Showing $\Bbb Z_4 $ and $\Bbb Z_2 \times\Bbb Z_2$ are the only abelian group with $4$ elements.

Question

We consider the two abelian groups

$\Bbb Z_4$ with addition modulo $4$.

$\Bbb Z_2 \times\Bbb Z_2$ with component-by-component addition modulo $2$.

a) Show there is no isomorphism between these groups.

b) Show that this groups are the only abelian groups( except isomorphism) with $4$ elements.

So I was able to solve a): all elements in $\Bbb Z_2 \times\Bbb Z_2$ have order $ \leq 2 $, but $1$ in $\Bbb Z_4$ have order $4$.

But how can I solve b) ? Do you have a hint? Maybe I can use a)?

Answer 1

Let $G$ be a group of order $4$, not necessarily abelian.

If $G$ has an element of order $4$, then $G$ is cyclic and isomorphic to $\mathbb Z_4$.

Otherwise, all elements have order at most $2$, that is, $g^2=e$ for all $g \in G$.

Let $a,b \in G$, with $a\ne e$, $b\ne e$, $a\ne b$. Then $G = \{ e, a, b, ab \}$ because $ab\ne e$ (since $b \ne a =a ^{-1}$). Also, $ba=ab$, that is, $G$ is abelian. Then $G$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$ via $a \mapsto (1,0)$ and $b \mapsto (0,1)$.

Answer 2

Actually, any group of order $4$ is abelian. Then if you're willing to use a sledgehammer, you could use the fundamental theorem of finite abelian groups.

Or by Lagrange, either there are three elements of order two, or else it's cyclic.