Give a reason why $\mathbb Z_4 \times \mathbb Z_2 $ and $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 $ are not isomorphic

Question

Give a reason why $\mathbb Z_4 \times \mathbb Z_2 $ and $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 $ are not isomorphic

In order for them to be be isomorphic, don't they have to have the same order?

Answer 1

If two groups $G_1$ and $G_2$ are isomorphic, then they have same same algebraic structure: orders of elements are preserved; in particular, if $\phi:G_1 \to G_2$ is an isomorphism then $\lvert a \rvert =\lvert \phi(a) \rvert$. Now you can easily see that $\mathbb{Z}_4\times \mathbb{Z}_2$ has an element of order $4$, but $\mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2$ has no element of order $4$; in fact, every non-identity element of $\mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2$ is of order $2$. Therefore, these two groups cannot be isomorphic.

Answer 2

Hint: can you find an element of order $4$ in any of these groups?

Answer 3

In order for them to be be isomorphic, don't they have to have the same order?

Yes, but groups of the same order need not be isomorphic!

One of your groups has an element of order $4$ whereas the other does not.

Use the following

Lemma: If $\varphi: G\to H$ is an isomorphism, then for any $g\in G$, the order of $\varphi(g)$ in $H$ is that of $g$ in $G.$

Can you prove this lemma?

Hint: It follows from the fact that $\varphi$ takes the identity element $e_G$ of $G$ to that of $H$, and preserves powers, i.e., $\varphi(k^n)=(\varphi (k))^n$ for all integers $n$ and all $k\in G$.