Prove that quotient group K/H is normal subgroup of quotient group G/H

Question

$H$ is the subset of $K$, $H$ and also $K$ are the normal subgroups of $(G, * )$. Prove that $H$ is the normal subgroup of $K$, and the quotient group $K/H$ is the normal subgroup of the quotient group $G/H$.

The first part seems to be trivial, or quite easy.

(I am going to use algebraic notation)

My proof is: We know that $H$ is the normal subgroup of $G$, it means that $ {\displaystyle \forall } g ∈G $ and ${\displaystyle \forall } h ∈ H $ : $g^{-1}hg∈ H$

$K$ is the normal subgroup of $G$, it means that $ {\displaystyle \forall } g ∈G $ and ${\displaystyle \forall } k ∈ K $ : $g^{-1}kg∈ K$

And we also know that $H ⊆ K$ whereas $H$ and $K$ are subgroups of $G$, they have to be subset of $G$ it means $H ⊆ K⊆G$.

Than $H$ has to be normal subgroup of $K$ because $ {\displaystyle \forall } g ∈G $ and ${\displaystyle \forall } h ∈ H $ : $g^{-1}hg∈ H$ and this statment is true also for $ {\displaystyle \forall } k ∈K $ and ${\displaystyle \forall } h ∈ H $ : $k^{-1}hk∈ H$ because $K⊆G$.

But what is probelm for me is to continue proof for the second part; quotient group $K/H$ is normal subgroup of quotient group $G/H$.

Thank you very much for any help.

Answer 1

The elements of $K/H$ are cosets which we can write as $kH$, and the elements of $G/H$ are cosets which we can write as $gH$. We want to show that $(gH)^{-1}kHgH \in K/H$. By the group law in $G/H$ we have $(gH)^{-1}kHgH = g^{-1}kgH$. Now $g^{-1}kg \in K$ because $K$ is normal in $G$, so $g^{-1}kgH \in K/H$.