Comparing $gH$ and $Hg$ for Infinite & Finite Groups

[alexmahone]

Suppose $G$ is an infinite group and $H$ is an infinite subgroup of $G$.

Let $g\in G$.

Suppose $\forall h\in H\ \exists h'\in H$ such that $gh=h'g$.

Can we conclude that $gH=Hg$?

What if $G$ and $H$ are of finite orders?

 

[topsquark]

Suppose $G$ is an infinite group and $H$ is an infinite subgroup of $G$.

Let $g\in G$.

Suppose $\forall h\in H\ \exists h'\in H$ such that $gh=h'g$.

Can we conclude that $gH=Hg$?

What if $G$ and $H$ are of finite orders?

I don't have an answer, just an observation. It is possible that given g and h that the h' might generate a subset of H, not all of H. (ie. \(\displaystyle \exists f: H \to H'\) is not an bijection.) In which case \(\displaystyle gH = H'g \subseteq Hg\) in general. I don't know if that helps or not.

-Dan

 

[Euge]

Suppose $G$ is an infinite group and $H$ is an infinite subgroup of $G$.

Let $g\in G$.

Suppose $\forall h\in H\ \exists h'\in H$ such that $gh=h'g$.

Can we conclude that $gH=Hg$?

What if $G$ and $H$ are of finite orders?

Hi Alexmahone,

I believe your statement is correct, but only for finite groups. Suppose $G$ is finite. The given condition implies $gH \subset Hg$. Since there is a bijection $gH \leftrightarrow Hg$, then $gH$ and $Hg$ have the same number of elements. Hence, $gH = Hg$.

 

[Euge]

Actually, the statement also holds when $G$ is commutative, infinite or not. The condition would not be needed in such a case.

 

[johng1]

Your problem rephrased slightly:
If $G$ is a group, $H$ a subgroup of $G$ and $g\in G$ with $gHg^{-1}\subseteq H$, does $gHg^{-1}=H$?

Euge has answered in the affirmative for some cases; also this follows if $g$ has finite order, easy proof.

I think an example is needed that shows this is not always true. To follow the example, you need to know a little about group presentations.

Let $G=<a\,,b\,:\, b^{-1}ab=a^2>,\,H=<a>\text{ and } g=b^{-1}$
Since conjugation is a homomorphism, $gHg^{-1}\subseteq H$. If this conjugation were an onto map, there would exist an integer $k$ with $ga^kg^{-1}=a$. But $ga^kg^{-1}=a^{2k}$ . Since $a$ has infinite order (loosely speaking there is no relation that allows $a$ to have finite order), we would get that $2k=1$, impossible. So $gHg^{-1}\neq H$.