# Prove that a finite Abelian group $G$ is not cyclic if and only if it contains a subgroup isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$

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 $$a_1, and $$a_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.

• You have not said what $p$ is in your question. Commented Jan 18, 2019 at 9:17

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.
Hint: Using the fundamental theorem of finite abelian groups, $$G$$ is a product of cyclic groups.