Exercise 3.1.29 (Sufficient conditions for $N \unlhd G$ if $G = \langle T \rangle$ and $N = \langle S \rangle$)

Question

Let $N$ be a {\bf finite} subgroup of $G$ and suppose $G = \langle T \rangle$ and $N = \langle S \rangle$ for some subsets $S$ and $T$ of $G$. Prove that $N$ is normal in $G$ if and only if $tSt^{-1} \subseteq N$ for all $t \in T$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $N =\langle S \rangle$ is a finite subgroup of $G$.

\bigskip
If $N \unlhd G$ then for all $g \in G$, $g N g^{-1} \subseteq N$, and since $S \subseteq N$,  $g S g^{-1} \subseteq N$. Moreover $T \subseteq G$, so for all $t \in T$, $ tS t^{-1} \subseteq N$.

\bigskip
Conversely, suppose that for all $t\in T$, $ tS t^{-1} \subseteq N$.

Since $N = \langle S \rangle$ is a finite subgroup of $G$, by Exercise 28, every $t \in T$ normalizes $N$, i.e.,
\begin{align}
\forall t \in T,\ t N t^{-1} = N.
\end{align}
Consider the normalizer $N_G(N)$. Then (1) shows that $T \subseteq N_G(N)$. By definition, $\langle T \rangle$ is the smallest subgroup of $G$ (for inclusion) which contains $T$. Since $N_G(N)$ is a subgroup of $G$ which contains $T$, then $\langle T \rangle \subseteq N_G(N)$, therefore $G =\langle T \rangle  \subseteq N_G(N)$, where $N_G(N) \leq G$, so
$G = N_G(N)$. This shows that $N \unlhd G$.

In conclusion, if $N$ is a finite subgroup of $G$ such that $G = \langle T \rangle$ and $N = \langle S \rangle$, then
$$N \unlhd G \iff \forall t \in T,\ tSt^{-1} \subseteq N.$$
\end{proof}

Exercise 3.1.29 (Sufficient conditions for $N \unlhd G$ if $G = \langle T \rangle$ and $N = \langle S \rangle$)

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Exercise 3.1.29 (Sufficient conditions for $N \unlhd G$ if $G = \langle T \rangle$ and $N = \langle S \rangle$)

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