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.]

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    $\begingroup$ 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). $\endgroup$ Aug 18 '20 at 8:14
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    $\begingroup$ @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 $kN=hnN=hN$ $\endgroup$
    – user6
    Aug 18 '20 at 8:56
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    $\begingroup$ @IgnorantMathematician you are right, what I should have said was that it wasn't immediately obvious. $\endgroup$ Aug 18 '20 at 13:10
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    $\begingroup$ 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. $\endgroup$
    – Shaun
    Apr 26 '21 at 20:18
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    $\begingroup$ @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. $\endgroup$
    – Andrew Kay
    Apr 28 '21 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 $k\notin H$.

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    $\begingroup$ $NH$ 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. $\endgroup$
    – Derek Holt
    Aug 18 '20 at 12:55
  • $\begingroup$ 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. $\endgroup$
    – Chris H
    Aug 18 '20 at 13:12

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