Theorem:LetGbe a finite group andpbe a prime. Ifpdivides the order ofG, thenGhas an element of orderp.

Proof 1:We induct onn= |G| and consider the two cases whereGis abelian orGis nonabelian. SupposeGis abelian. IfGis simple, then it must be cyclic of prime order and trivially contains an element of orderp. Otherwise, there exists a nontrivial, proper normal subgroup . Ifpdivides |H|, thenHcontains an element of orderpby the inductive hypothesis, and thusGdoes as well. Otherwise,pmust divide the index [G:H] by Lagrange's theorem, and we see the quotient groupG/Hcontains an element of orderpby the inductive hypothesis; that is, there exists anxinGsuch that (Hx)^{p}=Hx^{p}=H. Then there exists an elementh_{1}inHsuch thath_{1}x^{p}= e, the identity element ofG. It is easily checked that for every elementainHthere existsbinHsuch thatb^{p}=a,so there existsh_{2}inHso thath_{2}^{p}=h_{1}. Thush_{2}xhas orderp, and the proof is finished for the abelian case.

Could someone please explain for me using small words and step by stem everything after the word 2nd otherwise...

Does index [G:H] mean the number of (right) cosets of H in G? Also, where it says, it is easily checked...how?

We have gone over this proof in class 4 times. Also, I have gone to office hours twice. I understand it and follow when I'm there, but I can't replicate.