# Number of Subgroups of Finite Abelian Group of Order $2p$

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 $$\langle x \rangle$$ ?.

An abelian subgroup of order $$2p$$ is cyclic and therefore is $$$$ for some element of order $$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.
• 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 $2p$. – Albert B Dec 7 '19 at 17:58
• Yes it does have a generator - multiply the generators of $Z_p$ and $Z_2$. – S. Dolan Dec 7 '19 at 17:59