The meaning of $\bigcap_{x\in G} x^{-1}Hx$ and the proof for the fact that $N$ is a subgroup of $G$ such that $y^{-1}Ny=N$ for every $y\in G$ If $H$ is a subgroup of $G$, let $N=\bigcap_{x\in G}x^{-1}Hx$. Prove that $N$ is a subgroup of $G$ such that $y^{-1}Ny=N$ for every $y\in G$.
What does $N=\bigcap_{x\in G}x^{-1}Hx$ mean? I'm confused and I don't know how to utilize this condition.
 A: As none of the other answers actually tell you what $\bigcap_{x\in G}x^{1}Hx$ means, I thought I should tell you. You can then use their answers if you wish to solve the problem.

So, $\bigcap_{x\in G}x^{-1}Hx$ is the intersection of all subgroups of the form $x^{-1}Hx$. That is all it means.

For example, if $G=S_3$ and $H=\{1, (12)\}$ then you cycle through each $x\in S_3$ and form $x^{-1}Hx$. You then intersect these sets. So,
$$\begin{align*}
1^{-1}H1&=H\\
(12)^{-1}H(12)&=H\\
(13)^{-1}H(13)&=\{1, (23)\}\\
(23)^{-1}H(23)&=\{1, (13)\}\\
(123)^{-1}H(123)&=\{1, (23)\}\\
(132)^{-1}H(132)&=\{1, (13)\}
\end{align*}$$
Their intersection is the set $\{1\}$, as $1$ is the only element in each of these sets, so $\bigcap_{x\in G}x^{-1}Hx=\{1\}$.
Exercise: If I had chosen the subgroup $K=\{1, (123), (132)\}\leq S_3$ then $\bigcap_{x\in G}x^{-1}Kx=K$. Prove this, and understand why.
A: Hint:
\begin{align}
y^{-1}Ny&=y^{-1}(\bigcap _{x\in G} x^{-1}Hx)y\\
&=\bigcap_{x\in G}(y^{-1}x^{-1}Hxy)\\
&=\bigcap_{x\in G}((xy)^{-1}H(xy))\\
&=\bigcap_{x\in G}(g^{-1}Hg)
\end{align}
where  $g=xy$
Thus conjugating with $y$ just rearranges the terms of  $\bigcap_{x\in G} x^{-1}Hx$. (as one to one and onto)
Thus, the result.
A: Let $a, b \in N$, we need to prove that: $ab^{-1} \in N$. First we prove that $x^{-1}Hx$ is a subgroup of $G$ for any $x \in G$. For if $s, t \in x^{-1}Hx$, then $s = x^{-1}yx$, and $t = x^{-1}zx$ for some $y, z \in H$. So $st^{-1} = x^{-1}yxx^{-1}z^{-1}x = x^{-1}yz^{-1}x$. Since $y, z \in H$ and $H$ is a subgroup, we have: $yz^{-1} \in H$. So $x^{-1}yz^{-1}x \in x^{-1}Hx$. Thus: $st^{-1} \in x^{-1}Hx$, and $x^{-1}Hx$ is a subgroup of $G$. Now $a, b \in N$, so $a \in x^{-1}Hx$, and $b \in x^{-1}Hx$ for all $x \in G$. Since we just proved that $x^{-1}Hx$ is a subgroup of $G$, that means $ab^{-1} \in x^{-1}Hx$ for all $x \in G$. So $ab^{-1} \in \displaystyle \bigcap_{x \in G}x^{-1}Hx = N$, proving that $N$ is a subgroup of $G$. And for any $y \in G$, we have:
$y^{-1}Ny = \displaystyle \bigcap_{x \in G}y^{-1}x^{-1}Hxy = \displaystyle \bigcap_{x \in G}(xy)^{-1}Hxy = \displaystyle \bigcap_{z \in G}z^{-1}Hz = N$. We're done.
