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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.

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user362709
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2 Answers2

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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 $x^{3|y|} = x^9 \ne 1.$

user362705
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(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.

anon
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