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."
Best Answer
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.