$\begingroup$Given the following definition of normal subgroup
A subgroup $H$ of a group $G$ is said to be normal if, for every $g\in G$: $$gH=Hg$$
I've tried to show that $H\mathrel{\unlhd} G$ if and only if we have $gHg^{-1}=H$ for every $g\in G$:
(if) Let $H\mathrel{\unlhd} G$, $g\in G$; if $l\in gHg^{-1}$, there exists $h\in H$ such that $l=ghg^{-1}$ and since $gh$ = $h'g$ for some $h'\in H$, then $l=ghg^{-1}=h'(gg^{-1}) = h'\in H$. Let $h\in H$; we can write $h$ as $hgg^{-1}$, and since $hg = gh'$ for some $h'\in H$, we obtain $h = hgg^{-1} = gh'g^{-1}\in ghg^{-1}$. So $gHg^{-1}=H$.
(only if) Let $gHg^{-1}=H$, $g\in G$; if we take an $l\in Hg$, $l = hg$ for some $h\in H$, we note immediately that $l=hg=(gh'g^{-1})g=gh'\in gH$. What should I do to show that $gH\subset Hg$?
I'm also wondering if this mess is correct, and if it is possible a more elegant proof of this fact.
$\endgroup$2 Answers
$\begingroup$I think a more elegant proof is possible.
Clearly $gHg^{-1}=H\iff gHg^{-1}g=Hg\iff gHe=Hg\iff gH=Hg, \forall g\in G$, simply by multiplying by $g$ on the right (and simplifying $g^{-1}g=e$ and $He=H$).
$\endgroup$3$\begingroup$You have proved that $Hg\subseteq gH$. One might say "by symmetry" the other direction holds, or just prove it directly, using basically the same argument:
Let $m\in gH$ be arbitrary. Then $m=gh$ for some $h\in H$. Now $mg^{-1}=ghg^{-1}=h'$, for some $h'\in H$. Hence $m=h'g$, so $m\in Hg$. This proves that $gH\subseteq Hg$.
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