Show that $(\mathbb{Z}_{n} \times \mathbb{Z}_{n} , +)$ is not isomorphic to $ (\mathbb{Z}_{n^{2}} , +)$

Question

Show that $(\mathbb{Z}_{n} \times \mathbb{Z}_{n} , +)$ is not isomorphic to $ (\mathbb{Z}_{n^{2}} , +)$

I know that isomorphism preserves the order. In the first group there is no element of order $n^{2}$ but there is in the second group. So these groups are not isomorphic.

I think I am right. Is there another way to solve this problem?

Answer 1

Yes, you're correct (assuming $n>1$).

Suppose $\varphi: G\to H$ is a group isomorphism. Let $g\in G$ with order $m$. Then $\varphi(g)$ has order $m$ in $H$.

Another way to prove they're not isomorphic, although it's overkill, is to employ the Fundamental Theorem of Finitely Generated Abelian Groups.

Answer 2

Nit: You should state your assumption that $n>1$. If you don't want to make a big deal out of it, you can insert the assumption where it's used:

In the first group there is no element of order $n^2$, since $n>1$ guarantees $n^2>n$, but there is in the second group.

Better yet, add the assumption to the statement that you're proving.

Answer 3

More generally: $$\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \iff \operatorname{gcd}(m,n)=1$$

Therefore, if $n>1$, $\mathbb{Z}_n \times \mathbb{Z}_n \ncong \mathbb{Z}_{n^2}$.

Answer 4

The latter is cyclic; the first is not. This can be summed up by saying $\Bbb Z_{n^2}$ has an element of order $n^2$; whereas $\Bbb Z_n×\Bbb Z_n$ does not.