= Solution
\b[Definitions.] A <group homomorphism> $\varphi:G_1\to G_2$ preserves multiplication, $\varphi(xy)=\varphi(x)\varphi(y)$. A <group isomorphism> is a bijective <group homomorphism>. The <kernel of a group homomorphism> is
$$
\boxed{K=\ker\varphi=\{x\in G_1:\varphi(x)=e_2\}.}
$$
Preserving multiplication implies $\varphi(e_1)=e_2$ and $\varphi(x^{-1})=\varphi(x)^{-1}$: the first follows by cancelling $\varphi(e_1)$ from its square, and the second follows by applying $\varphi$ to $xx^{-1}=e_1$.
\b[Kernel subgroup.] The identity belongs to $K$. For $x,y\in K$, $\varphi(xy)=e_2e_2=e_2$, so $xy\in K$. Also $\varphi(x^{-1})=e_2^{-1}=e_2$, so $x^{-1}\in K$. Thus $\boxed{K\leq G_1}$.
\b[Normality.] For $g\in G_1$ and $k\in K$,
$$
\varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=\varphi(g)e_2\varphi(g)^{-1}=e_2.
$$
Hence every $gkg^{-1}$ is again in $K$, proving $\boxed{K\trianglelefteq G_1}$. Applying the same statement with $g^{-1}$ gives equality $gKg^{-1}=K$, and therefore $gK=Kg$.
\b[Group of cosets.] Define the product on the <left cosets> by
$$
\boxed{(aK)(bK)=(ab)K.}
$$
To check that this does not depend on representatives, replace $a,b$ by $ak_1,bk_2$. Then
$$
(ak_1)(bk_2)=ab(b^{-1}k_1b)k_2\in abK,
$$
because normality puts $b^{-1}k_1b$ in $K$. Thus the product is well defined. It is associative because multiplication in $G_1$ is associative; its identity is $K=e_1K$, and the inverse of $aK$ is $a^{-1}K$. Consequently the cosets form the <quotient group> $\boxed{G_1/K}$.
The map $aK\mapsto\varphi(a)$ is well defined and gives an isomorphism $G_1/K\cong\operatorname{im}\varphi$. 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.
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