Proving that every group of order $4$ is isomorphic to $\Bbb Z_4$ or $\Bbb Z^*_8$

Question

I want to prove at any group of order $4$ is isomorphic to either $\Bbb Z_4$ or $\Bbb Z^*_8$. I know that these two groups are not isomorphic to each other because they have different order, but I cannot go further to prove at any group of order 4 should be isomorphic to either of them. Help would be greatly appreciated.

Thanks!

Take a non-identity element $x$ and look at possible orders of $\langle x\rangle$. — William Stagner, Jan 28 '16 at 19:47
The group $\mathbb{Z}_8^*$ is not cyclic - see here. But of course it also has order $4$. — Dietrich Burde, Jan 28 '16 at 19:51
@DietrichBurde I know, but how do I show that any group of order 4 is isomorphic to either of them? — Frank Booth, Jan 28 '16 at 19:54
Using Cauchy's theorem with $p=2$ we see that every group of order $4$ is isomorphic to $C_4$, or to $C_2\times C_2=C_8^*$. — Dietrich Burde, Jan 28 '16 at 19:56

score 1 · Answer 1 · answered Jan 28 '16 at 20:00

Hint :

Let $e,a,b,c$ the elements of a group of order $4$ and $e$ the neutral element.

Then, either $a^2=e$ or $a^2=b$ or $a^2=c$. $a^2=a$ can be ruled out because this implies $a=e$.

Now, you have to show two things :

Knowing $a^2$ means knowing all products of the group.
$a^2=b$ and $a^2=c$ lead to isomorphic groups.

Proving that every group of order $4$ is isomorphic to $\Bbb Z_4$ or $\Bbb Z^*_8$

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