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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?

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Looks good to me. This answer must be 30 comments long.

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