Group element that normalizes finite subgroup that is generated by a subset of $G$

Question

Let $N$ be a finte subgroup of a group $G$, and assume $N = \langle S \rangle$ for some subset $S$ of $G$. Prove that an element $g \in G$ normalizes $N$ if and only if $gSg^{-1} \subset N$.

My question is about the forward direction. That is, we assume that $g$ normalizes $N$, i.e. $gNg^{-1} = N$, and we want to prove that if $s \in S$, then $gsg^{-1} \in N$. But I cannot see how to do this. In fact, if we let $N = \{e\}$ and $S$ be any nontrivial subset of $G$, then every $g$ normalizes $N$, but $gsg^{-1} \not \in N$ if $s \neq e$.

Answer 1

If $S$ generates $N$, then each element of $N$is of the form $s_1\ldots s_n$, where each $s_i$ is either an element of $S$ or the inverse of an element of $S$ or the identity element of $G$. In particular, $S$ is contained in $N$. Hence if an element $g$ of $G$ normalizes $N$ then in particular it normalizes every element of $S$.

Answer 2

The subgroup $N$ is the smallest subset of $G$ containing $S$. In particular, $S\subset N$.

Now, "$g$ normalises $N$" implies that $g^{-1}xg\in N$ for all $x\in N$. As $S\subset N$ this means that $g^{-1}sg\in N$ for all $s\in S$, and so $g^{-1}Sg\subset N$ as required.

(Note that the "$N$ is finite" assumption is not used here. It is required for the opposite direction though.)