Why does $N$ need to be normal? (Robinson exercise)

Question

In Robinson Ed. 2, Ex. 1.3.16 he asks: if $H\le K\le G$ and $N\lhd G$ and $KN=HN$ and $K\cap N =H\cap N$ then show $H=K$. My question is: does $N$ need to be normal in $G$?

Here's my attempted proof: It's enough to show $K\le H$. Let $k\in K$. Then there's an $h\in H$ such that $kN = hN$. So $h^{-1}k\in N$ but it's also in $K$, and therefore $h^{-1}k\in K\cap N = H\cap N$ and so $h^{-1}k\in H$ therefore $k\in H$.

I'm sure I went wrong somewhere. Where did I use normality?

[I'm returning to maths and group theory after many years, and I do remember always struggling with the intuitive meaning of normality in the past, even though the definition is simple.]

Answer 1

I believe the statement that theres an $h$ in $H$ such that $hN=kN$ need not hold, and we can see this in the following counterexample when $N$ is not normal. Consider $G=A_4$ the alternating group on $4$ elements, $N=\langle e,(132),(123)\rangle$, $H=\langle e,(12)(34)\rangle$, $K=\langle e,(12)(34),(13)(24),(14)(23)\rangle$. We have that $NH=NK=A_4$, since the size of this subgroup is at least $6$ and $A_4$ has no index $2$ subgroups. We also have $H\cap N=K\cap N=e$, since the orders are coprime.

In this example, if we picked $k$ to be $(13)(24)\in K$, then there would be no element $h$ in $H$ such that $h^{-1}k\in N$, since this would have to be the identity (because $N\cap K=e$), and $k\notin H$.