- #1
Swlabr1
- 15
- 0
Let $H\leq G$, where $G$ is some infinite group, and there exists some $g\in G$ such that the set $\{g^n: n\in\mathbb{Z}\}$ is a transversal for $G/H$. Then is $H$ normal in $G$?
I suspect not. However, I cannot seem to find a counter-example.
(By "a transversal for $G/H$" I mean that
1) $g^nH=g^mH\Rightarrow m=n$
2) if $hH$ is a coset of $G/H$ then there exists some $n\in\mathbb{Z}$ such that $g^nh^{-1}\in H$
so the powers of $g$ form a set of coset representatives for $G/H$, and no two of these representatives lie in the same coset.)
I suspect not. However, I cannot seem to find a counter-example.
(By "a transversal for $G/H$" I mean that
1) $g^nH=g^mH\Rightarrow m=n$
2) if $hH$ is a coset of $G/H$ then there exists some $n\in\mathbb{Z}$ such that $g^nh^{-1}\in H$
so the powers of $g$ form a set of coset representatives for $G/H$, and no two of these representatives lie in the same coset.)