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Question related to this already answered here [Verification]Let H be a subgroup of G and $N = \bigcap_{x\in G} xHx^{-1}$. Prove that N is a subgroup of G and that $aNa^{-1} = N$ for all $a \in G$
But I wanted to understand this question properly.
If H is subgroup of G
Let N=$\bigcap_{x\in G} $xH$x^{-1}$ Then N is Normal Subgroup.
I understand to it as subgroup Form fact that intersection of subgroups is again subgroup where $xHx^{-1}$ is subgroup .But I had not understand how to show it is normal .
Also Is this possible way to determine normal subgroup of any group from Finding subgroup first ?I know this is not computationally good thought but is it?
Any Help will be appreciated

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    $\begingroup$ This is done by showing $xH=Hx$. $\endgroup$ – JohnColtraneisJC Apr 30 '18 at 17:05
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    $\begingroup$ @ JohnColtraneisJC Is this possible like $g(xHx^{-1})g^{-1}$ $\in H $So $(gx)H(gx)^{-1}$ $\in N$ Thanks For Help $\endgroup$ – idon'tknow Apr 30 '18 at 17:11
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    $\begingroup$ @ JohnColtraneisJC Sorry but my question was different $\endgroup$ – idon'tknow Apr 30 '18 at 17:14
  • $\begingroup$ It is true that once you have found all subgroups, you can apply this theorem to find all normal subgroups. But there are much better ways, some discussion here: math.stackexchange.com/questions/3917/… $\endgroup$ – Mike Earnest Apr 30 '18 at 17:25
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You just need to show that $N$ is closed under conjugation, $g\in N\implies aga^{-1}\in N$. To show $aga^{-1}\in N$, you need to show $aga^{-1}\in xHx^{-1}$ for all $x\in G$. Multiplying on the left by $a^{-1}$ and the right by $a,$ this is equivalent to showing $g\in (a^{-1}x)H(a^{-1}x)^{-1}$. But we know this is true since $g\in N$, so letting $y=a^{-1}x$, you have $g\in yHy^{-1}$.

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