Let $H = \langle S \rangle$ be a subgroup of $G = \langle T \rangle$. Prove that $H$ is normal in $G$ if and only if $tst^{-1} \in H$ for all $s \in S$ and $t \in T \cup T^{-1}$. Here $T^{-1}$ denotes the set $T^{-1}=\{t^{-1} \mid t\in T\}$.
This is a result my text uses but does not prove, and I am curious to see how it should be done rigorously.
Consider the set $K$ of all $g\in G$ such that $gHg^{-1}=H$. Note that $K$ is a subgroup of $G$ (indeed, it is the kernel of the action of $G$ on the set of conjugates of $H$, by conjugation). By definition, $H$ is normal iff $K=G$.
Now suppose $tst^{-1}\in H$ for all $s\in S$ and all $t\in T\cup T^{-1}$. Since $S$ generates $H$ and conjugation by $t$ is a homomorphism, this implies $tht^{-1}\in H$ as well for all $h\in H$ and $t\in T\cup T^{-1}$. Fixing any $t\in T$, this says that $tHt^{-1}\subseteq H$. Letting $u=t^{-1}$, we also have $uHu^{-1}\subseteq H$, and conjugating both sides by $t$ we get $H\subseteq tHt^{-1}$. Thus $H=tHt^{-1}$, so $t\in K$. Since $t\in T$ was arbitrary, this proves $T\subseteq K$. Since $K$ is a subgroup, this means $K=G$ and so $H$ is normal.