$H\leq K \leq G$ and $K \trianglelefteq G \Rightarrow$ $K/H \trianglelefteq G/H$.

Question

Let $G$ be group and $H$ be a normal subgroup of $G$.

Statement

If $H\leq K \leq G$ and $K \trianglelefteq G$, then $K/H \trianglelefteq G/H$ holds.

I hear that this is proven by the second isomorphism theorem, but I don't know why.

The statement of the second isomorphism theorem is here.

Let $G$ be a group, $S\leq G$ and $N\trianglelefteq G.$

Then, $SN\leq G, S\cap N\trianglelefteq S$ and $(SN/N) \cong S/(S\cap N)$ hold.

In order to prove $K/H \trianglelefteq G/H$, how should I the second isomorphism theorem ?

Answer 1

This could be proven in another way. First, show that indeed $K/H$ is a subgroup of $G/H$. For normality, take $kH\in K/H$ and $gH\in G/H$, and note that $g^{-1}HkHgH=g^{-1}kgH$. Since $K$ is a normal subgroup of $G$, we have $g^{-1}kg\in K$, so $g^{-1}HkHgH\in K/H$, and hence $K/H$ is a normal subgroup of $G/H$.