QUESTION : Let $G$ be a group, let $X$ be a set, and let $H$ be a subgroup of $G$. Let $$N = \bigcap_{g\in G} gHg^{-1}$$ Show that $N$ is a normal subgroup of $G$ cointained in $H$.

MY ATTEMPT: I began by asking myself precisely what $\bigcap_{g\in G} gHg^{-1}$ means. I concluded that it must mean that if $g_1, g_2, g_3, ... , g_n \in G$ then $N$ might just be $$ g_1H{g_1}^{-1} \cap g_2H{g_2}^{-1} \cap g_3H{g_3}^{-1} \cap ... g_nH{g_n}^{-1}$$ which I figured is actually just $H$ because that's the only element that all those elements have in common.

Therefore I figured that $N=H$.

Now my problem comes in showing that $H$ is a normal subgroup of $G$. I've never been good at that.

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    $\begingroup$ $H$ need not be normal. And $H$ is not necessarily contained in all those subgroups. $\endgroup$ – Tobias Kildetoft Apr 3 '13 at 16:28
  • $\begingroup$ @TobiasKildetoft What? How can $H$ not be in all those subgroups? I did say that I think $N=H$ so if the question asks that I prove $N$ is normal and I prove that $H$ is normal, isn't that the same thing? $\endgroup$ – Siyanda Apr 3 '13 at 16:32
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    $\begingroup$ But $H\neq N$ unless $H$ is normal. If $H$ is not normal, there is no reason why you would have $H\subseteq gHg^{-1}$ for some arbitrary $g\in G$. $\endgroup$ – Tobias Kildetoft Apr 3 '13 at 16:34
  • $\begingroup$ @TobiasKildetoft I'm still super confused though. I think I may have to go back to doing some reading. Am I completely off with my attempt? $\endgroup$ – Siyanda Apr 3 '13 at 16:46
  • $\begingroup$ Well, you are correct in what it means (ie, that long line of intersections). But you seem to have misunderstood something if you think $H$ is always contained in $gHg^{-1}$. $\endgroup$ – Tobias Kildetoft Apr 3 '13 at 16:47

You already have several good answers, and have probably completed the question yourself, but I'd like to provide my point of view.

Begin with two trivial observations:

  1. Since one of the components of the intersection is $H$ conjugated by the identity, $N\subseteq H$.
  2. Any intersection of subgroups is a subgroup.

Together these facts give you that $N$ is a subgroup of $H$. Therefore, the main part of the proof is normality. Now, ask yourself the following question:

What are you gonna conjugate $N$ by so that the result isn't in $N$?

In particular, my claim is that when we conjugate $N$ by some $x\in G$, we are simply inducing a permutation of the components of the intersection, which of course does not change the resulting content. Can you see why this is true?

$\displaystyle N^x=\left(\bigcap_{g\in G}H^g\right)^x=\bigcap_{g\in G}(H^g)^x=\bigcap_{g\in G}H^{gx}=\bigcap_{k\in Gx}H^{k}=\bigcap_{y\in G}H^y=N$

It's easy to visualize: like a pinwheel, permuting the leaves does not change the bulb.

Now let's look at this from a different angle.

We know that we can't form a quotient group from $H$ unless $H$ is normal. But let's try anyway and see what we get.

Let $G$ act on the right coset space $G\backslash H$ by right multiplication. What is the kernel of this action?

Explicitly, this means "given the homomorphism $\theta:G\rightarrow \operatorname{Sym}(G\backslash H)$ by $\theta(g)=\theta_g$ where $\theta_g(Ha)=H(ag)$, what is $\operatorname{ker}(\theta)$?" or, more simply, "for which $k\in G$ does $Hak=Ha$ hold for all $a\in G$?

$Hak=Ha$ if and only if $Haka^{-1}=H$ if and only if $aka^{-1}\in H$.

If this is true for all $a\in G$, then in particular it is true for $a=\operatorname{id}_G$, so $k\in H$. Furthermore, the set of these $k$ must be normal in $G$ (as expected, since kernels are always normal). Since all such $k$ are contained in $\operatorname{ker}(\theta)$, we must then have that $N=\operatorname{ker}(\theta)$ is precisely the largest normal subgroup of $G$ contained in $H$.

In this way, we see that $G/N$ is the best we can do when trying to make a quotient group from a not-necessarily-normal subgroup $H$.

Hopefully this provides some motivation and/or intuition towards this problem and why it matters.

  • $\begingroup$ THANK YOU - this is the kind of answer I needed to see :) $\endgroup$ – Siyanda Apr 14 '13 at 8:28
  • $\begingroup$ @AlexanderGruber : The map $\theta$ you defined is an anti-homomorphism $\theta(gh)=\theta(h)\circ\theta(g)$, isn't it? See my comment below to the answer of Babak S. $\endgroup$ – Thibaut Dumont Apr 15 '13 at 13:59

Hint: $H\leq G$ and let $\Omega$ be the set of all $Ha$ where $a\in G$. Define an action like: $$(Ha)^x=Hax,~~ Ha,Hax\in\Omega;~~x\in G$$ By this action we see that the stabilizer of $Ha$ for example is $a^{-1}Ha$. Now try to show that the map $x\mapsto\bar{x},~~\bar{x}(Ha)=Hax$ is a homomorphism with the kernel $N$.

  • $\begingroup$ Thank you! That is an interesting method. So I don't have to use any of the information offered in the question? Or.... $\endgroup$ – Siyanda Apr 3 '13 at 16:41
  • $\begingroup$ @Siyanda: I am with Tobias in his leading comments. $\endgroup$ – mrs Apr 3 '13 at 16:42
  • $\begingroup$ @babakS. I'm still super confused though. I think I may have to go back to doing some reading. Am I completely off with my attempt? $\endgroup$ – Siyanda Apr 3 '13 at 16:46
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    $\begingroup$ Ah, this is my approach. Excellent answer! +1 $\endgroup$ – DonAntonio Apr 3 '13 at 16:48
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    $\begingroup$ @Siyanda, yes: you're completely off since your approach seems to be forcing on you to show something false, namely $\,N=H\,$ . This is not so, as you were already explained...perhaps your confusion is due to the fact that, for some reason, you seem to believe that H is contained in each of its conjugates, which is completely false. $\endgroup$ – DonAntonio Apr 3 '13 at 16:50

Your $N$ is normally denoted as $core_G(H)$, the largest normal subgroup of $G$ contained in $H$.

  • $\begingroup$ Note: I have also seen $\operatorname{core}_G(H)$ written as $H_G$ in some texts. $\endgroup$ – Alexander Gruber Apr 5 '13 at 4:15
  • $\begingroup$ Yes, good point, some texts use that $\endgroup$ – Nicky Hekster Apr 5 '13 at 17:02

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