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Is there a better (more direct or intuitive) proof for this proposition than I have come up with below? I am not sure whether it could be simiplified:

Let $G$ be a group with $H \leq G$. Then $K = \bigcap_{g \in G} gHg^{-1}$ is normal in $G$.

Let $a \in K$. Then $a \in gHg^{-1}$ for all $g \in G$. Therefore for all $g_1,g \in G$, $g_1ag_1^{-1} \in g_1gHg^{-1}g_1^{-1} = (g_1g)H(g_1g)^{-1}$ and so $g_1ag_1^{-1} \in K$ since $g_1g \in G$. Then $K$ is normal in $G$.

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    $\begingroup$ That's essentially correct, provided you observe that $G=\{g_1g\;;\;g\in G\}$. You have only noticed $\supseteq $. So it would maybe be slightly better to say: $g_1ag_1^{-1}\in g_1(g_1^{-1}gH(g_1^{-1}g)^{-1})g_1^{-1}=gHg^{-1}$ for every $g\in G$. Thus $g_1ag_1^{-1}\in K$ for every $g_1\in G$ and every $a\in K$. $\endgroup$ – Julien Apr 22 '13 at 19:23
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    $\begingroup$ For an element-free argument, you can also write $gKg^{-1}=g\left(\bigcap_{g'\in G}g'Hg'^{-1}\right)g^{-1}=\bigcap_{g'\in G}gg'Hg'^{-1}g^{-1}=\bigcap_{g'\in G}gg'H(gg')^{-1}=\bigcap_{g'\in G}g'Hg'^{-1}=K$. $\endgroup$ – Julien Apr 22 '13 at 19:26
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    $\begingroup$ You should also notice that $K$ is the largest (in terms of order) possible normal subgroup of $G$ that is also a subgroup of $H$ (and all of $H$'s conjugates). From a lattice prospective, that's kinda special. $\endgroup$ – Mud Apr 22 '13 at 19:33
  • $\begingroup$ @julien I think is there an 'element-free' argument is what I meant, but I didn't know how to say it! $\endgroup$ – user50229 Apr 22 '13 at 19:39
  • $\begingroup$ @julien If you write your element-free argument as an answer I would be happy to accept. $\endgroup$ – user50229 May 3 '13 at 13:04
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For every $g\in G$, we have $$ gKg^{-1}=g\left(\bigcap_{g'\in G} g'Hg'^{-1}\right)g^{-1}=\bigcap_{g'\in G} gg'Hg'^{-1}g^{-1}=\bigcap_{g'\in G} gg'H(gg')^{-1} $$ $$ =\bigcap_{g'\in G}g'Hg'^{-1}=K. $$ Note that the less trivial step is the second one. It is due to the fact that $x\longmapsto gxg^{-1}$ is injective for $\supseteq$. The inclusion $\subseteq$ is straightforward.

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  • $\begingroup$ Can you explain how the second equality makes sense? How do we put the $g$ and $g^{-1}$ inside the intersection? $\endgroup$ – BSplitter Mar 13 '17 at 3:56
  • $\begingroup$ While showing equality in the second step. I went as follows: Let $k \in \bigcap_{g'\in G} gg'Hg'^{-1}g^{-1}$, \rightarrow $k=gwg^{-1}$ where $w\in g'Hg'^{-1}$ for all $g' \in G$ \rightarrow $w\in \bigcap_{g' \in G}$ \rightarrow $k=gwg^{-1} \in g\left(\bigcap_{g'\in G} g'Hg'^{-1}\right)g^{-1}$. And inclusion $\subseteq$ is straightforward. Hence equality holds. Am I correct? Where am I using the fact that $x \mapsto gxg^{-1}$ is injective? $\endgroup$ – Saikat Oct 15 '20 at 6:22
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This is not much different in essence, but emphasizes the other important definition of normal subgroup as kernel of homomorphism.

Consider the action of $G$ on the cosets of $H$ given by multiplication. This is a homomorphism $\phi$ from $G$ to the symmetric group on the set $G/H = \{ gH : g \in G \} = \{ \{ gh: h \in H \} : g \in G \}$. As such, it has a kernel $K$, those $k$ such that $kgH = gH$ for all $g \in G$. This is precisely the $k$ such that $kg = gh_g$ for some $h_g \in G$ dependent on $g$, that is $k = gh_g g^{-1} \in gHg^{-1}$. In other words, $\bigcap_{g\in G} gHg^{-1} = \ker(\phi)$.

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Let $ a \in K$ this implies $ a \in qHq^{-1}\;\forall q \\Let\;q = g^{-1}r \;\;(notice \;there\;is\;no\;restriction\;on\;g\;or\;r) \\then \; a\in (g^{-1}r )H(g^{-1}r )^{-1} \\a\in (g^{-1}r )H(r^{-1}g) \\gag^{-1}\in rHr^{-1}$

Because r and g donot have any restriction, we can take any r and g, or, every g and r to give:

$\forall g \; \forall r \; gag^{-1} \in rHr^{-1}\\ \forall g \; gag^{-1} \in \cap_{r}\ rHr^{-1}$

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Let $n\in K, g\in G$. For any $b\in G,$ $n\in K=\bigcap\limits_{a\in G}a^{-1}Ha\subseteq (bg)^{-1}Hbg$. Then $n=(bg)^{-1}h(bg)$ for some $h\in H.$ Then $gng^{-1}=g(bg)^{-1}h(bg)g^{-1}=b^{-1}hb\in b^{-1}Hb$ for any $b\in G$. So $gng^{-1}\in \bigcap\limits_{a\in G}a^{-1}Ha=K$.

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Your $K$ is the kernel of the homomorphism $$G \to \mathrm{Sym}(G/H),\quad \tilde g \mapsto (gH \mapsto \tilde g g H),$$ hence normal.

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  • $\begingroup$ This answer was given already $\endgroup$ – user50229 Oct 17 '20 at 11:15

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