Questions related to group theory

Question

If $H$ and $K$ are two normal subgroups of a group $G$, such that $H \subseteq K$, then show that $K/H$ is a normal subgroup of $G/H$.

Here how do I prove that $hk=kh$ because in this question as we know $H$ is normal to $G$ so $hg=gh$ and $K$ is also normal to $G$ so $kg=gk$. And also $H$ is contained in $K$. Which means all $h$ belong to $K$.

Answer 1

The kernel of the natural projection $\varphi : G \rightarrow G/K$ contains $H$ because $H \subset K$, therefore it induces a map $\tilde{\varphi} : G/H \rightarrow G/K$ whose kernel is $\mathrm{Ker}{(\varphi)}/H=K/H$. As a kernel, $K/H$ is normal in $G/H$.

However, as said in comments, you should work a little bit more about normal subgroups before considering this proof.

Answer 2

By definition:

\begin{alignat}{2} K/H \unlhd G/H &\iff &&\forall g\in G,\forall k\in K, \space \space gH(kH)(gH)^{-1}\in K/H \\ &\stackrel{H\unlhd G}{\iff} &&\forall g\in G,\forall k\in K, \space \space gH(kH)(g^{-1}H)\in K/H \\ &\stackrel{H\unlhd K}{\iff} &&\forall g\in G,\forall k\in K, \space \space gkH(g^{-1}H)\in K/H \\ &\stackrel{H\unlhd G}{\iff} &&\forall g\in G,\forall k\in K, \space \space gkg^{-1}H\in K/H \\ \end{alignat}

and the latter holds because $K\unlhd G$.