# Statement(s) > [!NOTE] Statement 1 ($\ker \phi$ is a subgroup of $G$) > Let $G,H$ be [[Groups|groups]] and $\phi: G\to H$ be a [[Homomorphisms of groups|homomorphism]]. Then $\ker \phi$ is a [[Subgroup|subgroup]] of $G$ and is [[Normal Subgroup|normal]], # Proof(s) *Proof of statement 1*: We check that it passes [[Two-Step Subgroup Test|two-step subgroup test]]. First, by [[Homomorphism of Groups Preserves Identity|group homomorphism fixes identity]], $\phi(1_{G})=1_{H}$ so $1_{G}\in \ker \phi.$ Next, suppose $g_{1},g_{2}\in\ker \phi.$ Then $\phi(g_{1}g_{2})=\phi(g_{1})\phi(g_{2})=1_{H}1_{H}=1_{H}$ and so $g_{1}g_{2}\in \ker \phi.$ Finally, suppose $g\in \ker \phi.$ Then $\phi(g)=1.$ Now $\phi(g^{-1})=\phi(g^{-1})1_{H}=\phi(g^{-1})\phi(g)=\phi(g^{-1}g)=\phi(1_{G})=1_{H}$therefore $\text{ker}(\phi)$ is indeed a subgroup of $G.$ Let $g\in G,k\in\text{ker}(\phi)$ i.e. $\phi(k)=1_{H}.$ Then $\phi(gkg^{-1})=\phi(g)\phi(k)\phi(g)^{-1}=\phi(g)1_{H}\phi(g)^{-1}=\phi(g)\phi(g)^{-1}=1_{H}\in H$so $\text{ker}(\phi)$ is also normal. $\square$ # Applications **Consequences**: Since $\text{ker}(\phi)\unlhd G,$ $G/\text{ker}(\phi)$ is a well-defined quotient group. The [[First Isomorphism Theorem for Groups]] asserts that $\hat{\phi}:G/\text{ker}(\phi)\to \text{Im}(\phi)$ defined by $\hat{\phi}(g\;\text{ker}(\phi))=\phi(g)$ is an isomorphism (i.e. $G/\text{ker}(\phi)\cong\text{Im}(\phi)$). # Bibliography