How many non isomorphic groups of order 4 are there?

We will show that there are exactly two non isomorphic groups of order 4, namely the cyclic group of order 4, denoted by $C_4$ and the Klein four-group, denoted by $K_4$ .

Recall that:

$C_4=\lbrace x,x^2,x^3,x^4=e\rbrace$ , where $x\neq e$ .

and

$K_4=\lbrace e,x,y,z\rbrace$ where $x^2=y^2=z^2=e$ and $xy=z$ , $yz=x$ and $zx=y$ .

Prove that there are exactly two non isomorphic groups of order 4, namely the cyclic group of order 4 and the Klein four-group.

Let $G$ be a group of order 4. By Lagrange’s Theorem, it follows that the order of any element $g$ of $G$ divides the order of the group $G$ .

Hence the order of the elements of $G$ must divide 4.

Hence the order of the elements of $G$ are 1, 2 or 4.

The group $G$ has 4 elements. One and only one element has order 1, namely, the identity element (which is unique). Hence the remaining three elements must have order 2 or 4.

Consider the following two cases:
Case 1: The group $G$ has an element of order 4
Case 2: The group $G$ has no element of order 4

Case 1: The group $G$ has an element of order 4

Let $x \in G$ have order 4. Hence, by definition of order, $x^4=e$ where $x^1 \neq e$ , $x^2 \neq e$ and $x^3 \neq e$ .

Since $G$ is a group, it contains the identity element, hence $e \in G$ . We already know that $x \in G$ , and by closure $x^2,x^3 \in G$ .

Since $G$ has order 4 and $e,x,x^2,x^3\in G$ , then $G=\lbrace e,x,x^2,x^3 \rbrace$ which is the cyclic group $C_4$ .

Case 2: The group $G$ has no element of order 4

Let $G=\lbrace e,x,y,z\rbrace$ . Since $G$ has no element of order 4, it must follow that $x,y$ and $z$ have order 2. Therfore, $x^2=y^2=z^2=e$ .

Let us draw the Cayley table of this group.

Using the fact that the identity element leaves every element unchanged and that $x^2=y^2=z^2=e$ , we have:

$\begin{array}{c| c c c c} & e & x & y & z \\ \hline e & e & x & y & z \\ x & x & e & & \\ y & y & & e & \\z & z & & & e\end{array}$

By the axioms of groups, every element of the group must be included exactly once in every row and column of the Cayley table. Hence in row 2 column 3 one cannot put $x$ (because it is already used in row 2 column 1) and one cannot put $y$ because it is already used in row 1 column 3. Hence it must be $z$ .

$\begin{array}{c| c c c c} & e & x & y & z \\ \hline e & e & x & y & z \\ x & x & e & z & \\ y & y & & e & \\z & z & & & e\end{array}$

This procedure is repeated and it follows that the Cayley table can only be filled up in the following unique way.

$\begin{array}{c| c c c c} & e & x & y & z \\ \hline e & e & x & y & z \\ x & x & e & z & y \\ y & y & z & e & x \\z & z & y & x & e\end{array}$

Therefore $xy=z$ , $yz=x$ and $zx=y$ . This is thus the Klein four-group.

Hence the only two groups which have order 4 are the cyclic group and the Klein-four group, and their structure is different, hence they are non-isomorphic. $\square$