Centralizer of a group is the intersection of Stabilizers

Question

Let $C(S)=\{g\in G\mid gs=sg,\,\forall s\in S\}$. (This set is called the centralizer of $S$ in $G$.) I need to show that it is a subgroup by recognizing it as an intersection of stabilizers under some action of $G$. I know I have to use conjugation, but I don't have idea how to proceed. Any ideas or suggestions?

Answer 1

Let $G$ act on itself by conjugation. The stabilizer of $x \in G$ under this action is the set $G_x = \{s \in G : s \cdot x = sxs^{-1}=x \}$. The centralizer of $G$ in $G$ is the set $C(G) = \{c \in G: cg = gc \text{ for all } g \in G\}$. Note that we can rewrite the centralizer as $C(G) = \{c \in G: cgc^{-1} = g \text{ for all } g \in G\}$. Now show what $\bigcap_{x \in G} G_x$ is as a set and the solution should be clear.

Answer 2

We can rewrite this as $C(S)=\{g\in G\mid gsg^{-1}=s,\,\forall s\in S\}$ with $C(S)$ being the set of all elements in $G$ which commute with every other member. Now $1\in C(S)$, since $1h=h1$ $\forall h\in G$, and so then $\forall h\in S$. Now for two elements $a,b\in C(S)$ we have $\forall s\in S$, $asa^{-1}=s$ and $bsb^{-1}=s$. Now note $$a^{-1}(asa^{-1})a=(a^{-1}a)s(a^{-1}a)=1s1=s$$ and $$s=a^{-1}sa$$ Hence $s=a^{-1}sa=asa^{-1}$ and $C(S)$ is closed under inverses, that is if $a\in C(S)$ so is $a^{-1}\in C(S)$.

Now check it is closed under products. \begin{align} (ab)s(ab)^{-1}&=(ab)s(b^{-1}a^{-1})\\ &=a(bsb^{-1})a^{-1}\\ &=asa^{-1}\quad\,\,\,\,\text{since $b\in C(S)$}\\ &=s\qquad\quad\ \ \text{since $a\in C(S)$} \end{align} therefore $ab\in C(S)$, and $C(S)$ is closed under product taking. Hence $C(S)\subseteq G$.

The fact that $C(S)$ is a subgroup of $G$ follows by looking at how $G$ acts on its elements by group actions. The stabilizer of $s$ in $G$ is defined as the set $G_s=\{g\in G\mid g\cdot s=s\}$. To show it's a group first note $1\in G_s$ since $1\cdot s=s$. Now, if $a\in G_s$, \begin{align} s&=1\cdot s \\ &=(a^{-1}a)\cdot s \\ &=a^{-1}\cdot (a\cdot s)\\ &=a^{-1}\cdot s\qquad\quad\text{since $a\in G_s$} \end{align} and so $a^{-1}\in G_s$ also. Now check products: if $a,b\in G_s$ then \begin{align} (ab)\cdot s&=a\cdot (b\cdot s)\\ &=a\cdot s\quad\quad\,\,\text{since $b\in G_s$}\\ &=s\qquad\quad\ \,\,\text{since $s\in G_s$} \end{align} Hence $G_s$ is a subgroup of $G$.