# An attempted proof of Cauchy's theorem for abelian groups using composition series.

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).

• 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. Jan 22, 2019 at 17:51
• 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. Jan 22, 2019 at 18:58

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.
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.