Q:

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

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 $S$

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 $g=kl\in G$.

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

A:

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?

