Theorem: Let G be a finite group and p be a prime. If p divides the order of G, then G has an element of order p.
Proof 1: We induct on n = |G| and consider the two cases where G is abelian or G is nonabelian. Suppose G is abelian. If G is simple, then it must be cyclic of prime order and trivially contains an element of order p. Otherwise, there exists a nontrivial, proper normal subgroup . If p divides |H|, then H contains an element of order p by the inductive hypothesis, and thus G does as well. Otherwise, p must divide the index [G:H] by Lagrange's theorem, and we see the quotient group G/H contains an element of order p by the inductive hypothesis; that is, there exists an x in G such that (Hx)p = Hxp = H. Then there exists an element h1 in H such that h1xp = e, the identity element of G. It is easily checked that for every element a in H there exists b in H such that bp = a, so there exists h2 in H so that h2 p = h1. Thus h2x has order p, 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.