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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 $g^{-1}ng\in N$

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Note that 2) implies 1) easily if $G$ is finite, just by counting elements. But for infinite $G$ a better argument, such as Zev's, is needed. –  Greg Martin Oct 15 '11 at 9:12
    
Note also that it's easy to find counterexamples to the assertion that $N \subseteq gNg^{-1}$ for an arbitrary subgroup $N$: look in nearly any nonabelian group. For example, in the permutation group $S_3$, take $N$ to be a two-element subgroup such as $\{ (1)(2)(3),\, (1\ 2)(3) \}$. Then $N \not\subseteq gNg^{-1}$ for any of the four elements $g\in G\setminus N$. –  Greg Martin Oct 15 '11 at 9:15

3 Answers 3

up vote 4 down vote accepted

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 $N=gNg^{-1}$.

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I think Martin Isaacs stresses this point in his book Algebra: A Graduate Course immediately after he defines "normal subgroup of a group". –  Amitesh Datta Oct 15 '11 at 8:39

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, $N \subset gNg^{-1}$.

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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 $N\subseteq gNg^{-1}$ $\iff N=gNg^{-1}$.

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