Question

Let Ga finite abelian group. Prove that a)If pa primenumber divides G|, G has an element of order p b)If G2n with n odd, G has exactly oneelement with order 2

Answer 1

Gues) Given 6a inite abelian qaup the prave C Ca) IT P a pnme numbes clivides (G1, G thes element f rdes p Proof G is finite abelian, if 141- P then is Ordles p tsomophic to Zp G has an element Now Let has prepes non trivia subayaup H 14/H1 then eithes P IHI p 1) I P/IHI then pliG1 Case as elemert onf ardes p Hence Gas an case -) f P| 1G/H) geG) 3f 19+HI-P in G/H where <9s Zxp fors Some x 1311 191 then L-e then xgeG has arder p element of oreler P has G

(b) I 141-2n with element of odd, G far exactty n ane Crder 2 Pecof Case- Lei G doesn't has an element ordles twe e. at a a eG Let G has clement as,a, n and invese f ai-9i atat implies aeG of rder 2 GTs abellan But such thot <a= G hence each olivisor torm subqaup that 932up order der 21 n Cia1) ) Ghas subaraup Ghas element af ordler2 .This contradicts assumption element of odes 2. G has an and (a1- l/ 2 Let Case- 4,ьс4 then <e,a, b, ab fams but 12ea,abs1 4 subpadan is odd then but 41an 1n n Is assumption n te. auus element than ane Mare hence Can nat have of order 2.