Exercise 1.6.13 (If $\varphi$ is injective, then $G \simeq \varphi(G)$)

Proof. Let $e$ be the identity of $G$ and $e^{\prime }$ the identity of $H$.

• Since $e^{\prime }=\phi (e)$ (see Exercise 1), $e^{\prime }\in \phi (G)$, so $\phi (G)\ne \varnothing$.
• Let $c$ and $d$ be any elements in $\phi (G)$. By definition of the image $\phi (G)$, there are elements $a,b$ in $G$ such that $c=\phi (a)$ and $d=\phi (b)$. Then $\mathit{cd}=\phi (a)\phi (b)=\phi (\mathit{ab})\in \phi (G)$.
• Moreover, $c^{-1}={[\phi (a)]}^{-1}=\phi (a^{-1})\in \phi (G)$.

Since $\phi (G)\ne \varnothing$ and $\phi (G)\subset H$ is stable by product and by inverses, $\phi (G)$ is a subgroup of $H$.

Consider the map

obtained from $\phi$ by modifying the codomain of $\phi$.

• $\tilde{\phi }$ is surjective by definition of $\phi (G)$: if $h\in \phi (G)$, there exists some $g\in G$ such that $h=\phi (g)=\tilde{\phi }(g)$.
• $\tilde{\phi }$ is injective: if $\tilde{\phi }(g)=\tilde{\phi }(g^{\prime })$, where $g,g^{\prime }\in G$, then $\phi (g)=\phi (g^{\prime })$. Since $\phi$ is injective by hypothesis, $g=g^{\prime }$.

So $\tilde{\phi }$ is bijective. Moreover, since $\phi$ is a homomorphism, for all $g,g^{\prime }\in G$,

so $\tilde{\phi }$ is also a homomorphism.

This shows that $\tilde{\phi }$ is an isomorphism, so $G\simeq \phi (G)$.

If $\phi$ is injective, then $G\simeq \phi (G)$ □