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

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.]

• You didn't. (BTW I think your first statement is false, and what you should have is $hm=kn$ for some $h\in H$ and $m,n\in N$. And then all is OK). Aug 18, 2020 at 8:14
• @ancientmathematician I don't think OP's first statement is false. Given $k\in K$, $k=hn$ for some $h\in H, n\in N$ and then we see that $kN=hnN=hN$ Aug 18, 2020 at 8:56
• @IgnorantMathematician you are right, what I should have said was that it wasn't immediately obvious. Aug 18, 2020 at 13:10
• Which of Robinson's books are you referring to? Because neither his "A Course in the Theory of Groups (Second Edition)" nor his "Abstract Algebra (Second Edition)" correspond to this exercise. Apr 26, 2021 at 20:18
• @Shaun you're right, it should say Ex. 1.3.16. This is "A course in the theory of groups" second edition. I'll edit it. Apr 28, 2021 at 7:33

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 $$k\notin H$$.
• $NH$ has order 6, not 12, so we cannot possibly have $NH = A_4$. The fact that $A_4$ has no subgroup of order 6 is not relevant, because $NH$ is not a subgroup. Aug 18, 2020 at 12:55
• Oh my mistake, I was interpreting $NH$ to be the subgroup generated by $N$ and $H$ in the case when $N$ is not necessarily normal. Aug 18, 2020 at 13:12