# Proving there are only two groups of order $4$, up to isomorphism

I am trying to prove, for self-study, that there are two groups of order $$4$$ up to isomorphism. I have a characterization of the Klein four-group as $$\{e, a, b, c\}$$ where $$a,b,c$$ have order $$2$$ and the product of any two non-identity elements gives the third non-identity element. I also have proven that any finite cyclic group is isomorphic to $$\mathbb{Z}/4$$. I haven't proved that the Klein group is isomorphic to $$\mathbb{Z}/2 \times \mathbb{Z}/2$$, but I assume this is as simple as writing down its elements and verifying that they match the defining relations of the klein group.

Here is my attempt.

Let $$G$$ be a group of order $$4$$. By Lagrange's theorem, the order of any element of $$g$$ must divide $$4$$, so the possible order of $$g \in G$$ can be $$1$$, $$2$$, or $$4$$. The identity element is, of course, the unique element of order $$1$$. If $$G$$ admits an element of order $$4$$, $$g$$, then $$\langle g \rangle = G \cong \mathbb{Z}/4$$. Suppose that $$G$$ lacks an element of order $$4$$. Then the three non-identity elements $$a,b,c$$ of $$G$$ must have order $$2$$. That is, $$a^2 = b^2 = c^2 = e$$. Furthermore, $$ab \neq a$$ and $$ab \neq b$$; by cancellation, $$ab = a$$ would imply that $$b = e$$ and $$ab = b$$ would imply that $$a = e$$. Finally, $$ab \neq e$$ because, otherwise, $$a = b^{-1}$$, though because $$a$$ and $$b$$ have order $$2$$, they are their own inverses, and group inverses are unique. Therefore, $$ab = c$$. By exactly analogous arguments, interchanging labels, we find that $$ac = b$$ and $$bc = a$$. Therefore, the entire multiplication table is determined, and in fact $$G$$ is isomorphic to the Klein four-group.

How does this look?

• Mar 4, 2023 at 14:00