Clarification of a theorem from Dummit and Foote's Abstract Algebra

Question

The following is from page 82 of Dummit and Foote's Abstract Algebra, 3rd edition.

Theorem 6 ((1) and (5)) says that a subgroup $N$ of a group $G$ is normal if and only if $gNg^{-1}\subseteq N$ for all $g\in G$. By definition, a subgroup $N$ of a group $G$ is normal if $gNg^{-1}=N$ for all $g\in G$. So does this theorem mean $gNg^{-1}\subseteq N$ for all $g\in G$ if and only if $gNg^{-1}=N$ for all $g\in G$?

Or more specifically, dose $gNg^{-1}\subseteq N$ for all $g\in G$ imply $N\subseteq gNg^{-1}$ for all $g\in G$?

Answer 1

Hint: Since a group is closed under inversion, $gNg^{-1} \subset N$ for all $g$ iff $g^{-1}Ng \subset N$ for all $g$.

Answer 2

I'll write out the answer in more detail.

Assume (*) $gNg^{-1} \subset N$ for all $g$.

Let $h \in N$ be arbitrary.

Now fix $g$. By (*) $g^{-1}hg \in N$. So then $h = g(g^{-1}hg)g^{-1} \in gNg^{-1}$. Since $h$ was arbitrary, this implies that $N \subset gNg^{-1}$, so in fact $N = gNg^{-1}$

Answer 3

The whole crux of the matter lies in showing:

If $gNg^{-1} \subseteq N$ for all $g \in G$, then $N \subseteq gNg^{-1}$ for any $g \in G$.

The way I always like to do this is pick any old $n \in N$. Then pick any old $g \in G$. Then we need to find some $n'\in N$, such that:

$n = gn'g^{-1}$.

So I am going to start with $n$, and then pick $g^{-1}$, which is certainly SOME element of $G$. So by our assumption, it follows that:

$g^{-1}n(g^{-1})^{-1} = g^{-1}ng \in N$.

So, let $n' = g^{-1}ng$.

Then $n = ene = ene^{-1} = (gg^{-1})n(gg^{-1})^{-1} = (gg^{-1})n(gg^{-1}) = g(g^{-1}ng)g^{-1} = gn'g^{-1}$.