equivalent definition of normal subgroup Let $G$ be a group and $N$ a subgroup of $G$. I read that the following two definitions are equivalent:
1) $\forall g\in G$, $gNg^{-1}=N$
2)$\forall g\in G$, $gNg^{-1}\subset N$
does this mean that we have always $N\subset gNg^{-1}$ ? my guess is no because take $n\in N$ then if $n=gn'g^{-1}$ then $n'=g^{-1}ng$ and there is no reason that $g^{-1}ng\in N$
 A: No, it does not mean that we always have $N\subseteq gNg^{-1}$ for any subgroup $N$ and any $g\in G$.
First, note that it is clear that 1) implies 2). To see why 2) implies 1), note that if we know, for some particular $g\in G$, that $gNg^{-1}\subseteq N$, then letting $h=g^{-1}$, we have that $h^{-1}Nh\subseteq N$ (since $h^{-1}=g$). Thus, if we know that for every $g\in G$ we have $gNg^{-1}\subseteq N$, then also for every $g\in G$, we have $g^{-1}Ng\subseteq N$, because every element of $G$ is some other element's inverse. But $g^{-1}Ng\subseteq N$ is equivalent to $N\subseteq gNg^{-1}$ by left and right multiplying by $g$ and $g^{-1}$, respectively. Therefore, 2) implies that for all $g\in G$, we have both $gNg^{-1}\subseteq N$ and $N\subseteq gNg^{-1}$, hence $N=gNg^{-1}$.
A: Since 2) holds for all $g\in G$, it also holds for $g^{-1}$. Thus, $g^{-1}N(g^{-1})^{-1}=g^{-1}Ng\subset N$, that is, $N \subset gNg^{-1}$.
A: I redo Zev's exquisite answer for more space. 
No, it does not mean that we always have $N\subseteq gNg^{-1}$ for any subgroup $N$ and any $g\in G$.
First, note that by the definition of set equality, 1) implies 2). 
Why does 2) implies 1)? Note that $gNg^{-1}\subseteq N$ for some particular $g\in G$
$\iff$ Let $h=g^{-1}$ iff $h^{-1} = g$ $ \iff h^{-1}Nh\subseteq N$.  
Thus, if for every $g\in G$, we have $\color{blue}{gNg^{-1}\subseteq N}$,
then also for every $g\in G$, we have $g^{-1}Ng\subseteq N$,
because every element of $G$ is some other element's inverse. 
But by left and right multiplying by $g$ and $g^{-1}$, respectively, $g^{-1}Ng\subseteq N \iff N\subseteq gNg^{-1}$ . Therefore, 2) implies that for all $g\in G$, we have both $\color{blue}{gNg^{-1}\subseteq N}$ and $N\subseteq gNg^{-1}$ $\iff N=gNg^{-1}$.
