In a finite abelian group, the order of every element divides the order of the element with maximal order Can someone explain in detail Sean's proof for this same question in this thread? Specifically I'm confused about the following parts:
1) What does he mean when he says $ " $Suppose $ p $ is a prime dividing $ |b| $ to a higher power than $ |a|?" $
2) He says that we can write $ |a| = p^im $ and $ |b| = p^jn $ where $ j > i $ and $ p $ divides neither $ m $ nor $ n, $ but how can we prove that we can find such a $ p $ with that property? 
3) The proof reaches a contradiction at the end, but how does this imply that the initial theorem must be true?
4) This theorem is only true if the group is abelian, but I can't see where he uses this fact in his proof.
 A: 1) This means that $p^k$ divides $|b|$ for some natural number $k$ such that $p^k$ does not divide $|a|$.
2) This is restatement of (1).
3) In the prime factorizations of $|b|$ and $|a|$, every exponent occuring in $|a|$ is larger than the exponent that occurs in $|b|$. This implies that $|b|$ divides $|a|$.
4) The lemma in Sean's proof requires $G$ to be abelian: he argued that $(xy)^m = 1$ implies that $$(x)^{m|y|} = (x)^{m|y|} y^{m|y|} = (xy)^{m|y|} = ((xy)^m)^{|y|} = 1.$$ This already fails for the smallest nonabelian group $S_3$: let $x$ be the transposition $(1\, 2)$ and $y$ the cycle $(1\, 2\, 3).$
Here, $(xy)^3 = 1$ but $x^{3|y|} = x^9 \ne 1.$
A: (1) He means that $|a|=p^im$ and $|b|=p^jn$ for some $m,n$ coprime to $p$ and $j>i$.
(2) We don't prove that there is such a $p$ with that property, we're assuming it's true for the sake of reaching a contradiction.
(3) If there is no $|b|$ divisible by a higher prime power than $|a|$ is, then $|b|$ must divide $|a|$ (just compare their prime factorizations) for every element $b$.
(4) The lemma is used to say what the order of $a^{p^i}b^n$ is, and the lemma (while not stated) requires that $G$ be abelian.
