I want to prove that every group of order $4$ is going to be isomorphic to these two. I am trying to show that there exist only $2$ non-isomorphic groups of order $4$. I found the groups using Cayley Tables, (I think one is called the Klein group that I found, and the other one is a cyclic group generated that is isomorphic to $\Bbb Z_4$.)
To show that the two groups are not isomorphic to each other, I showed that one has $x^2=e$ for all $4$ elements, while the other one only has $2$. So, one has $|m| = 4$, and other has $|n|=2$. Thus, they are not isomorphic to each other.
I know there are many questions like this such as: Prove that there only two groups of order 4 up to isomorphism
I have found the $2$ groups already through trial and error and I also made their Cayley tables.
But, I want to prove that every other group of order $4$ is going to be isomorphic to these two. How do I show that every other group is isomorphic to them? I know that for example, $\Bbb Z_4$ is, but there are many possible groups, and listing out examples is not enough to validate that "All other order $4$ groups are isomorphic to these two."
 A: Let $G$ be a group of order four. Then $G$ is abelian.
Since $2\mid 4$, by Cauchy's Theorem, there exists an element $g\in G$ of order two. There exist three other elements of $G$, one of which is the identity $e$; call the other two $a,b$. If $ag=e$, then $a=ae=ag^2=(ag)g=g$, a contradiction, so either: $ag=a$, which implies $g=e$, a contradiction; $ag=g$,  which implies $a=e$, a contradiction; or $ag=b$. So let $ag=b$.
The orders of $a,b$ can be either two or four by Lagrange's Theorem.
Suppose $a$ has order two. Then $g=eg=a^2g=a(ag)=ab$. We have $e=g^2=(ab)^2=a^2b^2=eb^2=b^2$, so, since $b\neq e$, the order of $b$ is two. Thus $G$ must be isomorphic to $\Bbb Z_2^2$.
Now suppose $a$ has order four. Then $e, a, a^2, a^3$ are all distinct. We have either $a^2=g$ or $a^3=g$. If the latter, then $e=a^3a=ga=b$, a contradiction; hence $a^2=g$. But that leaves $a^3=b$. Thus $G$ must be isomorphic to $\Bbb Z_4$ (since it is generated by $a$).
A: The abelianess without any appeal to orders of elements, Cauchy's Theorem or Lagrange's or type of group: assume that $G$ has $4$ elements and is not abelian. Then we can find two non-identity elements $a,b$ that do not commute, so $ab \neq ba$. Note that this implies $ab \notin \{e,a,b\}$. Because $G$ is closed under the group operation we must have $G=\{e,a,b,ab\}$. Now $ba$ belongs to this set. Since $a$ and $b$ do not commute $ba \notin \{e,a,b\}$. Hence we must have $ab=ba$, a contradiction. So $G$ is abelian after all.
OK, so we have $\{e,a,b,ab\}$ is an abelian group of $4$ elements, so each of them does not equal one of the others. So were is $a^2$? Assume for the moment that $a^2 \neq e$. Can $a^2=a$? No, since then it would follow $a=e$. Can it be $a^2=b$? Yes, and it that case $ab=a^3$ and the group looks like $\{e,a,a^2,a^3\}$. Observe that $a^4$ must be the identity since if $a^4=a$, $a^4=a^2$ or $a=a^3$, the set reduces to less than $4$ elements. So the group is cyclic of order $4$ generated by $a$, in this case.
Similarly, if $b^2=e$, we would arrive at the group $\{e,b,b^2,b^3\}$, which is of course again cyclic of order $4$ and isomorphic to the one we already found.
We are left with the case where $a^2=e=b^2$. But then (using abelianess) $(ab)^2=abab=aabb=a^2b^2=e.e=e$ and now all elements of the group have order $2$. This one is isomorphic to $C_2 \times C_2$, a direct product of two groups of order $2$, also called the Klein $4$-group $V_4$.
So you see in this small case everything can still be figured out without using more sophisticated theorems.
A: It seems to me that you can also reason this way.
Let $G$ be a group of order 4.

*

*Every element of $G$ other than $e$ is of order 2 or 4.
In fact, if $a\in G$, then $|a|\leq 4$. Let $|a|=3$, and $b\notin\{e,a,a^2\}$. It is easy to see that $ab\notin\{e,a,a^2\}$ but then $|G|>4$. Contradiction.


*If $G$ has an element of order $4$, then $G$ is a cyclic group.


*If all elements of $G$ except $e$ are of order $2$, then the group $G$ is abelian ($ab=a^{-1}b^{-1}=(ba)^{-1}=ba$). Let $a$ and $b$ be two different non-identity elements of $G$. We have $G=\{e,a,b,ab\}$. This means that any two such groups are isomorphic.
It follows that every group of order 4 is either a cyclic group or the Klein four-group.
