> [!NOTE] Theorem (Group Homomorphism Fixes Identity) > Let $G$ and $H$ be [[Groups|groups]]. Let $\phi:G\to H$ be a [[Homomorphisms of groups|homomorphism]]. Then $\phi(1_{H})=1_{G}.$ *Proof*. We have $\phi(1_{G})=\phi(1_{G}1_{G})=\phi(1_{G})\phi(1_{G})$Since $\phi(1_{G})\in H$ it has inverse in $H$ denote $\phi(1_{G})^{-1}.$ Left multiplying the above equation gives $1_{H}=\phi(1_{G})^{-1}\phi(1_{G})=\phi(1_{G})^{-1} \phi(1_{G}) \phi(1_{G}) = 1_{H} \phi(1_{G})=\phi(1_{G})$