TITLE: Is finiteness necessary in this exercise?
QUESTION [5 upvotes]: This is from Dummit&Foote Abstract Algebra Chap 3.1 Problem 28.
Here is problem & solution of this problem.
Let $N$ be finite subgroup of $G$ and suppose $G=\langle T\rangle$ and $N=\langle S\rangle$ for some subsets $S$ and $T$ of $G$. Prove $N$ is normal in $G$ iff $tSt^{-1} \subseteq N$ for all $t \in T$.
https://crazyproject.wordpress.com/2010/03/01/a-criterion-for-finite-subgroup-normalcy-on-generating-sets-of-the-subgroup-and-group/
Finiteness of $N$ is used as: $tNt^{-1}=N$ from $tNt^{-1} \subseteq N$.
If $N$ is infinite, then $G$ is infinite, so $t^{-1}$ is not always expressed as finite product of elements of $T$. So we can't get $t^{-1}Nt\subseteq N$ from $tNt^{-1} \subseteq N$.
So we can't avoid line 
$tNt^{-1}=N$ from $tNt^{-1} \subseteq N$. without finiteness of $N$.
I think finiteness of $N$ is crucial in this proof, so I think there is counterexample of this excercise without finiteness.
Also hardness of taking counterexample is making $T$ s.t. $\langle T\rangle=G$. But, it should be $T \neq G$ because it implies normalness.  
I'm trouble to taking non abelian group $G$ and $T\neq G$ s.t. $\langle T\rangle=G$.
Can you find counterexample of this excercise or is that still true for without finiteness of $N$?

REPLY [3 votes]: This has been asked many times before. The standard counterexample is the group defined  by the presntation $\langle x,y \mid yxy^{-1} =x^2 \rangle$ with $T=\{ x,y \}$, $S =\{ x \}$. It can also  be defined as a subgroup of ${\rm GL}(2,{\mathbb Q})$.