Suppose that $H\trianglelefteq G$ and $H\leq K\leq G$. Show $H\trianglelefteq K$ and that $K/H\leq G/H$

Question

This is not for an assignment. I found this practice problem in the back of my abstract algebra book and I'm trying to figure it in preparation for an upcoming exam.

I've attempted the first part of the problem:

$H\trianglelefteq G$ means $\forall g∈G$, $gHg^{-1}$ $\epsilon$ $H$, therefore since $K≤G$ it also holds for $g∈K$. Implying that $H⊴K$.

For the second part, I'm not exactly sure where I should begin.

Answer 1

The solution to the first part is the one you gave. For what concerns the second part, there is an obvious group homomorphism $K \rightarrow G/H$ which is given by the quotient map. The kernel of this map is given by $K \cap H = H$, as $H \leq K$. Therefore, the map induces an injection, which is also a group homomorphism, $K/H \hookrightarrow G/H$, and therefore $K/H \leq G/H$.

Answer 2

For the second part, you can

1. Prove that $K/H$ is a group with a well defined multiplication - here, you will need the fact that $H$ is a normal subgroup of $K$.
2. Prove that $K/H$ is a subset of $G/H$.