> [!NOTE] Theorem > Up to [[Homomorphisms of groups|isomorphism]], there exactly two [[Groups|groups]] of order $4$: namely, the [[Cyclic Group of Order 4|cyclic of order four]] $C_{4}$ and the [[Klein 4-group]]. **Proof**: Suppose $G$ have four distinct elements: $1,a,b,c.$ Then either $G$ has an element of order $4$ or $G$ has no elements of order $4.$ In the first case, WLOG suppose $a$ is an element of order $4.$ Then $\langle g \rangle =G.$ WLOG suppose $b=a^{2}$ and $c=a^{3}.$ The Cayley table is as follows $ \begin{array}{c||c|c|c|c} \circ & 1 & a & b=a^2 & c=a^3 \\ \hline \hline 1 & 1 & a & b & c \\ \hline a & a & b & c & 1 \\ \hline b=a^2 & b & c & 1 & a \\ \hline c=a^3 & c & 1 & a & b \end{array} $ In the second case, by [[Order of Element of Finite Group Divides Order of The Group]], all the elements other than $1$ must have order $2$: that is, $a^{2}=b^{2}=c^{2}=1.$ By [[Group is Abelian if every square is the identity]], $G$ is abelian thus its Cayley table is as follows $ \begin{array}{c||c|c|c|c} \circ & 1 & a & b & c \\ \hline \hline 1 & 1 & a & b & c \\ \hline a & a & 1 & c & b \\ \hline b & b & c & 1 & a \\ \hline c & c & b & a & 1 \end{array} $