> [!NOTE] Lemma (First Isomorphism Theorem for Groups) \[MA268\] > Let $G,H$ be [[Groups|groups]] and $\phi:G\to H$ a [[Homomorphisms of groups|homomorphism]]. Define $\begin{align}\hat{\phi}: G/\text{Ker}(\phi)&\to \text{Im}(\phi)\\ g\text{Ker}(\phi) &\mapsto \phi(g) \end{align}$ where $\text{Ker}(\phi)$ denotes the [[Kernel of Homomorphism of Groups|kernel]] of $\phi$, which [[Kernel of Homomorphisms of Groups is Normal Subgroup of Domain|is a normal subgroup]] of $G$, $\text{Im}(\phi)$ denotes the [[Image of Homomorphism of Groups|image]] of $\phi$ and $G/\text{Ker}(\phi)$ denotes the [[Quotient Group|quotient group]] of $G$ modulo $\text{Ker}(\phi).$ Then $\hat{\phi}$ is a well-defined [[Isomorphism|isomorphism of groups]]. **Remark**: we may write $G/\text{Ker}(\phi)\cong\text{Im}(\phi)$ which reads '$G$ modulo the kernel of $\phi$ is isomorphic to the image of $\phi$.' ###### Proof Let $g_{1},g_{2}\in G$. Then $\begin{align} g_{1}\text{Ker}(\phi) &= g_{2}\text{Ker}(\phi) \\ \iff g_{2}^{-1}g_{1} &\in \text{Ker}(\phi) \\ \iff \phi(g_{2}^{-1}g_{1}) &= 1_{H}, & \text{where $1_H$ denotes the identity element in $H$}\\ \iff \phi(g_{2}^{-1}) \phi(g_{1}) &= 1_{H} \\ \iff \phi(g_{2})^{-1} \phi(g_{1}) &= 1_{H} \\ \iff \phi(g_{1}) &= \phi(g_{2})\\ \iff \hat{\phi}(g_{1}\text{Ker}(\phi)) &= \hat{\phi}(g_{2}\text{Ker}(\phi)) \end{align}$So $\phi$ is both well-defined (taking the forward implication) and injective (taking the backward implication). It is clear that $\hat{\phi}$ is surjective. So $\hat{\phi}$ is bijective. To see that $\hat{\phi}$ is a homomorphism, observe that $\begin{align} \hat{\phi}(g_{1}\text{Ker}(\phi)\cdot g_{2}\text{Ker}(\phi)) &= \hat{\phi}(g_{1}g_{2} \text{Ker}(\phi)) \\ &= \phi(g_{1}g_{2}) \\ &=\phi(g_{1})\phi(g_{2}) \\ &= \hat{\phi}(g_{1}\text{Ker}(\phi))\hat{\phi}(g_{2}\text{Ker}(\phi)) \end{align}$ $\blacksquare$ # Application(s) See [[Alternating Group has Index Two in Symmetric Group]]. # Bibliography