> [!NOTE] Theorem (Group with Exponent 2 is Isomorphic to Power of 2nd Cyclic Group) > Let $G$ be a [[Finite Group|finite group]] with [[Exponent of Group|exponent]] $2$ (i.e. for all $g\in G,$ $g^2 = \text{id}$). Then there exists some positive natural number $n$ such that $G \cong \underbrace{ C_{2} \times C_{2} \times \cdots C_{2}}_{\text{$n$ times}}$where $\cong$ denotes [[Homomorphisms of groups|group isomorphism]]. **Remark**: this yields $\# G = 2^n.$ ###### Proof It follows from [[Group of Exponent 2 is Abelian]] that $G$ is abelian. By [[Fundamental Theorem of Finite Abelian Groups]], there exist positive integers $b_{1},b_{2},\dots,b_{n}$ with $b_{1} \mid b_{2} \mid \dots \mid b_{n}$ such that $G \cong C_{b_{1}}\times C_{b_{2}} \times \cdots \times C_{b_{n}}.$ Write $C_{b_{i}}=\langle x_{i}\rangle$ where $x_{i}$ has order $b_{i}$. Let $g=(x_{1},x_{2},\dots,x_{n})\in G.$ Then by our assumption that $G$ has exponent $2$, $g^2=(\text{id}, \text{id},\dots,\text{id}).$ Therefore $x_{i}^2=\text{id}.$ Since $b_{i}>1$ we get $b_{1}=b_{2}=\dots=b_{n}=2.$ $\blacksquare$