 Solution

Definitions. A group homomorphism $\phi :G_1\to G_2$ preserves multiplication, $\phi (xy)=\phi (x)\phi (y)$. A group isomorphism is a bijective group homomorphism. The kernel of a group homomorphism isPreserving multiplication implies $\phi (e_1)=e_2$ and $\phi (x^{-1})=\phi (x{)}^{-1}$: the first follows by cancelling $\phi (e_1)$ from its square, and the second follows by applying $\phi$ to $x{x}^{-1}=e_1$.

Kernel subgroup. The identity belongs to $K$. For $x,y\in K$, $\phi (xy)=e_2{e}_2=e_2$, so $xy\in K$. Also $\phi (x^{-1})=e_2^{-1}=e_2$, so $x^{-1}\in K$. Thus $\boxed{K\le G_1}$.

Normality. For $g\in G_1$ and $k\in K$,Hence every $gk{g}^{-1}$ is again in $K$, proving $\boxed{K⊴G_1}$. Applying the same statement with $g^{-1}$ gives equality $gK{g}^{-1}=K$, and therefore $gK=Kg$.

Group of cosets. Define the product on the left cosets byTo check that this does not depend on representatives, replace $a,b$ by $a{k}_1,b{k}_2$. Thenbecause normality puts $b^{-1}{k}_{1}b$ in $K$. Thus the product is well defined. It is associative because multiplication in $G_1$ is associative; its identity is $K=e_{1}K$, and the inverse of $aK$ is $a^{-1}K$. Consequently the cosets form the quotient group $\boxed{G_1/K}$.

The map $aK\mapsto \phi (a)$ is well defined and gives an isomorphism $G_1/K\cong \mathrm{im}\phi$. If the original mapping is onto $G_2$, this becomes $G_1/K\cong G_2$. No surjectivity is needed for the kernel of a group homomorphism to be a normal subgroup or for the quotient group to exist.