#A42 INTEGERS 11 (2011 ON THE CONDITIONED BINOMIAL COEFFICIENTS Liqun To Shool of Mthemtil Sienes, Luoyng Norml University, Luoyng, Chin lqto@lynuedun Reeived: 12/24/10, Revised: 5/11/11, Aepted: 5/16/11, Pulished: 6/27/11 Astrt We nswer question on the onditioned inomil oeffiients rised in nd rtile of Brlotti nd Pnnone, thus giving n lterntive proof of n extension of Froenius generliztion of Sylow s theorem 1 Introdution In [2], Brlotti nd Pnnone proved the following extension of Sylow s theorem [6]: Theorem 1 Let G e finite group of order n, p prime dividing n, H sugroup of G of order p h Then for ny positive integer k > h suh tht n, the rdinlity of the set of ll the p-sugroups of G of order ontining H is ongruent to one modulo p In the speil se h 0, this result ws first proved y Froenius [3], nd redisovered y Krull [4] The proof in [2] is given y onsidering the olletion of ll the susets of G hving rdinlity nd ontining extly h right osets of H It is worth pointing out tht the ove result ws lso proved independently y Spiegel [5] using Möius inversion methods developed in Weisner s pper [7] As suggested nd finlly rised s question in [2], one should e le to show the result y onsidering the fmily of the susets of G hving order nd ontining t lest one right oset of H This leds to the following: Definition 2 ([2] Let,, e positive integers suh tht nd Let A e set of rdinlity prtitioned into susets ll of rdinlity The onditioned inomil oeffiient determined y, nd, denoted y is defined to e the numer of susets of A of rdinlity ontining t lest one omponent of the prtition,
INTEGERS: 11 (2011 2 The im of this pper is to nswer the question rised in [2], whih sks to prove the following: Theorem 3Let,, e positive integers suh tht nd Then (1 divides (2 If is power of prime p, then / 1 (mod p Remrk 4 If 1, the onditioned inomil oeffiient determined y, nd is just ( Then Theorem 3 follows from Lemms 5 nd 6 in the next setion Our method of proof of the min result is to express the onditioned inomil oeffiient expliitly s omintion of usul inomils nd then onsider the divisiility nd ongruene of these inomils (see (1 elow The method we use is very elementry We refer the reder to [1] (Chpters 2 nd 6, or other stndrd lger textooks for terminology nd nottions used in the pper 2 Proof of the Min Result We will need the following two results The first one n e verified diretly; the seond one is shown in [4], ut we still give proof here for the reder s onveniene Lemm 5 Let, e positive integers suh tht divides Then ( ( 1 1 Lemm 6 Let p e prime, s positive integer, nd g positive integer divisile y p s Then ( g 1 p s 1 1 (mod p Proof We prove the result y indution on s, the exponent of p For s 1, sine p g, we hve g i p i (mod p for 1 i < p, nd hene (g 1(g 2 (g (p ( 1 (p 1(p 2 1 (mod p As (p 1(p 2 1 is prime to p, we get g 1 p 1 (g 1(g 2 (g (p 1 (p 1(p 2 1 1 (mod p Suppose tht the result holds for exponents less thn s By strightforwrd omputtion we otin the equlity ( g 1 p s 1 ( g p 1 p s 1 1 ps 1 1 j0 p 1 i1 g (jp+i p s (jp+i Then the result follows y the indution hypothesis nd the ove rgument for the se s 1
INTEGERS: 11 (2011 3 Proof of Theorem 3 Let A e set of rdinlity prtitioned into susets ll of rdinlity The numer of susets of A of rdinlity ontining no omponent of the prtition is 0 n i < n 1+n 2+ +n ( ( n 1 n 2 (, n whih is the oeffiient of the term x in ((1 + x x The oeffiient equls Then ( r0 r0 ( ( ( r ( ( ( r r r r r ( 1 r ( 1 r ( 1 r 1 A r, (1 r1 where A r ( r ( ( r r For 1 r, y Lemm 5, we hve A r ( 1 r r 1 the other hnd, y Lemm 5 gin, we hve A r ( ( ( r r r ( 1 ( r r 1 ( 1 ( ( r r 1 Thus, ( ra r Therefore we hve tht divides A r So we otin r r ( ( r 1 r 1 r r ( ( r 1 r 1, hene ra r On ( r,r (,r divides A r A fortiori, This ompletes the proof of the first prt of the theorem For the seond prt, in view of Remrk 4, we my ssume tht,, p h re positive integers, where p is prime nd k > h 1 From (1, we hve
INTEGERS: 11 (2011 4 p h ( k h p r0 where B r ( ( m (m rp h r rp, nd for 1 r < h, h ( 1 r B r, m ( h,r divides B m r Hene h 1 divides B r, whih implies tht B r / mph 0 (mod p, for 1 r < h By Lemms 5 nd 6, ( ( mp h mph 1, 1 hene, By the sme rgument, ( mp h/ mph 1 (mod p ( m / mph h 1 (mod p Thus whether p is n odd prime or not, we hve B 0 + B h( 1 pk h 0 (mod p Putting the ove fts together, we get / mph p h 1 (mod p Remrk As mentioned t the eginning of the pper, we my prove Theorem 1 y onsidering F H (pk, the fmily of the susets of G hving order nd ontining t lest one right oset of H Let S H ( e the set of ll the p-sugroups of G of order ontining H Consider the right-multiplition tion of G on F H (pk Then F H (pk is prtitioned s union of orits: F H (pk l i1 O Ui Note tht St(U i divides U i, nd St(U i if nd only if O Ui ontins (unique sugroup K of order, nd hene onsists of right osets of K Sine K ontins some right oset of H, tully K H Thus we hve F H (pk l i1 G St(U i G ( S H ( + multiple of p
INTEGERS: 11 (2011 5 Applying Theorem 3 to the se G,, H p h, we otin F H (pk / mph / mph 1 (mod p p h Thus s desired S H ( 1 (mod p Aknowledgement This work is prtilly supported y the Sientifi Reserh Foundtion for the Returned Overses Chinese Sholrs, Stte Edution Ministry nd Reserh Seed Fund of Luoyng Norml University for Provinil or Ministry level projet The uthor is very grteful to the referee for helpful omments nd suggestions Referenes [1] M Artin, Alger, Prentie-Hll, 1991 [2] M Brlotti nd V Pnnone, Sylow s theorem nd the rithmeti of inomil oeffiients, Note di Mtemti 22 (2003, 83-87 [3] G Froenius, Verllgemeinerung des Sylowshen Stze, Berliner Sitz, (1895, 981-993 [4] W Krull, Üer die p-untergruppen, Arhiv der Mth, 12 (1961, 1-6 [5] E Spiegel, Another Look t Sylow s Third Theorem, Mth Mg 77 (2004, 227-232 [6] M L Sylow, Théoremès sur les groupes de sustitutions, Mth Ann 5 (1872, 584-594 [7] L Weisner, Some properties of prime power groups, Trns Amer Mth So 38 (1935, 485-492