$gAg^{-1} \subset A$ implies $gAg^{-1} = A$ I am trying to prove that $gAg^{-1} \subset A$ implies $gAg^{-1} = A$, where A is a subset of some group G, and g is a group element of G. This is stated without proof in Dummit and Foote. I know that $\| gAg^{-1} \| = \|A\|$, so I see it is true for finite A, but I am having trouble proving this fact for infinite A. Edit: Christian points out in the comments that this is not true if $A$ is an arbitrary subset. But the question of it being true for $A$ a subgroup remains unanswered.

I believe you're talking about Theorem 6 page 82 of the book,right? If that's the case, the assumption is $$\forall g\in G,\,gNg^{-1}\subset N$$ where $N$ is a subgroup of $G$. You want to prove that $N$ is a normal subgroup of $G$,i.e, $\forall g\in G,\,gNg^{-1}=N$.

Let $g\in G.$ By assumption, you have $gNg^{-1}\subset N$ so let's show that $N\subset gNg^{-1}$.

Let $n\in N$. To show that $N\subset gNg^{-1}$, we must show that $n\in gNg^{-1}$. We have:

\begin{align} n\in gNg^{-1} &\Leftrightarrow\exists m\in N,\,n=gmg^{-1}\\\ &\Leftrightarrow\exists m\in N,\,g^{-1}ng=m\\\ &\Leftrightarrow g^{-1}ng\in N \end{align}

The statement $g^{-1}ng\in N$ is true by assumption. Thus we have $n\in gNg^{-1}$ as desired.