Prove that a finite Abelian group $G$ is not cyclic if and only if it contains a subgroup isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$.
I am aware an answer exists here. I have been trying to work through the proof step by step and am running into some issues.
The reverse case is easy. Clearly if $G$ contains a subgroup $H$ isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$, (where $p$ is prime) it follows that $G$ cannot be cyclic, since every subgroup of a cyclic group is cyclic, and $H$ cannot be cyclic if it is isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$. The other direction is proving to be very challenging for me.
Suppose that $G$ is not cyclic. Since $G$ is finite, it is finitely generated and so, it is isomorphic to a group of the form $$\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times…\times\mathbb{Z}_{p_n^{r_n}}.$$ $p_i=p_j$ for some $i,j$ with $i\neq j$, since otherwise $\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times…\times\mathbb{Z}_{p_n^{r_n}}$ would be cyclic (and so too $G$ by existence of an isomorphism between the two). Without loss of generality, I assume that $i=1,j=2$. I am having trouble finding a subgroup isomorphic to $\mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}$. Clearly the set of elements of the form $(a_1,a_2,0,…,0)$ with $a_1<p_1^{r_1}$, and $a_2<p_2^{r_2}$ is a subgroup isomorphic to $\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}$, by means of the isomorphism $\phi:\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\to\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times\{0\}\times…\times\{0\}$, with $\phi:(a,b)\mapsto(a,b,0,…,0)$. However, I'm not sure how to use this kind of logic to find a subgroup isomorphic to $\mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}$.
A nudge in the right direction would be very much appreciated.
Best Answer
Since $G$ is not cyclic, $n>1$.
On the other hand, $p_1,p_2,\dots,p_n$ cannot be all distinct, because a direct product of cyclic groups with pairwise coprime orders is cyclic (Chinese remainder theorem).
Let $p$ be a prime that repeats. Since $\mathbb{Z}(p^n)$ (I use this more readable notation for the cyclic group of order $p^n$) contains a subgroup of order $p$, you are done.