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On the exten sion of Ferm at s theorem to matrices of orderBy J . B. MARSHALL.

(Received nth April, 1939. Read 5th May, 1939.)It is proposed to establish, by e lem enta ry m ethod s, a the oremfor ma trices analogou s to Fe rm at's Theo rem in the Theo ry ofNumbers . In Jordan ' s Traite des Substitutions (P ari s, 1870) pp . 127,128, the order of any given linear sub stitu tion or m atrix A withreference to any prime number p is determ ined , bu t the result givendepends on the particular characteristic equation satisfied by thematr ix A, and a general result applicable to all matrices of n rowsand n columns does not seem to have been published hitherto.

(1) F er m at 's Theorem in the The ory of N um bers is as follows:/ / p is a prime number and N is prime to p, then Np~1 1 isdivisible by p, i.e. NP^ 1 = 1 (modp).To extend this to matrices we have to answer this question:Ifp is a prime number, and A is a matrix of order n with integralelements , such that jA | is prime to p, what is the smallest value of qwhich will always satisfy the congruence Aq = / (modp)? He re / isthe unit ma trix, and the notat ion means (as in Turn bull and Aitken,Theory of Canonical Matrices (Lo ndo n, 1932), 22) th at ev ery e lem entof the matrix (A9I) is either zero or a multiple of p. Hence Aqmust be of the form

pa12pa 22 + 1 p023

On expanding the d eterm inant of this ma trix, we get \A9\ = 1 + amultiple of p, i.e. \Ag\~l (m o d p) . If |^41 = 0 (mod ^ ) , then|A | = 0 (modp), and therefore no relation of the form A9 = I (mod p)can exist.(2) If A = pB + C, then A2 = p2Bz + p (BG + CB) + C2, so thatA

2 = C

2(modp). Similarly A

q =G

q(modp) and th us we need consideronly those matrices whose elements are all taken from the numbers0 , 1 , 2 . . . . (p1). since any m atr ix with integ ral elem ents can beput in the form pB + C where C is such a matrix.

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86 J. B . M A R S H A L L(3) The set of matrices of ordern, whose elementsare all taken fromthe numbers 0, 1, 2 . . . . (p 1),and which aresuch that their deter-minants are prime to p, form a finite congruent group. The uni tma t r i xis the identical memberof thegroup. Theproduc tof anytwo members of the set is congruent (modp) with some memberofthe se t . Ev ery mem ber has itsreciprocal. Insymbols ,if A andBareany twomembers of the set, there exist members CandD such th atAB =C(modp) and AD=I (m od p). These propert ies are sufficientto es tabl ish thatthe set of matrices form a group .

The order of this group is (pn1)(pn p) {pnp2) (pnpn~1).This isestabl ished in Burns ide 's Theory of Groups (Cambridge 1911), 83,and may also be pr ov ed from first princip les by the followinge lementa ry method:L et urbe the vector given by the r th row of the m atr ix . Thenth e n e lements of u may each consist of any of the p numbers0, 1, 2 . . . . (p1),except th a t they may not all be zero since thiswould make them atrix s ingular. There aretherefore pn 1waysofselect ing this vecto r. W hen uxhas been selected, u2may be any ofth e pn possible vectors except those which are congruent with &xux( m o d p ) {kx= 0, 1, 2 .. . . (p 1)},since any of these vectors will makeall the minor dete rm ina nts formed from the firsttworows co ngruen twi th 0 ( m o d p ) andtherefore them atrix willbe s ingular (modp),andthis will not be thecase ifu andu2arel inearly indep ende nt . Thereare therefore pn p ways of selecting u2. Similarly uzmay be anyof the pn possible vectors except those which are congrue nt withk-yiii+ k2u2 ( m o d p ) {kt=0, 1, . . . . (p1) .k2= 0,1.. .. (p1 }andthe numbers of ways of selecting u3istherefore pnp2. The sameargument appl ies to all the other rows, and the tota l number ofmatr ices , i.e.theorder of thegroup, istherefore ass ta ted above.B y an elementary proposi t ion in the Theory of Groups then u m b e r qwhichwe are seeking must be a submult iple of theorderofthe group .(.4) By theCay ley-Ham ilton Theorem (Tu rnbullandAitken, page43)every matr ix of order n satisfies an ident i ty of the form

where pn=\A\. Removing or adding multiples of p this reduces tothe congruence An+X1An~1+X2 An~- + +Xn-i A + pi = 0(mo dp)w h e r e Xr= 0, 1,2 (p1), /x= 1, 2 (p - 1). Now if(x n + Xia; 1+ X2xn~2 + . . . . + \n_1 x + p) is a f a c to r (m od p) of

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EXTENSION OF FERMA T'S THEOREM TO MATRICES 87

{xq1), the above congruence must give A I=0 (mod2?),andhence if q is such that (xn+ \1xn~i + \2xn~2 + . . . . + /x) is a factor(modp) of xq1, nomatter what values aregiven to theA's and /x,then A qI=0 (modp) will betruefor allmatrices of order n withintegral elements.(5) By Fermat 's Theorem we know tha t (x+a){a=1, 2.. . .(p1)}isa factor (modp)ofa;* 11. Hence(x+a1)(x +a2){a1=1, 2, . . . .(p1)>a2= 1, 2, . . . .(p1), ax=^a2} must be afactor (xaodp) of a; 1 1.Hence (p 1)must be asubmultipleof thevalue ofq forn = 2.

Consider next the cases where th eexpression x2+ \1x + p isirreducible mod p), i.e. cannot be factorised. This introduces theconception of aGaloisField.

Let F(x)b e anarb itrary rational integral function with integralcoefficients, and letP (x) be a rational integral function of degree nirreducible modulo p (p being a prime number). If wedivide F(x)by P(x) weobtain a quotient Q {x)and a remainder which can bewritten in th e form / (x) + p.q (x) where / (x) is of th e formf(x)=ao + a1x + . . . . +cin-.xX'1'1,each a t belonging to the series0, 1, 2 (p - 1). ThenF (x)=f{x)+p.q (x)+P(x).Q (x).

Then f(x) iscalled the residueof F(x), moduli p andP(x). andwe writeF (x)s / ( ) (modd^j,P(x)). Since eachofthen a'smay takepvalues, there arep npossible residues. These residues form a field,called a Galois Field, of order pn. [L. E . Dickson, Linear Groups(Leipzig, 1901), 6.]

[Note. If P(x) is reducible (modp),or if p is no t a prime,th eresidues do notform a field, sincea tleast two non-zero residues canbe found whose product is 0(moddp, P(x)).](6) L e tP(x)be the irreducible (modp ) function x ^X^ xn~ land letF(x)bexT,sothatxr=f x)+p. q x)+(n+Xn_1x+. ..+X1xn~ l+xn)(b0+b1x+ . . .+br_Then/(cc) cannot bezero. For,ifso, wegetfrom theabove identity ,since /xis primetop,first that 60=0(modp)and hence thatb ltb2are all= 0(modp) and finally that bT_n=0 (mod p). B utb r_nmustbe 1. Hencef (x) cannot bezero.Since thenumber of residues isfinite, it must bepossible tofindtwo numbers rand ssuch that theresidues ofx rand x8are thesame.Let s be thefirst number greater than r such that theresiduesofxr and x are thesame.

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88 J . B. MARSHALL

Then x* xT=0 (moddp, P (x))xr(zs-T-l) = 0

Then, since xr^ 0 ,,x ~ T - 1 = 0

or, if sr= e, xe=\ ,,It follows that if t is any number x' and xt+e have the same residue,so that the residues are periodic with period e. This period eis asubmultiple of pn 1 (c/. Dickson, Linear Groups, 11). For as rtakes the values 0, 1, 2, . . . . , the residues of xT are 1, x, x2There are pn 1 non-zero residues altogether . We can form arectangular array of all these non-zero residues as follows:

1 x x2 xe~1 UXX UX2 Ux Xe~1u2 u2x u2x2 u2 xe-1

where, if the power to which x is raised in the first row is greaterthan (n1), it is understood that the residue (modd p, P (x)) istaken, and where u is any non-zero residue not included in the firstrow, u2 is any non-zero residue not included in the first two rows, andso on. Then all the residues in any line are different from each otherand from those in preceding rows. Hence, as all the pn .1non-zeroresidues must be included in the array , and there are ein each row,pn 1 is a multiple of e.

Also if xe=1 (moddp, P (x)),then xke=1 (moddp, P (x)). Hencexv ~x 1= 0 {moddp, P(x)} for all irreducible P (x) of degree n,ofthe form given. This is the same as saying th at P (x) is a factor(mod p) of x^ -1 1.(7) Now letP(x), no longer irreducible, be of the form {x + a)n(mod p)where a = 1, 2 . . . . {p I) . Then we have

x= a+ (x + a)= {p - a) + (x + a) (modp)

xp= (p a)p+ {x + a)v (modp),sincep is prime and therefore all the other terms are divisible by p.Hence xp =(pa) + (x+ a)p(xnodp) by Fermat's Theorem.Similarly, (x*)p= (p - a)*+ (x+ a)*' or XP';=(p - a) +(x +a)* -Similarly xp = (p a) + (x +a)p1 ,, , and so on.

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E X T E N S I O N OF FERMAT'S THEOREM TO MATRICES 89Henceif pris thelowest powerofp whichis greater than ore q u a l ton

xv' = {pa){modd p, (x+a)"},so that by Fermat's Theorem

xpr(v~i)= i {moddp, (x + a) },t h a t is(x+ a)nis a factor, modulop, of xpr^p-1) 1.(8) Now let P(x), of even degreen, be of the form (x2-f- Ax+ /x)"/2(modp) wherex2+ Xx+/xisirreducible (madp). Then,by p a r a g r a p h(6) above, x2+Xx-\-yi,is a factor (modp) ofxp'~1l. So we can wri te

*J>=-I= i -i-(^2+ xx +fj.).Q(x) (modp) ,where Q(x)is an integral function ofx of degreep23. Therefore(a^ '- i jp '= i + (X2+ Xx+ /*)p=.{Q{x) - (mod p),

andso on. Hence, if psis the lowest power ofp which ise q u a l to orgreater than \n,(XP*-1)P'= 1 ( m o d d p ,P (x)).

T h a t is,(xz+Xx+ /x)n /2 is a factor (modp) of x^-w - 1. The sfound here is evidently not greater than ther of the prev ious pa ra -graph, and therefore (x2+Xx+/x)"'2is a factor (modp)of a;(*"-"pr_ l.

Similarly, ifP (x)is of the form (a;3+ Xtx2+ A2a; + /x)ra'3, (mod^)where x3+X1x2+X2x+ix is irreducible (mod2>), t hen P(x) m u s t be afactor (mod p) of a J>~1)1)'' 1, and similar results can be ob ta inedwhen P(x)is a power of irreducible functions of higher degree.(9) Com bining the se resu lts wehave :

(a ) ifn= 2,P(x)maytake theforms (modp)of (i)(x+ a)2, (ii)(r + ax)(x + a2) {a1=j=a2}, or (iii) x2+Xx+ fx.(irreducible mod p).In case (i) P (x)is a factor (modp)of a;J>(:P~1) 1 sincep < n ^ p, \i\ 3-P-1

(iii^ r(p-'-i)_ 1Therefore in all three casesP (x)is a factor (modp)of(XP(P--U 1).

(6) If n= 3, P (:c) may t ake the forms (mod p) of (i) (x+ a)3,(ii) (z +a 2(.r + a2), (iii) (z + a i) (x +a2) (x + a3), (iv) (a; + aL)(a;2+ Aa;+ M , (V)x3+ X1x2+ X2x-\- ft (irreducible).In case (i) if p = 2 (so t ha t p < n

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90 J . B . MAR SHALLX P ( P - I ) _ 1 ; and (x + a2) is an independent factor of a:""1 1, P (x) mustbe a factor of xv{v~Vl 1. In case (iii), P (x) is a factor of xv~x 1.I n case (iv), since (x+ a) is a fa cto r of xv~x 1 and x2 + *-x + /x,which does not contain (x + a), is a factor of xp'~ x 1, P (x) must bea factor of x^'1 1. In case (v), P (x) is a factor of x ~x1.

Hence, in all cases, P (x) is a factor of xplq* 1 if p = 2, or ofXPI> 1 if ^ is an y o the r pri m e, wh ere q3 is the L.C.M. of p2 1 andp3 - 1, th a t is, q3= [(p2 - 1)(p3 - l)]/(p - 1).

(c) Similar arg um en ts show th a t , in general , when P (x) is ofth e n th degree, i t m us t be ei th er i rred uc ible , or made u p of factorswhich have appeared among the functions of lower degree, or bea power of some irreducible function of lower degree, and in allthree cases P (x) must be a factor (mod p) of x1 9,,1, wherepr is thelowest power of p ^ n, and where qn is the L.C.M. of

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EXT ENSIO N OP FERM AT'S THE OR EM TO MATRICES 91lowest value of qwhich makes Aq = I (m od 2) for all ma trice s is q= 6.This is also th e value obtain ed from th e ab ov e form ula by pu ttin gq = p (p21). Sim ilarly for n = 2,p = 3, 5, 7 the values of q arefound to be 24, 120, 336 re spe ctiv ely; for n = 3,p = 2, 3, th e value sof qare found to be 84 and 312 re sp ec tiv el y; an d for p 2, n = 4, 5the values of q are found to be 420 a nd 26040 resp ectiv ely. Thesame values of qare found from the fo rm ula .

MATHEMATICAL INSTITUTE,16 CHAMBERS STREET, EDINBUROH, 1 .