**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.13**

Let $G$ and $H$ be groups and let $\varphi : G \rightarrow H$ be a group homomorphism. Prove that the image $\mathsf{im}\ \varphi$ of $\varphi$ is a subgroup of $H$. Prove that if $\varphi$ is injective then $G \cong \mathsf{im}\ \varphi$.

Solution: By Exercise 1.1.26, it suffices to show that $\mathsf{im}\ \varphi$ is closed under multiplication and inversion. To that end, let $h_1, h_2 \in \mathsf{im}\ \varphi$. Then there exist $g_1, g_2 \in G$ such that $\varphi(g_1) = h_1$ and $\varphi(g_2) = h_2$. Since $$h_1 h_2 = \varphi(g_1) \varphi(g_2) = \varphi(g_1 g_2),$$ $\mathsf{im}\ \varphi$ is closed under multiplication. Now let $h \in \mathsf{im}\ \varphi$; then there exists $g \in G$ such that $\varphi(g) = h$. Now $$h^{-1} = \varphi(g)^{-1} = \varphi(g^{-1}),$$ so that $\mathsf{im}\ \varphi$ is closed under inversion. Thus $\mathsf{im}\ \varphi$ is a subgroup of $H$.

Suppose now that $\varphi$ is injective. We can then defined a mapping $\psi : G \rightarrow \mathsf{im}\ \varphi$ by $\psi(g) = \varphi(g)$; this $\psi$ is clearly a homomorphism and is bijective, hence an isomorphism. Thus $G \cong \mathsf{im}\ \varphi$.