I am trying to prove that $N_G(A)$ is a subgroup of $G$. I am given that $G$ is a group and $A\subseteq G$. Define $gAg^{-1}=\{gag^{-1}|a\in A\}$. Define the normalizer of $A$ in $G$ to the set: $$ N_G(A)=\{g\in G|gAg^{-1}= A\} $$ I have already proven that $N_G(A)$ is nonempty and that it is closed under the binary operation. The part I am struggling with is showing that $N_G(A)$ is closed under inverses. Here is what I have so far....
Let $x\in N_G(A)$. Then $xAx^{-1}=A$. Consider $x^{-1}A(x^{-1})^{-1}=x^{-1}Ax$. By substitution, $x^{-1}xAx^{-1}x=(1)A(1)=A$.
Is this a valid argument?
Lastly, I apologize if this a duplicate question. I tried looking and couldn't find something that helped me with my specific question.
The normalizer of a subgroup $S \subset G$ is the set of all elements $g$ satisfying $gS=Sg$ as a set.
We only need to check for product, inverse and identity.
For the identity $i$, $iS=S=Si$ as a set, is clear.
Let $f, g \in N_G(S)$. Then $fS=Sf$ and $gS=Sg$. Then, $$ fgS = f(gS) = f(Sg) = (fS)g=(Sf)g = Sfg $$
So $fg \in N_G(S)$.
Let $x \in N_G(S)$, then $xS=Sx$. Consider the set $x^{-1}S = \{ x^{-1}s : s\in S\}$. Note that $$Sx^{-1} = (x^{-1}x)Sx^{-1}=x^{-1}(xS)x^{-1} = x^{-1}(Sx)x^{-1}=(x^{-1}S)xx^{-1} = x^{-1}S$$
So $x \in N_G(S)$. Hence the normalizer is a subgroup of $G$.
The following is a very clever manipulation, one often used in group theory.
Clear out, the new definition is in town:
Identity : $iAi^{-1} = A$, obviously.
Product: Let $f,g \in N_G(A)$.then, $$fg(A)(fg)^{-1} = (fg)A(g^{-1}f^{-1}) = f(gAg^{-1})f^{-1} = fAf^{-1} = A$$
Inverse: let $x \in N_G(A)$. Then $xAx^{-1} = A$.
Then, $$x^{-1} A (x^{-1})^{-1} = x^{-1}Ax = x^{-1}(x A x^{-1})x = (x^{-1}x) A (x^{-1}x) = A$$
Do you see why equality is important? If $xAx^{-1}$ were just contained in $A$, then the equality $ x^{-1}Ax = x^{-1}(x A x^{-1})x$ can't be written! Therein lies the crux of the proof. Of course, we are done.