**Proof plan.** We construct the map $f : G/\ker(\varphi) \to \operatorname{im}(\varphi)$ by $f(g\ker\varphi) = \varphi(g)$, then verify it is well-defined, a homomorphism, and bijective in three separate claims. **Step 1: $\ker(\varphi) \trianglelefteq G$.** [claim: Kernel Is Normal] $\ker(\varphi)$ is a normal subgroup of $G$. [/claim] [proof] $\ker(\varphi)$ is a subgroup: if $g, h \in \ker(\varphi)$ then $\varphi(gh^{-1}) = \varphi(g)\varphi(h)^{-1} = e \cdot e^{-1} = e$, so $gh^{-1} \in \ker(\varphi)$, and $e \in \ker(\varphi)$ since $\varphi(e) = e$. Normality: let $g \in \ker(\varphi)$ and $x \in G$. Then $\varphi(x^{-1}gx) = \varphi(x)^{-1}\varphi(g)\varphi(x) = \varphi(x)^{-1} e \varphi(x) = e$, so $x^{-1}gx \in \ker(\varphi)$. [/proof] **Step 2: Well-definedness.** [claim: Well-Defined] The map $f : G/\ker(\varphi) \to \operatorname{im}(\varphi)$ given by $f(g\ker\varphi) = \varphi(g)$ is well-defined. [/claim] [proof] If $g\ker\varphi = g'\ker\varphi$ then $g^{-1}g' \in \ker\varphi$, so $\varphi(g^{-1}g') = e$, giving $\varphi(g)^{-1}\varphi(g') = e$, hence $\varphi(g) = \varphi(g')$. [/proof] **Step 3: $f$ is a homomorphism.** [claim: Homomorphism] $f$ is a [group](/page/Group) homomorphism. [/claim] [proof] \begin{align*} f(g\ker\varphi \cdot g'\ker\varphi) = f(gg'\ker\varphi) = \varphi(gg') = \varphi(g)\varphi(g') = f(g\ker\varphi) \cdot f(g'\ker\varphi). \end{align*} [/proof] **Step 4: $f$ is bijective.** [claim: Bijection] $f$ is injective and surjective. [/claim] [proof] Surjectivity: any $h \in \operatorname{im}(\varphi)$ satisfies $h = \varphi(g)$ for some $g \in G$, so $h = f(g\ker\varphi)$. Injectivity: if $f(g\ker\varphi) = f(g'\ker\varphi)$ then $\varphi(g) = \varphi(g')$, so $\varphi(g^{-1}g') = e$, hence $g^{-1}g' \in \ker\varphi$, giving $g\ker\varphi = g'\ker\varphi$. [/proof] Since $f$ is a bijective homomorphism, it is an isomorphism, completing the proof.