# Statement(s) > [!NOTE] Statement 1 (There are only two groups of order $4$ up to isomorphism) > 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]], $C_{2}\times C_{2}$. > [!NOTE] Statement 2 >Let $G$ be a [[Finite Group|finite group]] of order $4.$ Then either $G\cong C_{4}$ or $G \cong C_{2} \times C_{2}.$ # Proof(s) ###### Proof of Statement 1: Let $1,a,b,c$ denote the four distinct elements of $G.$ 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 of Exponent 2 is Abelian]], $G$ is abelian thus its Cayley table is as follows $\begin{array}{c||c|c|c|c}\circ & 1 & a & b & c \\\hline \hline 1 & 1 & a & b & c \\\hline a & a & 1 & c & b \\\hline b & b & c & 1 & a \\\hline c & c & b & a & 1\end{array}$ ###### Proof 2: Suppose $G$ is cyclic, then certainly $G\cong C_{4}.$ Otherwise, suppose $G$ is not cyclic.