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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.

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    $\begingroup$ $A$ is only a subset instead of a subgroup? $\endgroup$ – Vim Jul 1 '16 at 0:30
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    $\begingroup$ It's not true if $A$ is only a sub_set_. For a counterexample let $G$ be free over $g$ and $h$ and $A=\{g^nhg^{-n}\mid n\in\mathbb{N}\}$. $\endgroup$ – Christian Sievers Jul 1 '16 at 0:32
  • $\begingroup$ OK, thanks. how about if A is a subgroup? $\endgroup$ – David Warren Katz Jul 1 '16 at 0:36
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    $\begingroup$ This is not true if you only requiere that $gAg^{-1}$ is contained in A for one g. $\endgroup$ – Mariano Suárez-Álvarez Jul 1 '16 at 0:46
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    $\begingroup$ See math.stackexchange.com/questions/401592/… for a couple of examples. $\endgroup$ – Mariano Suárez-Álvarez Jul 1 '16 at 0:48
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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.

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Warning: WRONG HINT: Hint: if one assumes $A$ is a subgroup, then so is $gAg^{-1}$.

The above hint is not going to work! In fact as pointed out in the comments, the result is not true if one only assumes the inclusion holds for one $g$.

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