Let $G$ be a non-Abelian Group and $H$ is normal subgroup of $G$. Is it always true that a normal subgroup $K$ of $H$ is also normal in $G$? Justify your answer.

My answer is that, this is not true in general. I created a counter example. Consider on $A_4$,{ Group of even permutation of four symbols}, $K_4$ {Klein's four group} is normal subgroup of $A_4$, Now take any subgroup of $K_4$, subgroup will be normal in $K_4$ {being Abelian group }, But this subgroup will not be normal in $A_4$,as we know that proper normal subgroup of $A_4$ is only $K_4$.

Now my question:

1) am I correct?

2) Can we prove it instead of giving counter example?

3) Different counter examples are also invited.

Thank you.

  • $\begingroup$ A group where this is the case is called a T-group, (although this is more general than what you asked). See en.wikipedia.org/wiki/T-group_(mathematics) for example for some examples of T-groups. $\endgroup$ – Krijn Jul 14 '15 at 15:31

1) Yes you're correct

2) Giving a counterexample is a "proof"

3) to give a more natural reason, we have $K\triangleleft H$ means $hK=Kh$ for all $h\in H$. There is no reason to assume that this should extend to all elements of $G$; that is, you should not assume that $gK=Kg$ for all $g\in G$ just because it is true for all $g\in H$.

Also to add a bit, if I'm not mistaken, the group $K$ you are speaking of has a property called being "semi-normal" in $G$. My advisor told me a bit about this but I honestly haven't done any looking into it of my own; but maybe you could look up some stuff on this.

  • $\begingroup$ Thanx for quick reply, yeah we cannot assume that but there should be a support on not assuming such thing, isn't it? $\endgroup$ – Chiranjeev_Kumar Jul 13 '15 at 18:07
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    $\begingroup$ Explaining that this is what happens and there's no reason to assume that that happens is an excellent thing to do. Given that we're in a context where people are not sure whether this or that is a valid proof, we should emphasize that "there's no reason to assume" doesn't prove anything, we do need that example (because sometimes things work out in surprising ways, for reasons we didn't see at first). $\endgroup$ – David C. Ullrich Jul 13 '15 at 18:08
  • $\begingroup$ @David C. Ullrich: Exactly the same thing is in my mind, we can not assume, is not enough we should clear why we can not assume $\endgroup$ – Chiranjeev_Kumar Jul 13 '15 at 18:15
  • $\begingroup$ @Chiranjeev I'm not sure what your point is. We do have that counterexample - you found it. $\endgroup$ – David C. Ullrich Jul 13 '15 at 18:16

1)Yes, you are correct Chiranjeev.

2)No, I don't think it can be proved in general. For example in your counterexample $V_4$ is a normal subgroup of itself, and so it is clearly a normal subgroup of $V_4$. If you refined your statement by stipulating some extra conditions then maybe.


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