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"structure" on the cosets → normal?

Swlabr

New member
Feb 21, 2012
27
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
Feb 5, 2012
1,621
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
Feb 7, 2012
2,703
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.