Group homomorphism/Image/Subgroup/Fact/Proof

Let ${\displaystyle {}B:=\operatorname {Im} \varphi }$. We have ${\displaystyle {}e_{H}=\varphi (e_{G})\in B}$. Let ${\displaystyle {}h_{1},h_{2}\in B}$. Then there exist ${\displaystyle {}g_{1},g_{2}\in G}$ such that ${\displaystyle {}\varphi (g_{1})=h_{1}}$ and ${\displaystyle {}\varphi (g_{2})=h_{2}}$. Therefore, ${\displaystyle {}h_{1}\cdot h_{2}=\varphi (g_{1})\cdot \varphi (g_{2})=\varphi (g_{1}\cdot g_{2})\in B}$. Similarly, for ${\displaystyle {}h\in B}$ there exists a ${\displaystyle {}g\in G}$ fulfilling ${\displaystyle {}\varphi (g)=h}$. Hence, ${\displaystyle {}h^{-1}=(\varphi (g))^{-1}=\varphi (g^{-1})\in B}$.