 Solution

The image of a group homomorphism contains $e_{G_2}=\varphi (e_{G_1})$. If $u=\varphi (g)$ and $v=\varphi (h)$ lie in the image, then $u{v}^{-1}=\varphi (g{h}^{-1})$ also lies there. The subgroup test proves $\mathrm{im}\varphi \le G_2$.

The kernel of a group homomorphism is $K=\{g:\varphi (g)=e_{G_2}\}$. It is nonempty and closed under $k{h}^{-1}$ for $k,h\in K$, so it is a subgroup. For any $g\in G_1$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Replacing $g$ by $g^{-1}$ gives the reverse inclusion, provingFinally, if $K=\{e\}$ and $\varphi (g)=\varphi (h)$, then $\varphi (h^{-1}g)=e$. Thus $h^{-1}g\in K$ forces $g=h$, establishing injectivity. Conversely an injective homomorphism has trivial kernel, soNone of these arguments requires either group to be finite.