Groups of Order 4

Theorem
There exist exactly $2$ groups of order $4$, up to isomorphism:


 * $C_4$, the cyclic group of order $4$


 * $K_4$, the Klein $4$-group.

Proof
From Existence of Cyclic Group of Order n we have that one such group of order $4$ is the cyclic group of order $4$:

This is exemplified by the additive group of integers modulo $4$, whose Cayley table can be presented as:

From Group whose Order equals Order of Element is Cyclic, any group with an element of order $4$ is cyclic.

From Cyclic Groups of Same Order are Isomorphic, no other groups of order $4$ which are not isomorphic to $C_4$ can have an element of order $4$.

From Order of Element Divides Order of Finite Group, any other group of order $4$ must have elements of order $2$.

We have the Klein $4$-group, whose Cayley table can be presented as:

and is seen to have that property.

We have that Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic.

It remains to be shown that the Klein $4$-group is the only groups of order $4$ whose elements are all of order $2$ (except the identity).

Let the Cayley table be populated as far as can be directly established:


 * $\begin{array}{c|cccc}

& e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e &  &   \\ b & b &  & e &   \\ c & c &  &   & e \\ \end{array}$

Consider $a b$.

As $a^2 = e$, $a b \ne e$.

As $a e = a$, $a b \ne a$.

As $e b = b$, $a b \ne b$.

It follows that $a b = c$.

Hence we have:


 * $\begin{array}{c|cccc}

& e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c &  \\ b & b &  & e &   \\ c & c &  &   & e \\ \end{array}$

and the rest of the table is completed by following the result that Group has Latin Square Property.