Show that $c(S) = \{a\in G : a\in gSg^{-1}\ \forall g\in G\}$ is normal in $G$

Question

Question:

Let $G$ be a group, $S \leq G$, and define $$c(S) := \{a\in G : a \in gSg^{-1}\ \textrm{for all}\ g\in G\}$$ Show that $c(S)$ is a normal subgroup of $G$ contained in $S$

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Attempt:

I have been able to show that $c(S)$ is a subgroup of $G$:

Let $a$, $b$ $\in c(S)$. Then we have that $$ a\in gSg^{-1} \implies a = gs_{1}g^{-1}\ \textrm{for some}\ s_1\in S \\ b\in gSg^{-1} \implies b = gs_{2}g^{-1}\ \textrm{for some}\ s_2\in S $$

Proceeding with the subgroup test: \begin{align} ab^{-1} &= \left(g s_1 g^{-1}\right)\left(g s_2 g^{-1}\right)^{-1} \\ &= g s_1 g^{-1} g {s_2}^{-1} g^{-1} \\ &= gs_1{s_2}^{-1}g^{-1} \\ &\in gSg^{-1} \end{align}

So $c(S)\leq G$.

I'm struggling in particular with showing that $c(S)$ is normal in $G$. I have so far attempted to show that $c(S)$ is closed under conjugation by elements from $G$:

Let $b\in c(S)$. If $gbg^{-1}\in c(S)$ for all $g\in G$, then $c(S)\triangleleft G$:

\begin{align} b\in c(S) &\implies b = gsg^{-1}\ \textrm{for some}\ s\in S\textrm{, for all}\ g\in G. \\ &\implies gbg^{-1} = ggsg^{-1}g^{-1} = (gg)s(gg)^{-1} \end{align}

I'm not sure how to proceed from this point. I am also confused at the definition of $c(S)$ itself, I definitely do not have any intuition as to what the object $c(S)$ actually is.

Edit (In response to Chrystomath's comment):

Let $b\in c(S)$. Then $b = lhl^{-1}$ for fixed $l\in G$, $h\in S$. Let $k$ be any element of $G$: \begin{aligned} kbk^{-1} &= k(lhl^{-1})k^{-1} \\ &= klhl^{-1}k^{-1} \\ &= (kl)h(kl)^{-1} \\ &\in gSg^{-1} \end{aligned}

Writing $g=kl\in G$.

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My questions are:

• Was I correct in the way I showed that $c(S)\leq G$?
• How can I interpret the definition of $c(S)$?
• How do I show that $c(S)$ is normal in $G$?

Thank you

Answer 1

You have done the subgroup test for $c(S)$. Consider any element in $g^{-1} c(S) g$, wlog let it be $h =g^{-1} g' s g'^{-1} g$, where $s \in S$, according to the definition of $c(S)$. But it follows by definition that $h \in c(S)$, and $c(S) \lhd G$. However, it would seem that $S \subset c(S)$, not the other way. Have I missed something?