Intersection of conjugate subgroups is normal 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 $G$.
 A: 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, $\bigcap_{g\in G} gHg^{-1} = \ker(\phi)$.
A: 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.
A: 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:
$\forall g \; \forall r \; gag^{-1} \in rHr^{-1}\\
\forall g \; gag^{-1} \in \cap_{r}\ rHr^{-1}$
A: 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 $gng^{-1}\in \bigcap\limits_{a\in G}a^{-1}Ha=K$.
A: 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.
