$H$ and $K$ are normal subgroups of a group $G$ with $K \le H$, then $aK =H \iff a \in H$

Question

Let $H$ and $K$ are normal subgroups of a group $G$ with $K \le H$. How can I prove that $aK =H \iff a \in H$?

I‘m stuck with the part when $a \in H$ then $aK$ contains $H$.

Given $u$ in $H$, how can I show that $u \in aK$ ？

Thanks for helping.

Answer 1

This should work as a a counterexample: Let $G = \mathbb Z$, $H = 2\mathbb Z$ and $K = 4\mathbb Z$. Since $G$ is abelian, $K, H \trianglelefteq G$, and $K\le H$. Now we can choose $a \in H$, say $a = 2$, and consider $aK = 2 + 4\mathbb Z$. This doesn't contain $H$, as for example $4 \in H$ but $4 \notin aK$.

Answer 2

The statement is definitely incorrect. Take $K$ to be the trivial subgroup and $H=G$, for example. Then $a=G$ if and only if $a\in G$ is a nonsense. If we take $aK$ to mean $\langle a,K\rangle$ then it is still absurd, as it would imply that all groups are cyclic.

Nope, something's wrong here.