Abelian Non-Cyclic p-group I am trying to show that for a given prime $p$ and a finite Abelian $p$-group $G$ that if $G$ has an unique subgroup of order $p$, then $G$ is necessarily cyclic.
I am trying to prove the statement, by arguing by contradiction. Of course, the statement would be easy to prove if one knows the Fundamental Theorem of Finitely Generated Abelian Groups, but I was trying to prove the statement without invoking this Theorem, and I am having trouble proving the statement.
I know that by Cauchy's Theorem, one can find an element $x$ in $G$ of order $p$; how does one then produce an element $y\in G\backslash\langle x\rangle$ of order $p$?
 A: Let $G$ contain an unique cyclic subgroup of order $p^k$ and $a$ be an element of order $p^k$. Prove that it contains an unique cyclic subgroup of order $p^{k+1}$. 
Let $x,y$ be elements of order $p^{k+1}$ and $x^{p^k}=y^{p^k}=a$. Then $(xy^{-1})^{p^k}=1$, so $x=ya^m=y^{mp^k+1}$. Further use the induction.
A: I assume that $G$ is finite (sometimes the term "$p$-group" is used for infinite groups as well, but for infinite abelian $p$-groups the statement is false).
If it is undesirable to use FTFGAG, then I would proceed as follows.


*

*Introduce the endomorphism $\varphi: G \to G,\ g \to p g$. I write $G$ additively, so $pg$ means $g+g+\ldots+g$ ($p$ times).

*Introduce subgroups $G_i = \ker \varphi^i$. Clearly, $G_0 = 1$. Also, $G_1 \simeq \mathbb{Z}_p$ is the unique subgroup of order $p$. Since $G$ is finite, $G = G_i$ for large enough $i$. Let $n$ be the minimal such $i$.

*Note that $\varphi$ induces an injective homomorphism from $G_{i+1}/G_i$ to $G_i/G_{i-1}$ for every $i \geq 1$. It follows using induction that $G_{i+1}/G_{i} \simeq \mathbb{Z}_p$ for every $0 \leq i < n$.

*Now it is quite easy to see that any element from $G_n \setminus G_{n-1}$ generates $G_n$, so $G=G_n$ is cyclic.
