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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 $\langle x \rangle$ ?.

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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.

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  • $\begingroup$ May I ask how you know that an abelian subgroup of order $2p$ is cyclic, surely $G=\mathbb{Z}_p \times \mathbb{Z}_2$ has order $2p$ but isn't cyclic as it doesn't have an element of order $2p$. $\endgroup$ – Albert B Dec 7 '19 at 17:58
  • $\begingroup$ Yes it does have a generator - multiply the generators of $Z_p$ and $Z_2$. $\endgroup$ – S. Dolan Dec 7 '19 at 17:59
  • $\begingroup$ OK I understand now. $\endgroup$ – Albert B Dec 7 '19 at 18:00
  • $\begingroup$ Glad to be of help. $\endgroup$ – S. Dolan Dec 7 '19 at 18:01

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