Is it possible that a left coset of $H$ contains more than one right coset of $H$? Let $H$ be a subgroup of group $G$. 

Is it possible that a left coset of $H$ contains more than one right coset of $H$?

It is clear to me that the answer is 'no' if we deal with finite groups.
 A: We want to know if the following situation can occur:
$H$ is a subgroup of a group $G$, and there exist $x$, $y$, $z$ in $G$
such that $Hy\subseteq xH$ and $Hz\subseteq xH$ and $Hy\neq Hz$.
(I hope we agree which cosets are the right ones and which are the left ones
-- not that it really matters.)
$\newcommand{\NN}{\mathbb{N}}
$$\newcommand{\set}[1]{{\{#1\}}}
$$\newcommand{\defequiv}{\mathrel{\overset{\text{def}}{\Longleftrightarrow}}}
$
The condition $Hy\neq Hz$ is equivalent to $yz^{-1}\!\notin H$.
Consider the condition $Hy\subseteq xH$;
it implies that $y\in xH$,
and hence that $xH=yH$, $Hy\subseteq yH$, and $y^{-1}\!Hy\subseteq H$.
Similarly $Hz\subseteq xH$
implies that $xH=zH$, $Hz\subseteq zH$, and $z^{-1}\!Hz\subseteq H$.
Moreover, we must have $zH=xH=yH$, hence $z^{-1}y\in H$.
Conversely, if $y^{-1}\!Hy\subseteq H$, $z^{-1\!}Hz\subseteq H$,
$z^{-1}y\in H$, and $yz^{-1}\!\notin H$,
then it follows that with $x:=y$ (or $x:=z$)
we have $Hy\subseteq xH$, $Hz\subseteq xH$, and $Hy\neq Hz$.
Let $G$ be the free group with the free generators $a$ and $b$.
Define $H$ as the smallest subset of $G$ that satisfies the following conditions:
$$
\begin{gathered}
a^{-1}b\in H~, \qquad b^{-1}a\in H~, \\
\text{if $~h\in H~$ then $~a^{-1}ha\in H$}~, \\
\text{if $~h\in H~$ then $~b^{-1}hb\in H$}~, \\
\text{if $~h_1,h_2\in H~$ then $~h_1h_2\in H$}~.
\end{gathered}
$$
An easy proof by induction (on construction of elements of $H\,$)
shows that $H^{-1}\!\subseteq H$,
so it follows that $H$ is a subgroup of $G$.
Let $n\in\NN$, and $x_j\in\set{a,b}$, $\varepsilon_j\in\set{1,-1}$ for $1\leq j\leq n$,
and consider the following property $N$
of the word $w := x_1^{\varepsilon_1}\!\cdots x_n^{\varepsilon_n}$:
$$
N(w)~\,\defequiv\,\,
    \text{$\sum_{j=1}^k\varepsilon_j\leq 0\,$ for $\,0\leq k<n~$
        and $~\sum_{j=1}^n\varepsilon_j=0$}\,.
$$
It is easy to see that
if a word $w'$ is obtained from a word $w$ by a sequence of reductions
$x^{\varepsilon}x^{-\varepsilon}\to 1$,
where $x\in\set{a,b}$ and $\varepsilon\in\set{1,-1}$,
then $N(w)$ implies $N(w')$.
Using this, an evident proof by induction
(again on construction of elements of $H\,$)
shows that we have $N(h)$ for every $h\in H$
(presented as a reduced word).
Since $\lnot\, N(ab^{-1})$, we see that $ab^{-1}\!\notin H$.
We have an affirmative answer:
the subgroup $H$ of the group $G$, constructed above, and the elements $a$, $b$ of $G$
satisfy the conditions $Ha\subseteq aH$, $Hb\subseteq aH$ ($=bH$),
and $Ha\neq Hb$.
Remarks. $~$The example presented above is 'freely' constructed,
in the following sense.
Suppose
we have a subgroup $H_1$ of a group $G_1$ and elements $x$, $y$, $z$ of $G_1$
such that $H_1y,H_1z\subseteq xH_1$ and $H_1y\neq H_1z$.
Let $G_0$ be the subgroup of $G_1$ generated by $\set{y,z}$
and let $H_0$ be the smallest subgroup of $G_0$ that contains $y^{-1}z$
and is closed under the conjugations $g\mapsto y^{-1}g\,y$ and $g\mapsto z^{-1}g\, z$.
Then $H_0$ is a subgroup of $H_1$
and $H_0y,H_0z\subseteq yH_0=zH_0$ and $H_0y\neq H_0z$.
Let $\varphi$ be the homomorphism $G\to G_1$ that sends $a$ to $y$ and $b$ to $z$;
then $\varphi(G)=G_0$ and $\varphi(H)=H_0$.
$\qquad$To give an example,
let us for a few moments write the two groups in the Jeremy Rickard's answer
as $G_1$ and $H_1$,
and let $\varphi\colon G\to G_1 : a\mapsto y, b\mapsto xy\,$;
then $\varphi(G)=G_1$ and $\varphi(H)=H_1$.
$\qquad$This was the thinking leading to the present answer:
if there exists a subgroup of some group
so that some left coset of the subgroup
contains two different right cosets of the same subgroup,
then the 'free' construction will give such a subgroup.
$\newcommand{\suchthat}{\mid}
$We are (silently) assuming that the elements of the free group $G$
are represented by reduced words,
that is, by sequences of 'tokens' $a$, $b$, $a^{-1}$, $b^{-1}$
that do not contain any subsequence consisting of two opposite tokens.
The subgroup $H$ consists precisely of all elements of $G$ that have the property $N$.
We already know that $N(h)$ for every $h\in H$;
the converse, that every $g\in G$ having the property $N$ belongs to $H$,
is proved by induction on the length of the word $g$.
$\qquad$This result, that $H=\set{g\in G\suchthat N(g)}$,
can be given an amusing interpretation.
Let $P$ be the set of all sequences of parentheses "$($" and "$)$",
and write the empty sequence as $\varepsilon$.
Denote by $Q$ the subset of $P$ generated by the following closure rules:
$$
\begin{gathered}
\varepsilon\in Q~, \\
\text{if $\,q\in Q\,$ then $\,(\mspace{1mu}q\mspace{2mu})\in Q$}~, \\
\text{if $q_1,q_2\in Q\,$ then $\,q_1q_2\in Q$}~.
\end{gathered}
$$
The set $Q$ (computer scientists would call it a 'language')
consists of all sequences of properly nested parentheses.
Let us map each $g\in G$ to $\pi(g)\in P$,
replacing each token $a^{-1}$ or $b^{-1}$ with an opening parenthesis
and each token $a$ or $b$ with a closing parenthesis;
then $H=\pi^{-1}(Q)$.
If $G$ is a group
and some right coset of a subgroup $H$ is contained in a left coset of $H$,
then the left coset of $H$ is partitioned into right cosets of $H$.
Indeed, suppose that $Hy\subseteq xH$;
then $xH=yH$ and $H$ is a subgroup of its conjugate $yHy^{-1}$,
which is partitioned into distinct right cosets $Hx_i$, $i\in I$,
whence $xH=yH=(yHy^{-1})y$
is partitioned into the distinct right cosets $Hx_iy$, $i\in I$.
(This result is an extension of the concluding remark in the Jeremy Rickard's answer,
and of his response to my clueless comment about it.)
$\newcommand{\generd}[1]{{\langle#1\rangle}}
$$\newcommand{\ZZ}{\mathbb{Z}}
$Let the free group $G=\generd{a,b}$ and its subgroup $H$ be as above in this answer.
The left coset $aH=bH$ contains two different right cosets $Ha$ and $Hb$;
we have a hunch that these two are not the only right cosets contained in $aH$.
Precisely into how many right cosets of $H$ is its left coset $aH$ partitoned?
Inverting $aH=bH$ we get $Ha^{-1}=Hb^{-1}\!$,
thus $aHa^{-1}=bHb^{-1}=aHb^{-1}=bHa^{-1}$.
In particular we have $ab^{-1}\in aHa^{-1}$,
thus $(ab^{-1})^n\in aHa^{-1}$ for every $n\in\ZZ$,
where $(ab^{-1})^n\notin H$ if $n\neq 0$.
It follows that the left coset $aH$
contains countably infinitely many distinct right cosets $H(ab^{-1})^na$, $n\in\ZZ$.
Since $G$ is countably infinite,
we conclude that the left coset $aH$
is partitioned into countably infinitely many right cosets.
$\newcommand{\card}[1]{{\left|#1\right|}}
$Let $X$ be an infinite set,
and let $G$ be the group freely generated by the elements of $X$.
We are assuming that the elements of $G$ are represented by reduced words,
that is, by sequences of tokens $x\in X$ and $x^{-1}\in X^{-1}$
that do not contain any subsequence consisting of two opposite tokens.
In each $g\in G$ we replace every token belonging to $X^{-1}$ with "$($"
and every token belonging to $X$
with "$)$", and denote the result by $\pi(g)\in P$.
Then $H:=\pi^{-1}(Q)$ is a subgroup of $G$ with the following properties:
$Hx\subseteq xH$ for every $x\in X$;
$xH=yH$, but $Hx\neq Hy$ if $x\neq y$, for any $x,y\in X$.
Pick $x_0\in X$.
The left coset $x_0H$ contains $\card{X}$ right cosets $Hx$, $x\in X$;
since $\card{G}=\card{X}$,
the set of the right cosets of $H$ into which the left coset $x_0H$ is partitioned
has the cardinality $\card{X}$.
$\newcommand{\cardnum}{\mathfrak}
$Therefore, for every infinite cardinal number $\mathfrak{m}$
we are able to exhibit a group $G$, its subgroup $H$, and an $x\in G$,
such that the left coset $xH$ is partitioned into $\cardnum{m}$ right cosets.
What about a finite cardinal number $m\geq 2$?
In this case we generalize the construction from the Jeremy Rickard's answer:
we let $G:=\generd{x,y:y^{-1}x\,y=x^m}$ and $H:=\generd{x^m}$;
then the left coset $yH$ is partitioned into the $m$ right cosets
$Hx^ky$, $0\leq k<m$.
A: Yes, it's possible.
For example, let $G=\langle x,y: y^{-1}xy=x^2\rangle$ and $H=\langle x^2\rangle$.
Then $\langle x\rangle = H\cup Hx$, so $H=y^{-1}Hy\cup y^{-1}Hxy$, and so $yH=Hy\cup Hxy$.
You'll get a similar example from any group $G$ with a subgroup $H$ that is properly contained in some conjugate of $H$.
