# "structure" on the cosets → normal?

#### Swlabr

##### New member
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.)

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: "structure" on the cosets $\Rightarrow$ normal?

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.)
Hi Swlabr,

Does $$G/H$$ stands for the quotient group ? In that case all the other parts of the question are redundant since by definition of quotient groups $$H$$ should be normal in $$G$$.

Kind Regards,
Sudharaka.

#### Opalg

##### MHB Oldtimer
Staff member
Re: "structure" on the cosets $\Rightarrow$ normal?

What about the (non-normal) subgroup $\mathbb{Z}/2$ as a subgroup of the infinite dihedral group? More precisely, let $G = \langle u,t\ |\ u^2=1,\ utu=t^{-1} \rangle$. Let $H = \{1,u\}$, and take $g=t$ for the generator of the transversal.