Let $G$ be a finite abelian group and let $p$ be an odd prime. Prove that $G$ has $\frac{i_{2p}(G)}{(p−1)}$ subgroups of order $2p$, where $i_{2p}(G)$ is the number of elements of order $2p$ in $G$.
Clearly for each $x \in G$ with order $2p$ the cyclic subgroup $\langle x \rangle$ has order $2p$, for which there are $i_{2p}(G)$ elements, so there are $i_{2p}(G)$ subgroups of order $2p$. Then some of these must equal each other to gain the required result, $p-1$ in fact.
I am struggling to show this fact, I have an idea that any odd power of $x$ not equal to $p$ will generate the same subgroup, giving me the required $p-1$ identical subgroups. But not really sure if this is right or how to prove it. Also if this were the correct route to take how would I show that there aren't any subgroups of order $2p$ not of the form $\langle x \rangle$ ?.
Best Answer
An abelian subgroup of order $2p$ is cyclic and therefore is $<x>$ for some element of order $2p$.
The number of elements of order $2p$ in such a subgroup is indeed $p-1$ as you say and so these elements generate $\frac{i_{2p}(G)}{(p−1)}$ subgroups.