I'd like to be prove the second isomorphism theorem.
Let $H$ and $K$ be two subgroups of $G$ with $K$ normal in $G$. The subgroup $H \cap K$ is normal in $H$, $HK$ is a subgroup of $G$, and $K$ is normal in $HK$.
We want to show that $\frac{H}{H \cap K}$ and $\frac{HK}{K}$ are isomorphic using the map:
$$f: \frac{H}{H \cap K} \rightarrow \frac{HK}{K} $$ $$ : \overline{h} \rightarrow \overline{h \cdot 1}$$
My attempt:
$H \cap K$ is a normal subgroup of $H$, so the quotient $\frac{H \cap K}{H}$
is a group. Second, $HK$ is a subgroup of $G$ and $K$ is a normal subgroup of $HK$,
therefore $\frac{HK}{K}$ is also a group. So, the map $f$ is well-defined.
Let $x,y \in \frac{H}{H \cap K}$.
\begin{equation*} f(\overline{xy}) = \overline{xy\cdot1} = \overline{x \cdot 1 \cdot y \cdot 1} = \overline{x \cdot 1} \overline{y \cdot 1} = f(\overline{x})f(\overline{y}) \end{equation*}
Therefore, $f$ is a group-homomorphism. Besides,
\begin{align*} \text{ker}(f) &= \left\{ \overline{h} \in \frac{H}{H \cap K}: f(\overline{h}) = e_{HK/K} \right\} \\ % &= \left\{ \overline{h} \in \frac{H}{H \cap K}: \overline{h \cdot 1} = e_{HK/K} \right\} \\ % &= \left\{ \overline{h} \in \frac{H}{H \cap K}: \overline{h} \cdot \overline{1} = e_{HK/K} \right\} \\ % &= \left\{ \overline{h} \in \frac{H}{H \cap K}: \overline{h} = e_{HK/K} \right\} \end{align*}
So, $f$ is injective. But I'm stuck for the surjectivity. Any ideas?