Proof that if $G$ is an abelian group and $p||G|$ then there is only one Sylow p-subgroup where all of its elements have powers of a p.

Question

I tried to proof this but I'm not sure if it's fine or if I'm missing something. Any help or hints are appreciated. This is what I got:

By hypothesis we know that $p||G|$ then $|G|=pm$. Supposing that $p|m$ then we can factorize the prime p out of $m$ such that $|G|=p^{r}n$ and $p\nmid{n}$. By the Sylow theorem exists $H\leq{G}$ of order $p^{r}$. Suppose there exists another p-subgroup $K\leq{G}$ which elements have powers of a prime. It's clear that $K\leq{H}$. Sylow theorem also tells us that any two p-subgroups are conjugate $H=gKg^{-1}$, but remembering that $G$ is an abelian group so all of its subgroups must be abelian aswell. Then we get $H=K$ and there is only one Sylow p-group.

In the case $p\nmid{m}$ we have a Sylow p-group of order $p$ and we follow the previous steps to get the same result.

Answer 1

I read your question as asking "if $p \mid |G|$ and $G$ is Abelian, then there is only one Sylow $p$-subgroup". If this is not what you meant, please let me know and I'll update my answer.

The proof that you gave in the question is close, but is much more complicated than it needs to be. Your plan of using the abelian-ness of $G$ to get that $P$ is unique is a good one, and it will work. But your assertion that $K \leq H$ is not true in general, and we do not need the fact that all of the subgroups of $G$ are abelian. Also, I'm not sure what your case $p \not \mid m$ at the end is about. You case on $p \div m$ at the beginning, but there is no need to do so (unless I misunderstand the result you are proving). Here is a cleaned up version of your proof:

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Write $|G| = p^r m$ with $p \not \mid m$.

By the Sylow Theorems, we know that $G$ has a Sylow $p$-subgroup. That is, $G$ has a subgroup $P$ of order $p^r$, as you stated in your question.

The Sylow Theorems also tell us that every $P' \leq G$ of order $p^r$ is conjugate to $P$. For some $g \in G$,

$$gPg^{-1} = P'$$

Again, as you note in your question, since $G$ is abelian, we have $P' = gPg^{-1} = P$. Thus, every $p$-Sylow subgroup must be equal to $P$, and so $P$ is the only one.

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I hope this helps ^_^

Answer 2

You don't have to make this distinction about whether $p$ divides $m$ or not; the whole point here is that you can define

$$H:=\{g\in G\mid\text{ord$(g)$ is a power of $p$}\}$$

and under the assumption that $G$ is abelian, you can prove this is a subgroup of $G$. Then you'll find that $|H|$ is a power of $p$, so $H$ must be contained in some Sylow $p$-subgroup $P$. But using Lagrange's theorem you can easily prove that $P\subseteq H$, so in fact we must have $H=P$.