> [!NOTE] Theorem > Let $G$ be a [[Groups|group]]. If $a^{2}=1$ for all $a\in G$ (i.e. $G$ has [[Exponent of Group|exponent]]) then $G$ is [[Groups|abelian]]. ###### Proof Let $a,b\in G.$ Then $\begin{align} ab &= (bb) (ab) (aa) & \text{since }bb=1=aa \\ &= b ((ba)(ba)) a &\text{by associativity} \\ &= ba & \text{since } (ba)^{2}=1 \end{align}$ **Proof**: Multiplying $(ab)^{2}=1$ by $(ab)^{-1}$ gives $ab= (ab)^{-1}$Now by [[Inverse of the Product of Group Elements]], $ab=b^{-1}a^{-1}=ba$since multiplying $a^{2}=1$ and $b^{2}=1$ by $a^{-1}$ and $b^{-1}$ respectively gives $a=a^{-1}$ and $b=b^{-1}.$ Application: Follows from [[Group of Exponent 2 is a Power of 2nd Cyclic Group]].