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Let $S = \mathcal{P}(G)$ $S$ be the collection of subsets in the group $G$. Let $G$ act on $S$ by conjugation. So, $g: B\to gBg^{-1}$ where $B \subset G$. This text claims that it is easy to check that the normalizer of $A$ is precisely the stabilizer of $A$, under this action.

i.e $N_G(A) = G_A$

Isn't the normalizer the set of elements where $gA = Ag$ is satisfied? In the stabilizer, the elements satisfy $Ag = A$ , i.e the subset of $G$, $g$, acts as an identity. I suppose the stabilizer is the normalizer when the group $G$ is non-abelian. This isn't specified, though.

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By definition, the stabilizer $G_A$ is the set $\{g\in G:g\cdot A=A\}$. Since the action $\cdot$ is conjugation, $g\cdot A=gAg^{-1}$. So this can be written as $$ G_A=\{g\in G:gAg^{-1}=A\}=\{g\in G:gA=Ag\}=N_G(A). $$

I suspect the following might be the source of confusion: the way you're using $gA$ in the definition of the stabilizer is shorthand for the image of the action $g\cdot A$ in this context, not the "coset" $gA$ in $G$.

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  • $\begingroup$ Wow I can't believe I forgot to account for the operation >_< Thanks $\endgroup$
    – Obliv
    Jun 20, 2016 at 14:19

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