6
$\begingroup$

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 $G$ conitained 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.

$\endgroup$
  • 3
    $\begingroup$ $H$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $N$.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ Ah, this is my approach. Excellent answer! +1 $\endgroup$ – DonAntonio Apr 3 '13 at 16:48
  • 1
    $\begingroup$ @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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.