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I came up with this proof for the abelian version of Cauchy's Theorem (if a prime $p$ divides the order of an abelian group then it has a subgroup of order $p$). I'm hoping someone could please check it's correct, and then answer a few questions about it. I hope this kind of question is appropriate here.

If $G$ is cyclic then the result is obvious, so assume otherwise and proceed by induction on $\lvert G \rvert$. Since $G$ is finite we can take a composition series $$G \supset G_1 \supset G_2 \supset \ldots \supset \{e\},$$ and since $\lvert G \rvert$ is the product of the orders of its composition factors there must be a factor $\frac{G_i}{G_{i+1}}$ with order divisible by $p$. But then $\lvert G_i \rvert = \lvert \frac{G_i}{G_{i+1}} \rvert \lvert G_{i+1} \rvert $ is also divisible by $p$. If $G_i \neq G$ then the result follows via induction, so assume $G_i = G$, and pick $x \in G \setminus G_{1}$. By simplicity of the first composition factor, $G_{1}$ is a maximal proper subgroup, so it must be that $\langle x \rangle G_{1} = G$, so $\lvert G \rvert = \frac{\lvert x \rvert \lvert G_{1} \rvert}{\lvert \langle x \rangle \cap G_{1} \rvert }$. Then $\lvert x \rvert = \lvert \langle x \rangle \cap G_{1} \rvert \lvert \frac{G}{G_{1}} \rvert$, so $p$ divides $\lvert x \rvert$ which completes the proof.

Questions:

  1. Is the proof correct? It feels too easy which makes me worried. If it is correct, have I added any unnecessary complications that could be removed to simplify it? If it's not correct can it be salvaged?

  2. Is there any way to make this work for non-abelian groups? I'm pretty sure I only used commutativity in the second last sentence, in the non-commutative case $G_1$ is a maximal normal subgroup and $\langle x \rangle$ isn't necessarily normal so the proof doesn't go through. Is there some way to work around this? All I can think of is replacing $\langle x \rangle$ with the normal subgroup generated by $x$, but I don't know how to show that it's proper (if it even is).

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    $\begingroup$ You can compare your proof with a "book proof" here of Cauchy's Theorem for abelian groups, which is may be a bit easier than yours. $\endgroup$ – Dietrich Burde Jan 22 at 17:51
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    $\begingroup$ You've used the dreaded word "obvious". Is it really obvious (to you or to the intended reader) or just familiar? Remember that there's a plethora of would-be theorems out there that seem obvious but turn out to be wrong. $\endgroup$ – Shaun Jan 22 at 18:58
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Your proof is fine. You can actually simplify it a bit since you don't even need to use induction at all once you have your composition series. Indeed, let $i$ be such that $p$ divides $G_i/G_{i+1}$. Then there is an element $x\in G_i/G_{i+1}$ of order $p$ (since $G_i/G_{i+1}$ is simple and in particular cyclic). Now pick $y\in G_i$ whose image in $G_i/G_{i+1}$ is $x$. The order of $y$ is a multiple of the order of $x$, and so the order of $y$ is divisible by $p$ and we're done.

If you try to make a similar argument for nonabelian groups, it would reduce to proving Cauchy's theorem for simple groups. This is a nontrivial reduction, but I don't think it actually really helps: I don't see how it would be any easier to prove Cauchy's theorem for simple groups than for arbitrary groups, except in that it handles the abelian case for proofs of Cauchy's theorem that treat the abelian and nonabelian cases separately.

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A different way of reaching the conclusion is as follows.

Let $\phi$ be the Frobenius map $x \rightarrow x^{p}$. As $G$ is abelian it is easy to see that $\phi$ is a homomorphism. Clearly $im(\phi) \subseteq G_{1}$. Hence $ker(\phi) \neq 1$. Since the non-trivial elements in $ker(\phi)$ are elements of order $p$, the result follows.

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