fl is not diagonal, and ju(N) = u(D) )a(N/fl) 0 .
Let N = al&-,qam. Then Di = dl~d M and N/f
E,9l-6''F where =i a~ and (dij(67L) = 1. If p is
a prime such that p 2 Jai then p 2 Id i or P *~ Then if
i -l M-1
P= 7,( Cfi E) 1 ,r21N and P [ or
j=1j=iil
P2 Ill/fl Hence p(N) = u(fl) pQ4M/) .Assume, then, that
a= Pil*Pi2* ...** where p. :f Pi p. if j 0 k
18




19
D =P 1P2*P tP tl* where P. 4 P. if i j
and JPi 1 0 ip. if ij t and if j > t + 1 there exists i 14 t such that IP il = 1P. And N/f QiQ2'"" *Q q' Qq+i* Q. where 4: if i *j;
if i q 1 11:1Pi) V \P.\ ;and if j g~jl there exists
an i 1< q such that 1P.1 = P. Then )a(D) ja(M/D)= (-1Jp~ + Ni = N' = Pl tQl Q*P *t+
QQ+1 ... Then for any i, j, k, 1 IP ) fl'%\
)Pil + Qk1 and J~k1~t'~l (by assumption) ; if ijj c q + t
then I P9 1Q since (D,N/D) = Im and if i q + t +1
there exists j q + t such that IP il = 1P I or kil
*Therefore )a(N)'= (-l)q+t = u(D) p(M/D)
Proposition 2: Let N 9 Z (ailm + Km ) where
i~fi
if i j j Then rj.u(D) if il l=
DIN Lo if INI #1
D)NI
Proof: Let s =1 Then DIN DIN and u(D) *0 imply
that Di dI M where dia 1 *Hence
__ DTa D u() Ii al -1
DI1*I ~ l fa 1
DRd.Ia 1m d




20
1 if IN Mk 1
DIM 0 if )w MI 1
DIR
Assume the theorem true for all s e- k 2 .By lemma 1, if DIM DIE and yi(D) j0 ,then D= k k-i
d ~I' where d~ia- If M, =a~ +1
k-i
G(9k anid NM 1(ailm + Km9 clearly )i(D)
k-i D jni
LuD)and if M2 2 -1 Q 0(a k Mk + Km) k n
M / (ak + ) then,. l() aD
(a k + ik DIN2 DI
D2 2IN
k k-i
For each D = di define ID 2 Od I m )I
il1i=l 1 mk
k-l
and ID2 =' Im) d k, Then D = D 1D2 and
(D,)D2) =Im u ~ i~ = tm~(Dl)p(D 2) u K~(Dl))
DIN D1lm N1M
DIR D2JR1 D 11 M1
D21 N2




21
1 if IM 1 2 = 1
- 'U(E) liI%(F) = 2f1Y9=11=1
EI1 I2 0 otherwise
EM1 2
But IM11 = IN21 = 1 iff IM! = 1 Hence if the theorem is true for all s k 1 it is true for s = k Theorem 8: Let M N (m,p) such that if i t j then (ailaj) = 1 If f,g both map a(m,p) into the integers
g(M) = o f(D)C f(EN) = / u(D)g(M/D)
D\M ENU D\M
where M s UN (mod M,p) is the unique factorization of M into a unit (U) times a diagonal (N). Proof: Assume g(M) = f(D). Then, since u(D) = 0
DIN
unless D is diagonal:
u(D)g(M/D) = E u(D) f(E)
DIM DI E IM/D
EM
f f(E) uD
EIN DIMN/E
DkM
= f(E)
E tM
E




22
f f(EN)
E1U
since, by proposition 2, p(D) =0 unless II'M/Ek
D)M/E
Conversely, assume If3 f(EN) p j(fl)g(M/fl)
EJ.U D\N
If (D) =f(EF) DIX FIX I
F diag
F-=F,) ~(TF/D) g(D)
FtN DIFU
F d.iag
since )a(UF/D) = 0 if UF/D is not diagonal
f (D 4 (UF/Dl)g(D)
W1 FG=M/U DlFU
FTJ/D diag
g(D) 7, u((N/GJD)
DIM D!N/G
N/fl diag GIN
(G aiag)
By proposition 2, Ljaq/GD) = 0 unless =1
so




23
(fl) =g(D) =g(N)
DtIM DIM
N/l diag
I M/D'i= 1
since M/fl diagonal and \N/D\, = 1. imply M D. Def. 10: D is a A-divisor of N(D IAN) if N/fl is diagonal. Theorem 9: Let N F_ (m,p) such that (aija)= 1 whenever i /j .If f,g map (m,p) into the integers.
f (M) g g(D) g(N) = y(N/D) f(D)
DIAN DIN
Proof: Assume f(M) = Z g(D) .Then
D in,
since if D 11AN DIN then )i(M/D) =0
DIN D1A EA
AM
DJM
By proposition 2, EF g(/) =0 unless tN/E\ 1
E 7 g(.E) = g(M)
AM
W.NEI= 1




since EjIM and IM/f =l1 imply MN=E.
Conversely, assume g(M L u.(M/D)(Dl)
Then
g(D p(2 D/E)f(E)
f& (E) 1
E 1ANi ElmN
EINAH
applying proposition 2. u( 0 unless JN/H\ 1
Therefore IEI = 1 and
P1u(M/D)f(D) 7-1' f(E) f(N)
IE) = 1
In case m 1 (i.e., N is a 1-square matrix),
theorems 8 and 9 both reduce to the Mdbius inversion formula.




CHAPTER IV
THE FUNCTIONS n(A), f(A), n*(A) AND f(A)
Convention: For the remainder of the paper, A, (m,p)
will be a one block Jordan form, i.e., A = aIm + J Def. 11. In definition 1 a, a complete residue set (mod A,p) was defined, with n(A) the number of elements in 6 .
011 = { E 0j:EIX and ElA implies E is a unit is a partially reduced residue set (mod A,p). The number of elements in 11 is denoted by 1(A) Lemma 2: Let D range over the A-divisors of A For each d define D = X : X ranges over ll
Then UD D is a complete residue set (mod A,p) and if D X D' then qD l = *
Proof: Let B B = B1 ~ B2 ... Bs where Bi
biIn + J If (bi) refers to the greatest common
1 i
divisor of the set b : i = 1,...,s j let d =
((bi),a) Define ci, i = 1,...,s by bi = doi and D = a In Define C=C1 ... s suchthat C
n= m 1 C Cs~ ,, ~ ta
bi
c iIn + J. Then ((ci), ) = (() ) = 4 = 1 so
that C ,1 Further C A = [(c + ) ~ *..
s1 1
(csn n )] = 30~ (dciIn + dJn)
s s i=1 i
25




26
s
= (dciln. + Jn.) -- B (mod A,p)
i=l 1
so that BE JD and DD is a complete residue set (mod
A,p).
'A
To prove the second assertion, assume B C D1
C2 A (mod A,p), where D1 and D2 are A-divisors of C2 D22 A, and C01,02 in partially reduced residue sets (mod D1,p) and (mod D2,p) respectively. In particular, both A/D1
and A/D2 are diagonal. Let A/D1 = Im and A/D
2 m Then
A
(1) c D5 = [(1li + Jk) ... a k+ ik )]16m
i 1 r r
r
-- (i 1Ik. + ) (mod A,p)
i=l 1 i
(2) c2 A2 E l + s... O +I
s
- (S+ 2J4) (mod A,p)
i=l
Let d 2d2 = a. Then D =d Im + Jm and
S11 2 2 m
D2 d2Im +m Since C1 is in a partially reduced
residue system (mod DlP), ((i1),dl) = 1 Similarly, ((Bi)d2) = 1 .
Now, since B = 0 2 *A (mod A,p) we know
S1 2 2
that in (1) and (2) r = s and the blocks of (2) can be




27
rearranged so that (cclk i 2I + )
for all i Assume that this rearrangement has been made.
Then ((ai j),1dl1) = 1 and ((Pi 2)'2d2) 2 but a 1 i62 for all i and 1 d = 2 d2 = a Therefore 1 = 2 and d = d2 Therefore i = ai
for all i Therefore D = D2 and C1 = C2 and the
representation of B is unique.
As an immediate consequence of this theorem one obtains
Theorem 10: (*(D) = n(A) .
D IAA
By applying theorem 9 to theorem 10 one obtains Theorem 11: L?*(A) = U2Ea(A/D)n(D) .
DIA
Cor. 11.1: If a is prime
*(A) = n(A) s(m)
Proof: If a is prime, DIA,u(A/D) / 0 then
D = A or D = Im + Jmp u(A/A) = u(Im) = 1
u(A/(Im + jm)) = u(alm) -1
(p*(A) = (A/D)n(D) = n(A) n(Im + Jm) = n(A) s(m)
D /A
mm
Cor. 11.2: If a = p m 1 define B = pm-Im + m
Then *(A) = n(A) n(B) m
Then #*(A) = ulA) ulB)




28
Proof: If a = pm DIA, and )p(A/D) 4 0 then D = A or D = B u(A/A) = 1 and u(A/B) = 1 The corollary follows.
Convention: Let A* = aI m Def. 12: Define Qj,= X : (X,A) = Im and
-- m
X= XCL :Im 6 X 5 A*j [ is a complete residue set (mod A*,p)]. The number of elements in a* is denoted by n*(A); in Z', by V(A) Lemma 3: Let D range over the A-divisors of A For each D, define = X : X W 01 Then U 1 is
-) DD D D
a complete residue set (mod A*,p) and if D 4 D then
JD D Ll
Proof: Let X i* Then X = x1 x2G... Oxm Let U= ((xi),a); &d = a; yi6= xi Define D = dIm + Jm
x./
and Y = ylOY2. O*ym Then ((yi),d) = (( i ) = 1 ,
m m
hence Y And Y = yi i= X
i=1 i=l
Therefore X cD and UDD is a complete residue (mod A*,
p).
To prove the second assertion, assume X = Y1 A
k 1
2 A where D1,D2 are A-divisors of A and Y 2 are Y2 D22 l2
in reduced residue sets (mod D1,p) and (mod D2,p), respectively.




29
Then Y 1 Dyi1Y = L( y ; Di = d%1 + J
i:=l i=l
A/D1 = & I d&-= a; D2=d1m+J A/D 2 =&9I ;d6
a. Yl~ A Y 2 A Yi2Hence,
=
by rearranging the diagonal of' Y2 A one obtains
2 D2
yi = yLC5 or all i *But (YlDl) = (Y2,D2) = I
Therefore ((y1i),dl) = ((y2i),d2) =1. But ((61y11), 6ydl) 1~i ((6'2y2i) I S2d2) =652 1 Thr2oeS
Hence di = d2 and y1i = y2i f'or all. Therefore the representation f'or X is unique.
As an immediate consequence of' this lemma, one obtains Theorem 12: n *(A) = C, q (D)
Di 1J A
Applying theorem 9 to theorem 12, one obtains: Theorem 13: Y-(A) = PiAA/D)n*(D)
Cor. 13.1: If' a is prime
(f (A) =n* (A) 1
Proof': If' a is prime, fl/A and p.(A/D) # 0 then Dl=A or D=Im + .m uAA = 1 and juA(m+ Jm
- 1. q(A) = ).(A Di) n*(D) = n*(A) n*(Im + J )
-n*(A) n(I) m n*(A) 1




30
Co132'Ia =p mn-ll J
Cor. 13.2: If a = p m > 1 define B = p -m + Then L(A) = n*(A) n*(B) Proof: If DIA (A.D) 0 then D = A or D= B y(A/A) = 1 u(A/B) = -1. The corollary follows. Theorem 14:
(i) *(A) = s(rl)s(r2)...s(rt)
mlIr .. mtIr = ((1
1y
a > ml m2 ... > m > 1
(ii) n(A) = Dis(rl)s(r)...srt)
mlIrl ... mtIr 1 t
a > m 1> m2> ... > mt 1
Proof: Let Y E1 (in equation (ii), YE ), Y = UX
where this is the unique factorization of Y into a unit
(U) and a diagonal (X) Then, since. E1Y, EA imply E
is a unit, EIX and EJA imply E = Im Hence Xzd(in
(ii)), X F a, X diagonal imply XZ6Z,*).
For each unit U E a there exists a unique o Z(m) associated with U i.e., r-4- U iff U = (I + J )G...
n1 n1
def
(I + Jn) and o n1 + n2 +...+n Thus if
s s
X M I I~~atr Ei)(~ tha
r (9 Mtr (ii, such that
= !t




31
m > m2> ... > mt there exist s(rl)s(r2)...s(rt) distinct elements of 0- (in (ii),t,) which have the
same diagonal elements as X The theorem follows.
Theorem 15: n(A) = D s(m!)s(m)...s(mt)
Dj A 6l m 1D a mt
D =dI + Jm d 1 4F2 > 1
Ill In1 2 t
Proof: Apply theorem 14 (i) to theorem 10.
Theorem 16:
a X11V xnln-1 x21
n (A) = i 1 l 1 1)
nlml+ ... +ntmt =a 11=1 x12=1 x21=1 x22=1
1 ml < m2.., mt
x
tn t-1
S1
xtn =1
Proof: Let A. = air + J Then X ? 1* implies X = x G)... x where xl >, x2 >,... xr i.e.
/a xl1 r-1
n*(Ar) = a 1l 1
x1=1 x2= xr =1
Let U E 6 such that IU = 1 Then U =




32
n n2 n
(1) a J() ) (2)),,) where J(i)
= Jm for some m. and there are n. terms in the it
1 121
summation, mi : m. for i j
r
Now if V ~ I Jq rq = m ,it is clear there
are n*(Ar) distinct matrices X E 6/ such that X = VY
(mod A,p), for some diagonal matrix Y where YiX and VIX Hence if U satisfies (1), there are n*(An )-n*(An2
*.. n*(Ant distinct X (1 such that in the unique factorization of X into a unit times a diagonal, the unit obtained is U .
The theorem follows. Theorem 17:
*(A) = (A n) m + .. +n m 11 ..
DA nlml + ...+ ntm=m 1
D=dl 111Kt
D=dm+Km 1 m m2< ... mt
x lq-1 d x tq -1
xnl =1 x21=1 xtnt =
Proof: Substitute theorem 16 in theorem 11, noting that u(A/D)= 0 unless K m =




CHAPTER V
THE FUNCTIONS k(A), nk(A), k (A) AND nk*(A)
If M = e P(a + J) Mk is the product of M
i=1 i i
times itself k times. Let 1 q -' mI Then in finding the q% elementary divisor of XIm consider the minor formed by taking the first q element of the first gq rows. This is of the form
k
X al
1-a1
O X -alk
9
k
* O
O 0 0 k alk
and clearly its determinant is QL-alk)q Then consider the minor formed by taking the first q elements of the 2nd through (q+l)st columns. This has the form
kal k-1
X-alk kal k-1
0 (%-a lk) kalk-1
0 0
S k k-1
0 O (-a) ka1
k
and clearly its determinant is of the form (%-alk )Q(X) + (kalk-1 )q Hence, provided (k,p) =lthe greatest common 33




34
divisor of these two minors is a non-zero constant, and the elementary divisor is 1.
By repeating this argument for mI < q K m2 and, in general, mi 4 q < mi+1 we see the elementary divisors
kl km2 .. kmt
of M are (X-alk )l (X-a2 ) ... (X-at k )mt and the rest are 1. Therefore, the Jordan form of Mk is
t
Zl(aikIm + m
i=l i m m
t
Notation: Mk will be represented by L; (aik mi+ Jmi)
and .(p) will be picked so that (p,k) = 1. Def. 13; For B,C F (m,p), B and C are relatively k-prime [(B,C)k Im] iff D Ea(m,p),Dk,/B and Dk/0 implies D Im Def. 14:
(1) A k-complete residue set (mod A,p), Ok is a
complete residue set (mod Ak,p). The number of elements in ak will be denoted by nk(A).
(2) A k-reduced residue set (mod A,p) is
gk' X_ 'Fk :: (Ak'X)k m The number of elements in
(k is denoted by fk(A).
(3) A k-partially reduced residue set (mod A,p) is k = 9 2k : Dk X and DkfAk implies D is a unit
The number of elements in 2k is denoted by k(A).




35
(4) ak is a k-complete residue set (mod A*,p)
[A* = alm]. The number of elements in is denoted by
nk (A)
Lemma 4: Let D range over the A-divisors of A. For each
D, define D= x A : x/9' Then UD is a kcomplete residue set (mod A,p) and if D X D, DD, D O
s
Proof: Let Y& Ck Y = YiIm + Jm) Let e =
((yi),ak)k [ek/(yi) ek/ak and if fk/(y.) and fk/ak then fle]. Define x such that x iek = Yi for all i ,
and d such that dkek = ak Define D = alm + Jm and
S
X -2 (xIm + Jm ). Then if CID and CIX by lemma 1,
i1 2.i in
s
S= l + K where old cfx and Om. Km i Jm
m M.m.m
id3=1 1 2 a i
for all i But (dk,(xi))k = = 1. Theree e k
fore c = 1, and C must be a unit. Hence XE k
Further X D- (XiIm m ) emi 1S
S
s
(ekxi m + e Jmi + m) = Y (mod Ak,p)
In order to show the second assertion assume
In order to show the second assertion, assume




36
k k s
Y x A~Z where Y = (Im +m ) D = dim
DE i=l
+ Jm and E = Im +m Let de = a = E Then since
Y g (Y),dk = 1; and since YE (Yi ), k = 1.
By lemma 1 and theorem 5, X and Z must have the forms:
s S
X= )(XiIm + Jm. ) Z = zo (z I + J) where
i=1 m mi i= im m
k k +
x Yi and zilyi for all i X = (ek m
k s
ek Ak ( (zi e kI + k k
em = Z A = i ( i + km i), (mod Akp)
i. Z i=1 i. 2.
k
Hence by some rearrangement of the indices we have x.ek =
k k k k
z = Yi for all i ((yi),dk k = ((zi ) = 1.
Therefore (,d) = 1. But di S. Therefore d Similarly, (e,j) = 1. But flde Therefore jd Therefore
g= d, and E = D Then e = e and x. = z. for all i and X = Z Therefore the representation for Y is
unique.
The lemma follows.
As an immediate application of this lemma, one obtains
Theorem 18: nk(A) = k*(A)
Applying theorem 9 to theorem 18 gives




37
Theorem 19: C *(A) U])(A/D)nk(D)
D IA
Lemma 5:: Let D range over the A~-divisors of A. For each
D~~~ ~ ke n )A XE
DX dDenn 4 :. X. Then UDis a complete
residue set (mod A*,p) and if D /D' then &%flD DI =-1 Poof: Let Y E Then Y /Gi Le (),k
i =1
e, de=a and x e k= yi for all i. Define X= ES ~x and D) d.I m+ J M Then (xidkk= ik
i-ie e
=1 so xA ,and X A __(klm
~Zekx i =Y' and hence YE 0~) i
Assume Y = XA z whr X= Zxi Z
E ~ 2 G=iI d.Im + Jm E = &Im + Jmand ((xi),d kk= i=l )
k )k = 1*Let d~e = =a .Then, by some reIn m
arrangements of the indices, Y =C xiek ZIz.E k
But ek =((e k x), ekk~ )k ((Cekzi), ek&Sk k hee
fore D) = E. Hence if D D)', Y O Theorem 20: nk*(A) E ZZ (A
DI A




Proof: The proof follows immediately from lemma 5. Theorem 21: Lk(A) = u(A/D)nk*(D)
D)A
Proof: Apply theorem 9 to theorem 20.
t
n
Def. 15: If M = l (aIm + Jm) define a(M) = i s(m)
i=1 i=1
Theorem 22:
(i) nk(A) = ac(X)
XEok
(ii) k (A) = a(X)
Proof: If XE k (in equation (ii), X& 2) x =
s
2 x I where x. 4 x then there are a(X) distinct
i I im 1 i+1
matrices Y in E (in (ii), Y in ) having unique
factorization into a unit times a diagonal where the diagonal is X The theorem follows.
If k = 1, theorem 22 reduces to theorem 14.




CHAPTER VI
THlE FUNCTIONS nl k(A)2 V(,)(A)2 *() AND n()A
k
Def. 16; Let A (k) = 10A (A (k) is an ink-square matrix). Define n (k) (A) = n(A (k));, (7(k)(A) = ((A (k) ); f(k)*(A)=
Theorem 23:
DIAA
(ii) q (A) = ,(A/D f *k(D)
(i) (k) DIA nk
Proof: Define B = aImk + Jm Then it is clear that
=* (A (k)) hence "n'* (B) = n(k)*(A) and ((k)(A) = C(B). Thus, using theorems 12 and 15 and lemma 1, we get
(i) nk*(A) = n*(B) = Z (E) =( ~jk(k))
E 1B D 1'A
U( ((k)=E)
(k) ) j
DJDjA
(D (k)=E)
39




40
If k = 1 theorem 24 reduces to theorems 12 and 13.
def m )
Def. 17: If r a + a2+ ...+ar = mk we define fm
= 1 if there exist k elements ,..., d of Z(m) such
that 0- may be partitioned into 1 + + ...+k m()
= 0 if there is no such partitioning. Theorem 24:
(i) n(k)(A) Z m
P
SGx iI a=XE jk)
i=1
"O= C + ... + C
(ii) (k*(A) = m()
P
ag iI X &X
i1 (k)
r= 1 +* +* +p
Proof: Define B = aImk +mk If X (k), then X E
mk Jmk(k)' (in equation (ii), if X e then XE But if
XE ( (in (ii), X Z then XE 6?k) (in (ii) X E )
iff there exists a permutation matrix P such that
P
P A(k), i.e., iff X = @(xi + ) = al ... +
i=1
a and m(-) = 1. Notation: (a) is the 1-square matrix Then n((a)) Notation: (a) is the 1-sauare matrix, Then n((a))




41
n*((a)) =a and q((a)) =1((a)) =Ca),the Euler e
function,
Theorem 25:
(i) n (A) = n()(
(ii) q(A = ua/d.)n ((d))
(i ) I'A) =dja (mn)
p
(iii) n(A) = Z: n n()((a))
pVIA n. j=l i) i=1
V.=I m+Ji
1 i i.
(iv)~~i ?*() F1 ((a/d))n n(d)
(iv ~p VIA. i=1. dia (.
V=v
Proof:
(i) n (mn) ((a)) = n( (a)(in)) n(A*) =*A
(ii) By theorem 13, Yf(A) = 5,p(A/fl)n*(D) .But DIA
DIA, i(A/D) 4 0 implies D =d.Im + jmwhere d.Ia .Then
A/D = (a/d)I m and u(A/D) =u((a/d)) .By part (i),
n (D) = n~(in)((d.)) .' Hence (-P(A) Z (ad) ()
nl (D ) n (U(a/d))n ((d)) = m
DIA ((m)




42
(iii) and (iv) Let V6 a (in (iv), 62), be a unit,
p n.
i.e., V = V where V. = IM + JM mi < mi+ .
iImi +
n
Let p = 1. Then V =Z V1 where V 1= I + J
Define B = aln + Jn and = XE L: X = VP (mod A,p)
PX, P diagonal, VIx] [in (iv), q such that X iff
XE and X EX ] Then there is a one-to-one correspondence between the elements of and X (in (iv) and ( ) given by n n
Givi
i=1 i=1
Thus (iii) and (iv) following in case p = 1.
Assume the theorem true for all p < k Let V =
k
G 1 Vi where V. = Im + Jm Let V = nk Vk
k-1 n.
and W = fG V Let q = nkmk and define B =
i=l
alq + J C = (aI + J) +Iq ; D = aImq + Jm-q and
Q Q Q q m-q m-Q .m-q E = Iq + D Then there is a one-to-one correspondence between the elements of 0 and ( (in (iv), 9 and )
given by X E-+ X + Im-q and between the elements of 6and (in (iv), A/and (ST) given by Y I + Y .
But if ME 62 (in (iv), CV), M = VF where FINM F diagonal and VIM since nk n. for i k M can be written




uniquely as the product of a matrix in (in (iv),
by a matrix in c (in (iv), &' ). Therefore n(A) = n(C)n(E) = n(B)n(D) [in (iv), *(A) = (*(C) (E) =
1(B) (*(D)] and the theorem follows.




CHAPTER VII
THE FUNCTIONS pQp(A) AND Y(A)
Let m be fixed. Define =aI m: a 1 /and
= fM + K: ME and Omb K Jm Let be the
set of all Jordan forms over the integers. Def. 18: Let PC I be such that if Methen M ; MM2 (mod M,p) iff M1 F and M2~ '. If there is a A satisfying this same condition such that for any Z E there exist unique matrices P F( Q6) such that Z PQ
(mod Z,p), C and Q will be called factor sets of and together they will form a conjugate pair. Throughout,
and will be a conjugate pair.
Def. 19: ?(R(M) = (M/D)
U DIM
Theorem 26: Let A, B, M6 ; AIM, BMIN, and AB M (mod
N,p): If (A,B) ==Im then J 6(M) = u (A)ue(B). Proof: Applying theorem 7 and proposition 1:
'U (() =( B)(/D
DN IE2
E2B
E1,E2
44




45
=-p= 211- =a(A/E1) 2 p(B/E2)
ElA E 1 2 E 1A E21B
E21 B E 1 E 2V
E1,E2
= y(A)uQ(B)
In the case = ( this reduces to proposition 1.
Theorem 2D ) (A/D) 1 if tAl = 1
DTheorem 27: DA 0 otherwise
Proof: 'y e(A/D) 1 (As)
DIA D\A EIA/D
= p((El
E 1 A DE =A/k
= D u(E1) since there exists only one factorization
El1A
of AA, into the product of an element from by an
m 1 if UAd = 1
element from And L i (E1) =i by
proposition 2. E1 JA0 otherwise
Theorem 28: Let f,g map into the integers, A =
t
aiJm + where 0 K J and if i / j ,
i=l i i mi mi i




46
t
(a,a) = 1. Let A I 2 aiM + J Then
ii
Wi f (A) FiEg(A/D)--: 'Z g(A/V)
DIA V 1A
DIXK Ivi =1
DIA
(ii) g (A) L"f(D)yi, (A/fl) g L~(A/D)
DIA D 1A
f (A/V)D6 lf=1
Proof:
(i) Assume f(A) = 3g(A/D) for allA D \A
DIX
L 7nf (D)j;, (A Di) g 23(t (D/E) )yi (A/D)
DIA DI1A E \D
( ZZ-z g (N))j2u (M)
DM SA EN SD
E
- 1g(N) 21 )1(M)
NIA D
N NE D




N)A ME A/N
NIX EIT
= g(N) (AN
N IA EIAAN E
NIX I
=~~ E i (A/I =
g(N) 1otferise by theorem 27
NIX
F,2 g(A/V)
V 1A
(ii) Assumae g(A) f (D))i, (A/D) fLor all A I.
D A
DIX
(A/l) f (B.2 y N
2f(E) (N)
E'IA DM 5 Ae
DI D2




48
- f(E) A F-(N)
EA DNIA
E
EA DIA
D
= f(E) if by theorem 27
EIA 0 otherwise
Eli
S2 f(A/v)
VIA
IVI =1
If Q = and = Imj, (i) reduces to the sufficiency condition of theorem 8, while (ii) reduces to the necessity condition of theorem 9.
Corollary 28.1: If AE
f(A) = 3 g(A/D) 4- g(A) = f(D) (A/D)
DIA DIA
D' D Z'
D* D 9
Proof: If AE VIA and IVI = 1 then V = Im Hence
the corollary is immediate from theorem 28.
Def. 20: Let A = alm + Km,0m4% m < Jm X E I = s s
(x_ iim m.+ m Km The greatest common
divisor (XA) of X and A is the matrix G divisor (YA) of Y and A is the matrix G =




49
S
((xi,a)Im + Jm) From lemma 1 we see that GIA, i=l 1 i
GiX and if HlIA, HIF, then HIG Def. 21: Let A = aIm + KmO m4 K J A 6-reduced
m)m m m- m
residue system (mod A,p) is given by 6'= {XE: (X,A)E6. The number of elements in a( will be denoted by py(A) Lemma 6. If AE ,2 = [D : DfA, De DE for each
D /L define D = [XD : X is from a E-reduced residue system (mod A/D,p) Then (9 D is a complete residue
D&O
system (mod A,p) and if D / D1 then n,
Proof: If X Et let (X,A) = M = DE where DE(, EE .
Then ( ) = E E Clearly, then, is in a c-reduced
D~ D D
residue system (mod A/D). Then X I- D X And since the D
factorization of M into DE is unique if (X, )= E1 1T -1)=
D D
where D1 then (X,A) =DIE1 = M and E El D D1
From this lemma it immediately follows that Theorem 29: If A
EZ ~?(A/D) n n(A)
DIA DEq
Letting = and = {Imj in theorem 29, for
A = aIm + Jm we get theorem 12. Theorem 30: If A 7/




50
( (A/D)n(D)
D IA
Proof: Apply corollary 28.1 to theorem 29. Corollary 30.1: If A = aIm + JmA* = alm
9(A*) = C*(A) = A n(D)i(A*/D) DIA
DE*
Proof: Let 6= Imi and in.theorem 30.
Notation: CO& X 'E :(xI =I 1 Lemma 7: Let A = aI m + ; QC ; 1= D : DIA, D o-i.
For each D 4, define J D = N DX : N ; X ranges
over a 6-reduced residue set (mod A/D,p) Then J
D D
is a complete residue set (mod A,p) and if D / D/ then D D
Proof: Let X eCa, (X,A) = Y VM such that IV = 1 VIY and N is diagonal. Then N = DE where DE EER
and (,A) = VE since VE and Et Therefore
X/D is in a e-reduced residue system (mod A/D,p). Moreover, from the uniqueness of D and E and since all units are in G, this factorization must be unique.
From this lemma it immediately follows that




51
Theorem 31: If A = aIm + Jm and co
DA4(A/D) = n(A)
D \A
Applying theorem 29(a) to theorem 31 gives Theorem 32: If A = aIm + Jm and C
LvA/V) = n(D) (A/D)
YV\A D \A
ye




BIBLIOGRAPHY
Cohen, Eckford. "A Class of Residue Systems (mod r) and
Related Arithmetical Functions, I," Pacific Journal of
Mathematics, IX (1959), 13-23.
Cohen, Eckford. "Some Totient Functions," Duke Mathematical Journal, XXIII (1956), 515-522.
Dickson, Leonard E. History of the Theory of Numbers.
New York: Chelsea Publising Co.,T92.
Gantmacher, F. R. The Theory of Matrices. New York:
Chelsea PublisHing Co., 1959.
52




BIOGRAPHICAL SKETCH
Marijo O'Connor LeVan was born on October 27, 1936, at Detroit, Michigan. In June, 1954~, she graduated from Catholic High School of Pensacola. In June, 1959, she received the degree of Bachelor of Science from Spring Hill College, Mobile, Alabama. In September, 1959, she entered the Gr aduate School at the University of Alabama where she worked as a graduate assistant in the Department of Mathematics through August, 1961. She received the degree of Master of Arts from the University of Alabama in June, 1961. She entered the Graduate School of the University of Florida in September, 1961. From September, 1961 to June, 1962, she worked as a graduate assistant in the Department of Mathematics. From June through August, 1962, she held a National Science Foundation Summer Teaching Fellowship. From September, 1962, until the present she worked as an instructor in the Department of Mathematics.
Marijo O'Connor LeVan is married to Jerome Herbert LeVan. She is a member of Pi Mu Epsilon and the American Mathematical Society.
55




This dissertation was prepared under the direction of the chairman of the candidate's supervisory committee and has been approved by all members of that committee. It was submitted to the Dean of the College of Arts and Sciences and to the Graduate Council, and was approved as partial fulfillment of the requirements for the degree of Doctor of Philosophy. August 8, 1964
ean, Colle 1f rts and Sciences
Dean, Graduate School Su prvisory Committee: Airman
C-V




Full Text

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GENERALIZED TOTIENTS OVER CERTAIN CLASSES of n-SQUARE MATRICES By MARIJO O'CONNOR LeVAN A DISSERTATION PRESENTED TO THE GRADUATE COUNOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA .August, 1964

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ACKNOWLEDGMENT To the members of her committee, and especially to Dr. John E. Maxfield, chairman, the author wishes to express her gratitude for their assistance and friendship. ii

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ACIGJOWLEDGMENT. INTRODUCTION. Chapter TABLE OF CONTENTS . . I. II. III. ELEMENTARY THEORElVT.S. DIVISION AND FACTORIZATION. THE FUNCTION p(M) ..... Page ii 1 3 7 18 IV. THE FUNCTIONS n(A), ~ (A), n*(A) AND ~ *(A). 25 V. THE FUNCTIONS Cf k(A), nk(A), (f k (A) AND nk (A) . 33 VI. THE FUNCTIONS n(k)(A), Lj} (k)(A), cf (k)*(A) A1l---:D n(k) (A). 39 VII. THE FUNCTIONS (J (A) AND c.f g (A) BIBLIOGRAPHY BIOGRAPHICAL SKETCH iii 44 52 53

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INTRODUCTION Euler first investigated the totient in 1760 in connection with his generalization of Fermat's theorem. 1 He defined tf(m) as the number of integers not exceeding m and relatively prime to m. Among the more elementary (and interesting) properties of the totient are (1) tjJ(pm) m m-1 where is prime = p p p (2) if ( a, b) = 1' then
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2 is m. C. Jordan~' 2 considered the function Jk(m) of the number of distinct k-tuples from a complete residue set (mod m) whose greatest common divisor was prime to m. E. Cohen 2 considered the function Lfk(m) the number of elements in a complete resude set (mod mk) which have no klli power divisor in common with mk. He proved 2 that these three functions are equal. This paper is a result of seeking the number of unordered sets of k integers whose greatest common divisor is prime to m. It generalizes the totient to certain k-square matrices, preserving properties (1), (3) and(~). In Chapter V it further generalizes Cohen's (and hence von Sterneck's and Jordan's) function to matrices. Chapter VII generalizes Cohen's paper "A Class of Residue Systems (mod r) and Related Arithmetical Functions I. II 1 L. E. Dickson, History of the Theort of Numbers (vol. l; New York: Chelsea Publishing Co.,952) pp. 113-155. 2 Eckford Cohen, "Some Totient Functions," Duke Mathematical Journal, XXIII (1956), 515-522.

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CHAPTER I ELEMENTARY THEOREMS Let p be an odd prime and j(p) the field of integers (mod p) represented by the least non-negative residue set. For each natural number m, define 0((m,p) as the set of all m-square matrices over j(p) whose characteristic equations have m roots. Following the usual notation, let Im be the m-square identity matrix and Om them-square zero matrix. Define J m+l (~) = 6 I 0 .. ,~ A matrix M in &cm,p) will A(t)B is the direct sum of A and B), each = a.I + i m. s i: i=l m. = m J_ a and whenever s ]_ a. = ]_ a. l, m. f m. l. J(m,p) is the set of all regular J.+ J_ J.+ Jorden forms in tK,(ra,p) Throughout, M will be a (fixed) element of ,,c0(m,p) with the additional requirement that 1 f a. 'p-1 for all i ]_ Let j (M) be the subset of dG(m,p) consisting of ,v all matrices T having a Jordan form T which will satisfy 3

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5 Theorem 2: If B C and D E (mod 1'1,p) then there exist matrices s and T in j (M) such that (i) s D T E (mod 1'1,p) (ii) B + s C + T (mod 1'1,p) (iii) BS SB CT TC (mod 1'1,p) Proof: Let X = (x .. ) Y = (y .. ) Z = ( z .. ) and l.J l.J l.J W = (w .. ) l.J be the Jordan forms of B, C, D and E, respectively. Then there exist non-singular matrices P and Q such that and -1 C = QYQ Define s = PZP1 and By definition S and D have the same Jordan form, hence are congruent (mod 1'1,p). Similarly, T = E (mod 1'1,p). Hence applying the transitivity proved in theorem 1, Since we know x .. = y .. (mod a.) l.J l.J l. for all i,j Further z D E = W (mod 1'1,p) so that z w .. (mod a.) for all i,j l.J l.J l. Hence B + s = PXP-l + PZP-l = P(X + Z)P-l P(x .. = l.J + ) -1 z .. p lJ P(y .. ) -1 + w .. p lJ lJ (mod 1'1,p) = P(Y + W)P-l Q(Y .. W) Q-1 (mod 1'1,p)

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= QYQ-1 + QWQ-1 = C + T The proof of the third part of this theorem will be found in the proof of lemma 1. 6

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CHAPTER II DIVISION AND FACTORIZATION Def. 2: Let B,C EJ(m,p) such that I B C (p-2)I m m + Jm. Then Blc (mod C,p) iff there exists DEJ (m,p) such that BD = C (mod C,p) and \BD\ \C \ Lemma 1: Let M be as in the last section and B t ,J (m,p). s Then BIM (mod M,p) iff B [:4>Bi where i=l and J m. J. B. = biI + J. mi Proof: Let B satisfy the conditions of the lemma. s Define D = L $Di where Di = d 1 I KI such that m. + m. i=l J. J. bidi = ai and K + KI = Jm. (matric addition). If mi mi J. s BD = F = (fij) then F = G~Fi where f .. J.J. = b.d. J. J. = i=l kiai for all i f .. 1 = 0 if mii+l = 0 f .. 1 is J.l.+ J.J.+ b. of di if mii+l = 1 and for all other i,j f .. l. l. J = 0 Since (bi,p) = (d 1 ,p) = 1 for each i, in each block, Fi, there exists at least one kxk minor of F. l. whose determinant is a non-zero scalar (namely the minor 7

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formed by taking the first k elements of the 2nd through (k+l)st columns). Also, the elementary divisors of m. (/\.-k.a.) 1 l. l. But F s = 7t i=l m. F = (/\.-k. a.) 1 l. l. l. Hence F. l. m. (k.a.) 1. l. l. are k-1 ones and and s C = 7t i=l Therefore, ki = 1 for all i, and the Jordan form of Fi is M. for all l. B IM (mod M,p). i 1 Therefore, BD = F M (mod M,p) and 8 Conversely, assume BIM (mod M,p). ., Then there exists D f'. ) (m,p) such that Im~ D M and BD = s F;; M (mod M, p) where )'(t)(a.I + L-J l. m. 1 l. l.= F Then f 11 = a 11 since f 11 Further, B = (bij) E. ,,J (m,p) that t 11 = b 11 d 11 Hence a.J ) l. mi I al and and D = b11I al Let F = (f .. ) l.J fll = al (mod al) ( d ) A ( m, p) l.J imply Now if ml::> 1 f 12 j O since F 1 and M 1 must have the same Jordan form in J-(p) But r 12 = b 11 d 12 + b 12 d 22 Hence either b 12 = 1 in which case b 22 = b 11 or d 12 = 1 in which case d22 = dll or both. But al = f22 = fll = d22b22 hence if either b22 = bll or d22 = dll the other equality must also hold. By repeating this argument we see that for i = 1,2, ,m 1 m 1 1 times and b .. l = 1 l.l.+ 1 F. R. Gantmacher, The Theor~ of Matrices (vol. l; New York: Chelsea Publishing Co., 1 59), pp. 139-151.

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9 or 0 for i :::i:: 1 2 m -1 1 Moreover, since mmlml+l = 0 we know that f m 1 m 1 +1 = 0 and since 1 ,f. bll 1'p 2 1 ..:: ... dll p 2 and 1 -s bll dll p 2 hence 1~ bll + since d l = 0 .. mlml+ Hence we have described B 1 = b 1 I + K ml ml where b 1 / a 1 B = Bl + BI By repeating this argument on the remaining s 1 blocks we can similarly describe B' and the lemma follows. Also, since the proof in no manner depends on the location of the units on the super diagonal it is clear that BD ~DB~ M (mod M,p), Returning to part iii of theorem 2 we have (in the notation of that theorem): SB= (PZP1 )(PXP-l) = P(ZX)P-l PXZP-l = BS SB= PZXP-l QZXQ-l; QYWQ-l = CT and CT= (QYQ1 )(QWQ-l); QWYQ-l =TC Theorem 3: If B,C e )cm,p) Blc (mod C,p) and q is any odd prime such that'c .. q 2 for all i, then Bfc l.l. (mod C,q). Proof: Since r C e .Jcm,p)' C ... L@ Ci where Ci = i=l

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10 and c. < p 2 J. for all i But, by hypothesis, ci q 2 for all i lemma 1, if B/c (mod C,p) Therefore CE .J (m,q) Applying then B = r 2() Bi where Bi = i=l and 0 <: K < Jm. Then m ... m. .. J. J. J. bi f Ci f q 2 for all i and therefore B E_&(m,q) Then again applying lemma 1, B,C e )(m,q) and B having the form of the lemma show that Bio (mod C,q). If J (m) is defined as the set of all distinct Jordan forms over the natural numbers, theorem 3 allows the following restatements of definition 2 and lemma 1: Def. 2'. If Im~ B C where B,C t. J(m) B/C iff there exists D J (m) such that BD F ;; C (mod C,p) for any odd prime p such that c 1 :p 2 for all i and r F iLa@ (ciim. + ciJ ) or B = PTB'D where B'/c (P a 1 l. mi l.= permutation matrix). s BIM iff (b.I + K ) where bi l a 1 Lemma 1. B = l. mi m. i=l J. and om f K 6 Jm for all i i mi i s Def. 3: Let B IM t B = L@ (biim. + K ) such that i=l J. mi q q K 2:e Jn. where z: nj = mi and nl n < .. nq = 2' .. mi j=l J j=l

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11 s Then M/B is defined to be the matrix D = z~ (diim + i=l i K I') where bidi and K I' J = a. + Km. = m. l. m. mi l. l. l. It follows from lemma 1 that BM B = M (mod M,:p), but, unless m. = 1 l. for all that BC= M (mod M,:p) i, there is at least one C such and M C t B (mod I:'I,:p). Def. 4: Let B,C e) (m) Then B and C are relatively prime [(B,C) = Im] if the only matrix which divides both B and C is the identity matrix. Note that if (B,C) = Im then one of B and C m-2 must be diagonal, or r 2 [2 EJ) I 1 would divide both. i=l Convention: Whenever the determinant of a matrix is taken, it will be taken in the ring of integers. Def. 5: Let B J(m) Then (1) Bis a unit if \B\ = 1 (2) Bis a :prime if B is not a unit and DlB implies D = I or D = B. m Theorem 4: Let PE )(m) Then P is :prime iff P is diagonal and /p/ = :p,a :prime. F~oof: Assume P is :prime. Then since P is not a unit, for any unit V :I: Im V { P Hence P must be diagonal,

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12 m i.e., P = 'C.,0x 1 Suppose for i :f: j, xi.;. l and xj i 1. i=l Define Q = Then Q /= p since q .. =} l but, by lemma 1, JJ QjP. Hence there cannot be two distinct xii l. Let Define R = m i-1 L!_.Gl ()a@ k=l m {t)l k=i+l and xi f 1 x 1 =ab. i-1 S=L~l@b$ k=l ~01 Then RS= P. Hence one of k=i+l R and s must be p and the other, Im Therefore x. = ab implies one of a or b is x. and the other 1, l. l. i.e. x. l. is a prime. Hence if p is prime, Pis diagonal with one element a prime number and the others all 1, i e I P I = xi a prime Conversely, let P be diagonal and lP\ = p a prime. Then P must have p as one entry and 1 as all other diagonal entries. Assume p is in the first row. Then, m-1 by lemma 1, if Q / P Q = x [: @ 1 where x \ p But i=l then x = 1 or x = p. Hence Q = Im or Q = P. Def. 6: if D lr-i E = D or A non-unit and for all E = Im D is a primitive divisor of M ~)(m) E f. J (m) if E In and E /r-i then

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13 s Theorem 5: Let M t. J(m) M = S'@ (a.I G 1. m. Then i=l l. M has a unique factorization into matrices V and D such that V is a unit and D is diagonal and VD M (mod M, p). Further, if, whenever i j (a~aj) = 1 then D has a unique factorization . r n. V ( Tt P. 1.) 1 l. where each Pi is a primitive divisor of M and = M (mod M,p). s Proof: s Let M = L Mi i=l where :n. = a.I + Jm J. J. m. l. l. s l.= Define V = L@ (Im. + J m. ) and D = L,(f)a.I l. mi Then by lemma i=l l. l. i=l 1, V/M and D/M, so this is a factorization of M, s Jmi) ) provided VD M (mod M,p). VD = ( L, (Im. + 1 l. ( t$ airm.) l.= s s = [ (a. I + ai Jm.) L(f> (a. I + Jm.) = l. m. l. m. 1 l. i=l l. l. 1 l. l. l.= l.= (mod M,p) since (a. k ,p) 1 for all i and for all k = l. ,,..., M (mod M,p) where "' v/:n Suppose VE'= V is a unit, E / M ,.J and E is diagonal, Then, by lemma 1, V = s s M LGJ (I m. l. + K ) m. l. where 0 < K J m.' m. m. and E = Ltt> d.I l. m. l. J. l. 1 l. l.=

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14 s s where d. ( a. l. l. ,-J VE = LG3 (d.I + d.K ) = 1. m. 1. m. i=l l. l. (d.I + K ) (mod M,p) since 1. m. m. l. l. k (di ,p) = 1 for all i and ,._, k Hence if VE= M (mod M,p), d.I + K = a.I + Jm. 1. m. m. 1. m. l. l. l. l. for all i; i.e., d. = a. l. l. and K = J for all i. m. m. l. l. ,v Therefore V = V and E = D and the factorization is unique. Suppose ( ai, aj) = 1 for all i = j Let n. nil ni2 ir. l. Then if i j pil pil pi2 Pir. = pik = l. k and 1 For each j ,k define pjk = j-1 s L(;)Im. pjk 1 m. (t) L,I Then pjklM m. l. J 1 l. i=l l.=J+ lemma 1, and pjk is primitive by the same lemma. j-1 s n.k Le, LeTm. since pj~ = I aiim. if m. l. l. 1 l. r. i=l l.=J+ s l. r TC TC pjk = D we can renumber the pjk to k=l j=l re s ult of the theorem. Def. 7: For m 1 define ( def Z(m) =zr. 2 n. l. i a. = l. for all by Further, s z r. l. i=l get the each = n. l. is a natural number and r= m}, i.e. o is a partition of m. Define s(m) as the cardinality of the set Z(m) and -u(r) as the number of terms in the series tr""

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15 a..t l. Theorem 6: Let and . Pt l. where (a. ,a.)= 1 if i + j Then the number of divisors J. J of M is s .n (1 + cx.il)(l + cx.i2) (1 + cx.it.)s(mi) J..=l l. Proof: Applying lemma 1, Bf M if and only if B = s (b.I + K ) where b. la. and 0 L. K f Jm. J. m m. J. J. m. m. 1 l. J. J. J.. l. J..= By theorem 5, B may be uniquely factored into the product of a unit and a diagonal matrices, U = s and D = L $ b.I l. mi i=l By applying lemma s l. l. 1, there are TT s(mi) distinct u which may be chosen and, i=l ti for each i there are TT (1 + d .. ) choices for j=l l.J each U. The theorem follows. Def. 8: If D[M, and (D,M/D) =Im, then D is a canonical divisor of M. Theorem 7: If D is a canonical divisor of M and there exist unique El E 2 t:. 4(m,p) such that E1E2 (mod M,p), E 1 /D and E2lM/D. bi for E \r1 = E Proof: Since (D,M/D) =Im, one of them must be diagonal. We may assume this is D. Then, applying lemma 1, we have

PAGE 19

16 s s D = L, (J) i=l d.I 1. m. 1. M/D = "@(&'.I L.J 1. m. +L ),O .{L fJ. m. m. ml.. m. 1 1. 1.= 1. 1. 1. s where v 1 = I' Et) ( I + J ) L.J mi m. i=l 1. and we have and s N 1 = J7a,a.I /_...) 1. m. 1 1. 1.= s N2 = L~I 1. m. i=l 1. Therefore L mi = J m. 1. and d. b. = a. for all i Also, since D is a canonical divisor l. 1. l. of M ( d &' ) = 1 1. 1. Let EJM Then by lemma 1 and theorem 5, E UP s (mod M,p) where

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17 s and P = l e.I e. la. and this factorization is '----l l. m l. l. 1 l. l.= unique. Since e.\a. d.6. = a. l. l. l. l. l. and (d. ,b.) = 1 there l. l. exist unique b.,~. such that l. l. s b. ~1 l. ~. I s-. l. l. Define E 1 = Li @b.I l. m. i=l 1 = e 1 b.\d. and l. l. (~.I + K ) Then E 1 \n and E 2 \M/D and E 1 E 2 = i m. m. l. l. b.K ) E (mod M,p) l. m. The uniqueness of El and E2 l. follows from the uniqueness of u and p and of bi ~l. and

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CHAPTER III THE FUNCTION p(M) De f 9: Let ME )(m,p). Then (1) if M =Im, (M) = 1 (2) if M = P 1 2 ... PtPt+l Ps such that Pi is prime for all i ; Pi f Pj if ii j ; if i,j t then /Pi/ f jpj\ if j t + 1, there is an i t such that j P.j = IP. I then )l(M) = (-l)t 1. J (3) for all other M )l(M) = 0 Proposition 1: If D is a canonical divisor of M, then (M) = (D) (m/D) Proof: If M is not diagonal, then either D or M/D is not diagonal, and (M) = (D) p(M/D) = 0. Let M = a 1 @ .. $am. Then D = d @"@d 1 m and M/D = a 1 e .. ,e where d. 6. = ai and (d. ') = 1. If p 1. 1. 1. 1. is a prime such that p2 /a. 1. then P2 }di or p2 ( 6i Then if m-1 C l or j=l j=i+l P 2 Jtt /D. Hence (M) = p(D) (M/D) Assume, then, that 18

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19 D = P 1 2 PtPt+l 8 where Pi* Pj if i j ; and l P. t ; I P I if i, j t and if j ?;. t + 1 there J. J exists i t such that IP.I = \P.l And 1'1/D = J. J Q. Q. if i j J. J Ql. Q2 ... QqQq+l ~ where if i,j f q /Pi\ :;= \ Pj\ and if j qi+l there exists an < J. .... q such that IP-I J. Then p(D) (1'1/D) = = IP. I J (-1f+q l'1 = 1'1 l'1 p p Q D = 1 t 1 .. QqPt+l 0 QQ+l Then for any i, j, k, 1 \ P} ,= \P j\ /PilTIQkl and }Qkj~IQ 1 \ (by assumption) ; if i,j q + t, then IP.I =i: IQ.\ since (D,1'1/D) = Im and if i q + t + 1 J. J there exists j q + t such that IP.\ = IP. l or fQ.\ = J. J J. /Qj f Therefore (1'1)= (-l)q+t = (D) (1'1/D) s Proposition 2: Let 0 K "J m. m. m. l. J. J. if i =I Then l'1 = L@ (a.I + Km.) l. m. 1 l. l. J.= = if \ l'1 \ = j J-l(D) [~ Dll.'1 if 1'11 :/:Proof: Let that D = dI m D/M s = 1 Then where a la 1 C p(D) = D=dI d\alm DIM' nlH and Hence where (a. ,a.) = 1 J. J 1 1 p(D) 0 imply

PAGE 23

i.e. L {l if IM\ = 1 p(D) = D I M O if IM\ t 1 D l M 20 Assume the theorem true for all s k 2. By lemma 1; if D /M DIM and p(D) =/ 0 then D = k L~d.I i m. i=l ]. where d. \ a. ]. ]. k-1 k-1 SIG) (a.I L i m. ]. + K ) m. ]. (f)I and M 1 1 = ')7 $ (a.I + K ) clearly mk l__; i m ml. 1 l. )l(D) = D/M, l.= k-1 L)l(D) DIM' and if M 2 = [: G5--Im. l. 1 DIH' 1 i=l Mi= 2 k For each D = ')I ~d.I G \.!I i m k-1 and D 2 = ( G i=l 1 l. l.= D\M define DIM, ( akI + Km ) and mk k L )l(D) DIM' 2 D IFI' 2 and

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21 (E) L = [: if t Mll :: l M 2 \ = 1 = p(F) EM' F\M 2 otherwise 1 E\M 1 Fl I\~ 1 But lr1 1 1 = lr12! ;;: 1 iff IM 1 = 1 Hence if the theorem is true for all s k 1 it is true for s = k Theorem 8: Let M .J (m,p) such that if i f j then (a.,a.) = 1. If f,g both map ~(m,p) into the integers J.. J g(M) = f(D) :> f(EN) = C ;i(D)g(M/D) D\M EIU D\M where M UN (mod M,p) is the unique factorization of M into a unit (U) times a diagonal (N). Proof: Assume g(M) = L f(D). Then, since (D) = 0 D\M unless D is diagonal: 2 p(D)g(M/D) = D1M = = L:. u(D) DIM L f(E) E\M Lf(E) E\M I~{= 1 EIM/D E\M f(E) L p(D) D\M/E D\M

PAGE 25

= f(EN) E/U 22 since, by proposition 2, L (D) = 0 unless I M/E \ = 1 D)M/E Conversely, assume L f(EN) = E IU (D)g(M/D) D \M L f(D) = L L f(EF) DIM FIM EIU F diag = C L (UF/D)g(D) F\M DIFU F diag since (UF/D) = 0 if UF/D is not diagonal C f(D) = G G p(UF/D)g(D) D M FG=M/U D\FU FU/D diag = L g(D) D\M M/D diag (G diag) By proposition 2, L11MIGVD) = 0 unless g \ = 1 so

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23 z:, f(D) = L g(D) = g(M) DIM D \M M/D diag \ M/D\ = 1 since M/D diagonal and \M/D \ = 1 imply M = D. Def. 10: D is a 6-divisor of M(D 1 6 M) if M/D is diagonal. T h eorem 9: Let M j(m,p) i -J. j If f g map J ( m p) f(M) = g(D) ;:> g(M) = Dl 6 M such that (a. ,a.)= 1 whenever J. J into the integers. G p(M/D)f(D) DIM Proof: Assume f(M) = L g(D) Then D \ M L..,. y(M/D) f(D) = C )l(M/D)f(D) DIM D\ 6 M since if D f 6 r1 DIM t;hen ;i(M/D) = O p(M/D)f(D) g(E) = C g(E) L g((M/E)/D) E \ 6 M D\M/E D\M By proposition 2, L g(M~E) = 0 unless \ M/E \ = 1 = C g(E) = g(M) E\ 6 M lM/E \= 1

PAGE 27

since E 1 6 M and \ M/E \ = 1 imply M = E Conversely, assume g(M) = L p(M/D)j(D) DIM Then = f (E) G (M/H) Et 6 M EiM E EJM/H (DH=M) 24applying proposition 2. cM~H) = 0 unless IM/H\ = 1. Therefore I El= 1 and Bu(M/D)f(D) = DIM f(E) = f(M) E 1 6 M IE\ = 1 In case m = 1 (i.e., M is a 1-square matrix), theorems 8 and 9 both reduce to the M5bius inversion formula.

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CHAPTER IV THE FUNCTIONS n(A), f (A), n*(A) AND lf(.A) Convention: For the remainder of the paper, A~J. (m,p) will be a one block Jordan form, i.e., A= aim+ Jm. Def. 11. In definition 1 (1, a complete residue set (mod A,p) was defined, with n(A) the number of elements in a,. Q 11 = { XE "1:E Ix and El A implies E is a unit 5 is a partially reduced residue set (mod A,p). The number of elements in Q;ll is denoted by \f (A) Lemma 2: Let D range over the t.-divisors of A For each d define q D = { X X ranges over .z911 ] Then UD ~D is a complete residue set (mod A,p) and if D 1D I then Q' n n CD/ = 0 Proof: Let B t (1, b.I + J.. n. J.. J If n. l. divisor of the set Define Define c.I + J J.. n. n. J.. ]. Then B = Bl ff) B2 tf) BS where B. = ]. (b.) refers to the greatest common ]. [ bi = l,.~.,s J i let ci, i = l, ,s by b. = de. J.. J.. C = c 1 $ Cs such that b. C(ci), ~) = CCcr), d = and c. = J.. 'so D = that Cf )} 11 25

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26 s = L., G (dciin. + Jn_) l. l. i=l B (mod A,p) so that A,p). BE JD and lJ 0 D D is a complete residue set (mod To prove the second assertion, assume A B Cl Dl c 2 ~ 2 (mod A,p), where D 1 and n 2 are 6-divisors of reduced residue sets (mod D 1 ,p) In particular, both A/D 1 A/D 1 = 6 1 Im and A/D 2 = A, and c 1 ,c 2 in partially and (mod n 2 ,p) respectively. and A/D 2 are diagonal; Let ~2Im Then A (1) Cl D 1 = [(cxlrk. l. + Jk_) (cxrik l. r + Jk )Jol 1 m r r L._, 0 (cxi 01 Ik. + Jk.) (mod A,p) i=l l. l. (2) C2 A [(~lI+ J,_)$ @(~sI.,s + J_es)J[~Im] = D2 s L~ (~i~2 1 1 l. l.= (mod A,p) Let 6 1 d 1 = 5 2 d 2 =a. D 2 = d 2 Im + Jm. Since c 1 is in a partially reduced residue system (mod D 1 ,p), ((cxi),d 1 ) = 1 Similarly, ( ( (3 i) 'd2) = 1 Now, since B C A C A (mod A,n) we know = 1 D 1 = 2 D 2 .t' that in (1) and (2) r = s and the blocks of (2) can be

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27 rearranged so that (ai6lik. + Jk_) = ([3ia2rj. + JJ.) l l l l for all i. Assume that this rearrangement has been made. but aio"l [3i6"2 for all i and o 1 dl = o2d2 = a = Therefore 6"1 = a2 and dl = d2 Therefore [3 = a. l l for all i Therefore Dl = D2 and Cl = c2 and the representation of B is unique. As an immediate consequence of this theorem one obtains Theorem 10: G
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Proof: 28 If a= pm_ D\A, and p(A/D) l O then D = A or D = B. u(A/A) = 1 and u(A/B) = 1 The corollary follows. Convention: Let A*= aI m 2 1>1 __ rxE. /"J L: Define u.., t I..(_, ( X, A ) = Im f and Im X A*~ [ a.* is a complete residue set (mod A*,p)J. The number of elements in a_* is denoted by n*(A); in 11./, by
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29 Then yl = A/D 1 = 6"1Im a. y A 1 D 1 m LYli y2 = i=l d,~= a D2 = m = L G1 Y1i61 = i=l m GY2i Dl = dlim + Jm i=l d2Im + Jm A/D 2 = ~2Im d202 m = G 0 Y2is2 i=l Hence, by rearranging the diago na l of one obtains = Therefore ((y 1 i),d 1 ) = ((y 2 i),d 2 ) = 1. But ((6j_y 1 i), 6 1 d 1 ) = cf 1 and ( (6 2 y 2 i) 6 2 d 2 ) = o 2 Therefore b 1 = o 2 Hence and for all. Therefore the representation for X is unique. As an immediate consequence of this lemma, one obtains Theorem 12: Applying theorem 9 to theorem 12, one obtains: Theorem 13: c.p (A) = G p(A/D )n (D) DIA Cor. 13.1: If a is prime <.f(A) = n*(A) 1 Proof: If a is prime, D/A and p(A/D) # 0, then D = A or D =Im+ Jm. p(A/A) = 1 and p(A/(Im + Jm) = 1 Lf (A) = L ;u ( A D) n ( D) = n (A) n ( Im + J m) DIA = n*(A) n(I) = n (A) 1 m

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32 where = J for some m. and there are n. terms in the illi m. l. l. l. summation, m. i m. for it j l. J r Now if V = C I~J rq = m, it is clear there q q are n*(A) distinct matrices XE Q_,, such that X = VY r (mod A,p), for some diagonal matrix Y, where v/x. Hence if U satisfies (1), there are Y!X and n*( .A. )n*(.A nl n2 n*(A ) distinct nt X f Cl, such that in the unique factorization of X into a unit times a diagonal, the unit obtained is U The theorem follows. Theorem 17: ~*(A) = r p(A / D) Dt( ) Proof: Substitute theorem 16 in theorem 11, noting that u(A/D)= 0 unless Km= Jm.

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CHAPTER V If is the product of M times itself k times. Let the q~ elementary divisor of 1 q ..( m 1 Then in finding AI Mk, consider the minor m formed by taking the first q element of the first qi rows. This is of the form 0 A -a 1 k 0 k O O .. 0A-a 1 and clearly its determinant is Q.-a 1 k) q Then consider the minor formed by taking the first q elements of the 2nd through (q+l)st columns. This has the form 0 0 0 0 0 ( k) k k-1 A-a 1 a 1 and clearly its determinant is of the form (A-a 1 k)Q(A) + (kal k-l)q H d d (k,p) 1th t t ence, provi e = e grea es common 33

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35 (4) ak* is a k-complete residue set (mod A*,p) [A* = aI J. The number of elements in /} is denoted by m v"k nk (A). Lemma 4: Let D range over the ~-divisors of A. For each D, define CD = [x : X tft; j : Then uDc>D is a D complete residue set (mod A,p) and if DI D 1 cD~ ~D' = s Proof: Let y & y G~.I + = i=l J. mi k [ek/(y.) k k and if ((yi),a )k e /a J. then f /eJ. Define xi such that x.e J. and d such that dkek k Define = a s X'= L,G>(xiim + Jm_) Then if c/n i=l i s C = &r + Km. l m. J.= J. J. for all i But fore c = 1, and Ak Further X :Jc= D J. where c/d k (d '(xi) )k C must be { .t@cxiim. J.=l i s c/x. J. k = ( ~' e a unit. + J ) m. J. Jm_) Let k e = J. fk/(yi) and fk/ak' k for all i = y. l. D = arm+ Jm, and and c/x by lemma 1, and om.~ K f Jm. mi y. J. ""'1c e J. J. ) k 1, ThereIi) / / XE Uk. s = 2=@ i=l LBl(Yi Im + J ) = Y (mod Ak ,p) i=l i m. l. In order to show the second assertion, assume

PAGE 39

A k y -;; X :Jc 1) Ak ?! Z ::le where E 36 s y = ~GJ (Yi 1 m. + 0 m.) l.=l l. l. E = G'rm + Jm Let de = a = [ E:" Then since By lemma 1 and theorem 5, X and Z must have the forms: s s X = C (x. I + Jm.) 1 l. m. l.= l. l. z = L~ (z.I + J ) 1 1. m. m l.= l. l. where x. \y. 1. l. and z. \y. l. l. for all i Ak s ekJ ) C fr) (z. f kr + k k = z= E Jm.)' (mod .A ,p). m. zk 1 l. m. l. l.= l. l. Hence by rearrangement of the indices have k some we x. e = l. k z.~ = y. for all l. l. i k ((y.),d )k = ((z. J. l. k k ) d )k = 1. Therefore (f,d) = 1 ar ly, ( e &' ) = 1. 1. But d[ SE~ Therefore d IF. Simi8'/d: Therefore But cf l de Therefore [' = d, and E = TI Then E: = e and x. = z. l. J. for all and X = Z. Therefore the representation for Y is unique. The lemma follows .As an immediate application of this lemma, one obtains Theorem 18: .Applying theorem 9 to theorem 18 gives i

PAGE 40

37 Lemma 5: Let D range over the 6-divisors of A. For each D define ,JD [x : XE4'.f. Then uJI"n is a complete residue set (mod A*,p) and if D -;i D 1 then c} D(Jj n' = [l Proof: Let y c_ ~ k de= a and k = e Xie m G~x. and D = dim+ 1 l. l.= = 1 and hence Assume m m Then y = (f)y. 1 l. l.= Let = y. l. for all i Define X = y. k k Jm Then ( (xi) ,d )k = CC-),\:)\( e e m Ak = Z :le where E X = L@x. z = 1 l. l.= L 6> z. D = dI + J m E = 6 I + Jm and i=l l. m m Let de = tE = a Then, by some rem m arrangements of the indices, Y = GE) x. ek = L 1 1. 1 l.= l.= k k kk k k k k z. E l. But e = ( ( e xi) e d ) k = ( ( E z 1 ) $ ) k = k Therefore D = E. Theorem 20: Hence if nk (A)= D -;6 D 1 yef )D1 L Y,k(A) D\ 6 A

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Proof: The proof follows immediately from lemma 5. Theorem 21: lf k(A) = L p(A/D)nk (D) D ) A Proof: Apply theorem 9 to theorem 20. t If M = L@ (a 1 I + Jm.) define Def. 15: -. 1 m. l.= l. l. Theorem 22: (i) nk(A) = C a.(X) ,,. XE~ (ii) lf k (A) = L a.(X) xea~ a.(M) = Proof: If XE {2,k (in equation (ii), X 0-{ ) X = s 38 z: @x.I i=l i mi where x. x. l l. l.+ then there are a.(X) distinct matrices Y /J. /") // in v'k (in (ii), Y in~ ) having unique factorization into a unit times a diagonal where the diagonal is X. The theorem follows. If k = 1, theorem 22 reduces to theorem 14.

PAGE 42

CHAPTER VI THE FUNCTIONS n(k) (A),
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40 If k = 1 theorem 24 reduces to theorems 12 and 13. Def 17: def If rr .;::. a 1 + a 2 + + ar = mk we define [ (o) m = 1 if there exist k elements eTj_, ... of Z(m) such that er may be partitioned into = 0 if there is no such partitioning. Theorem 24: (i) n(k) (A) J'(r) m p L_, x. I =X E {lk) 1.a. ( J.= J. o= a,.,+ +a. .L p (r) m Proof: Define B = aimk + Jmk : If X E. a(k), then X E 1J (in equation (ii), if X f tz(;) then XE 63 11 ). But if X f (i3 (in (ii), X Z. ~ 11 ), then x E a(k) (in (ii) Xe. atk)) iff there exists a permutation matrix P p T (k) r P XP A i.e., iff X = L.,$(x.I 1 J. a.. J. = J. a.p and 6 m ( o-) = 1 such that +Jex,_)' o=a.1+ + J. N otation: (a) is the 1-square matrix. Then n((a)) =

PAGE 45

42 a ll (iii) and (iv) Let Ve (in (iv), 0.,), be a unit, p i.e. V = ~@ i=l V. = I l. m. m. < m. 1 l. l.+ l. n Let p = 1. Then V = Cfev 1 where v 1 = Iq + Jq Define B = ain + Jn and X = (XE~ : X = V P (mod .A,p) PIX:, p diagonal, vlx] [in (iv), (f. such that xt Lf iff a" X and X f X J Then there is a one-to-one correspondence between the elements of ~* and X (in (iv) and {f) given by n Uc.~ i=l i n Lec.Vl 1 l. l.= Thus (iii) and (iv) following in case p = 1 .Assume the theorem true for all p < k. Let V = where Let k-1 n. and w = LG v. Let q = nkmk and define B = i=l l. aiq + Jq ,C = (aiq + J ) + I ; D = aI + Jm-q and q m-q m-q a,' E = I + q D Then there is a one-to-one correspondence between the elements of and {; (in (iv), (B'' and t;'') given by X X + I m-q and between the elements of/)_, and (in (iv), ,.dl11 and t 11 ) given by Y Iq + Y But if ME a (in (iv), {}./), M = VF where F\M F diagonal and vlM, since nk p ni for i pk, M can be written

PAGE 46

4-3 ft ,1 uniquely as the product of a matrix in (in (iv), C) n by a matrix in
PAGE 48

45 = = )1 {? (A)u-@(B) In the case Er = 1 this reduces to proposition 1. r, {lo if \A\ = 1 Theorem 27: L p 0 (A/D) = DIA u otherwise Proof: = = DE Q L pe7(A/D) = DfA D 2.~ D\A D !~ 22, DE= A/E 1 D t: (t EE. {f = C u(E 1 ) since there exists only one factorization E 1 \A of A/JS. into the product of an element from q by an r, [l if IAI = 1 element from And J1(E 1 ) = by proposition 2. E 1 jA O otherwise I Theorem 28: Let f,g map 1 into the integers, A = t [ ai Jm. + Km. where 0 K "'Jm. and if i ;t j m .... mi 1 l. l. l. l. l.:::

PAGE 50

= L g(N) N)A NIA = g(N) NIA NIA = 2 g(N) NIA NIA L )LD (M) ME ;; A/N EIA E (A/N) E IA/Np G E EIA EEQ {: if IA/N\ = 1 otherwise =Li g(A/V) VlA jv( =l '+7 by theorem 27 (ii) Assume g(A) = L f(D)po;;i(A/D) for all Ae i D \A u L g(A/D) = DIA DIA DE (Q DIA EN q M f (E))l~ (N) EIA = E f f(E) Dl1r;., A. )l cJ' (N) ElA EN-;; M DlA D cq?

PAGE 52

49 s L((x.,a)I 1 l. m. l.= l. From lemma 1 we see that G\A, Glx and if HIA, HIF, then HIG. Def. 21: Let A = aim + Km, Om~ Km~ J~ A (:?-reduced residue system (mod A,p) is given by a(j)= {x [{L: (X,A)cflJ. The number of elements in {L 6> will be denoted by if& (A) Lemma 6. If A E. }* _<'.9 = [n : D fA, D c tQ DE~* J for each D E 19-define (; D = { XD : X is from a if-reduced residue system (mod A/D ,P) f Then U CD is a complete residue DfA9 system (mod A,p) and if D /= D 1 then 1 con~rJ' = Proof: If X ttl, let (X,A) = I'1 = DE where D E E c (y Then (~ ~) = E (?. Clearly, then, is in a @-reduced X residue system (mod A/D). Then X =DD. And since the f t t f M t DE 1. f ( X A ) E 1 ac or1.za ion o n in o 1.s unique :-r, :I = D D where D 1 then (X,A) = D 1 E 1 = M and E = El D = Dl From this lemma it immediately follows that Theorem 29: If A e f* C lf (JCA/D) = n(A) DIA D zc!t Df ~* Letting 6> = t and Q= { rml in theorem 29, for A= aim+ Jm we get theorem 12. Theorem 30: If A f. )

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'f ~(A) = }l l'.O(A/D )n(D) D lA CJ DE i* Proof: Apply corollary 28.1 to theorem 29. Corollary 30.1: If A= arm+ Jm' A*= arm (f (A*) = Cf (A) = L n(D)p(A* /D) D\A DE~ Proof: Let @ = [ Im} and = } in. theorem 30. Notation: JJ= fx~ 1 :IX I = 1~ 50 Lemma 7: Let A = aim + Jm ; cfl C ~; /J..= [ D : DI A, D Q_ r. For each D E. /)-, define d D = [ N :::; DX : N e. 1 ; X ranges over a Er-reduced residue set (mod A/D,p) J. Then ~~JD is a complete residue set (mod A,p) and if D j D / then jD {) JD, = Proof: Let X E Cl (X ,A) = Y = VI1 such that \V \ = 1 VlY and M is diagonal. Then M = DE where and VE ? since VE Er and Therefore X /D is in a @-reduced residue system (mod A/D,p). Moreover, from the uniqueness of D and E, and since all units are in@, this factorization must be unique. From this lemma it immediately follows that

PAGE 55

BIBLIOGRAPHY Cohen Eckford. "A Class of Residue Systems (mod r) and Related Arithmetical Functions, I, 11 Pacific Journal of Mathematics, IX (1959), 13-23. Cohen, Eckford. "Some Totient Functions," Duke Mathemati cal Journal, XXIII (1956), 515-522. Dickson, Leonard E. Histor? of the Theor~ of Numbers. New York: Chelsea Publisning Co., 19 2. Gantmacher, F. R. The Theory of Matrices. New York: Chelsea Publishing Co., 19"59. 52

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BIOGRAPHIC.AL SKETCH Marijo O'Connor LeVan was born on October 27, 1936, at Detroit, Michigan. In June, 1954, she graduated from Catholic High School of Pensacola. In June, 1959, she received the degree of Bachelor of Science from Spring Hill College, Mobile, Alabama. In September, 1959, she entered the Graduate School at the University of Alabama where she worked as a graduate assistant in the Department of Mathematics through August, 1961. She received the degree of Master of Arts from the University of Alabama in June, 1961. She entered the Graduate School of the University of Florida in September, 1961. From September, 1961 to June, 1962, she worked as a graduate assistant in the Department of Mathematics. From June through August, 1962, she held a National Science Foundation Summer Teaching Fellowship. From September, 1962, until the present she worked as an instructor in the Department of Mathematics. Marijo O'Connor Levan is married to Jerome Herbert Levan. She is a member of Pi Mu Epsilon and the American Mathematical Society. 53

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This dissertation was prepared under the direction of the chairman of the candidate's supervisory committee and has been approved by all members of that committee. It was submitted to the Dean of the College of Arts and Sciences and to the Graduate Council, and was approved as partial fulfillment of the requirements for the degree of Doctor of Philosophy. August 8, 1964 Dean, Graduate School


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