> [!NOTE] Theorem ($\text{Im } \phi$ is a subgroup of codomain) > Let $G,H$ be [[Groups|groups]]. Let $\phi:G\to H$ be a [[Homomorphisms of groups|homomorphism]]. Let $\text{Im }\phi$ denote the [[Image of Homomorphism of Groups|image]] of $\phi.$ Then $\text{Im } \phi$ is a [[Subgroup|subgroup]] of $H.$ *Proof*. We check that it passes [[Two-Step Subgroup Test|two-step subgroup test]]. (Identity) First, by [[Homomorphism of Groups Preserves Identity|group homomorphism fixes identity]], $\phi(1_{G})=1_{H}$ so $1_{H}\in \text{Im } \phi.$ (Closure) Next, suppose $g_{1},g_{2}\in \text{Im} \phi.$ Then there exists $g_{1},g_{2}\in G$ such that $\phi(g_{1})=h_{2}$ and $\phi (g_{2})=h_{2}.$ Then $\phi(g_{1}g_{2})=\phi(g_{1})\phi(g_{2})=h_{1}h_{2}$ and so $h_{1}h_{2}\in \text{Im }\phi.$ (Inverses) Finally, suppose $h\in \text{Im } \phi.$ Then there exists $g\in G$ such that $\phi(g)=h.$ Then $\phi(g^{-1})\phi(g)=\phi(g^{-1}g)=\phi(1_{G})=1_{H}$Multiplying both sides on the right by $\phi(g)^{-1}$ gives $\phi(g^{-1})=\phi(g^{-1})\phi(g)\phi(g)^{-1} = \phi(g)^{-1}= h^{-1}$Thus $h^{-1}\in \text{Im }\phi,$ as required.