# 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:

$\quad \begin {array} {r|rrrr} \struct {\Z_4, +_4} & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \hline \eqclass 0 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \eqclass 1 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 \\ \eqclass 2 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 \\ \eqclass 3 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 \\ \end{array}$

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 $4$.

$\Box$

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:

The Klein $4$-group can be described completely by showing its Cayley table:

$\quad \begin {array} {c|cccc} & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end {array}$

and is seen to have that property.

$\Box$

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

$\Box$

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.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $4$ - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 41 \zeta$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Groups with four elements