Does a left coset inclusion in a right coset imply they are equal? Let $H$ be a subgroup of a group $G$ and $x$ be an element of $G$.
If $G$ is finite, $xH\subset Hx$ implies $xH=Hx$ since $xH$ and $Hx$ have the same finite cardinality (that of $H$).
What happens when $G$ and $H$ are infinite? Can a left coset be strictly included in a right coset?
 A: An easy example is mentioned in Van der Waerden’s paper Ein Satz über Klasseneinteilungen von endlichen Mengen, Hamburger Abhandlungen, 5 (1927) 185–188. Take $G$ to be group of all functions $x \mapsto rx + s$, where $r,s \in \mathbb{Q}$, $r\ne 0$ (with composition as a group operation). Take $H$ to be the subgroup of functions $x \mapsto x + 2m$ where $m \in \mathbb{Z}$. Let $f(x) = 2x$ and let $g(x) = 2(x+1)$. Then
$$ fH \cup gH = Hf = Hg$$ where the union is disjoint.
A: Consider the groups $\mathbb{Z}$ and $\mathbb{Q}$ under addition and let
$\phi:\mathbb{Z}\rightarrow Aut(\mathbb{Q})$ be the mapping where $\phi(n)(r)=n\cdot 2^{r}$. Then $\phi$ is a group homomorphism. We can therefore define the semidirect product $\mathbb{Q}\times_{\phi}\mathbb{Z}$ by letting $(r,n)(s,m)=(r+\phi(n)(s),n+m)=(r+s\cdot 2^{n},n+m)$ and where $(r,n)^{-1}=(\phi(-n)(-r),-n)=(-r\cdot 2^{-n},-n)$. Then $\mathbb{Z}\times\{0\}$ is a subgroup of the group $\mathbb{Q}\times_{\phi}\mathbb{Z}$. On the other hand, $(0,1)(r,0)(0,-1)=(0,1)(r+2^{0}\cdot 0,-1)=(0,1)\cdot(r,-1)=(0+2^{1}\cdot r,0)$ making $(0,1)(\mathbb{Z}\times\{0\})(0,-1)=2\mathbb{Z}\times\{0\}$ a proper subgroup of $\mathbb{Z}\times\{0\}$
A: In the free group on two generators $x, y$, if we take $H$ to be the subgroup generated by elements of the form $x^n y^m x^{-n}$ for $n \geq 0$, then $x H x^{-1}$ is strictly contained in $H$ because $y \notin x H x^{-1}$. 
