# 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.

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 $H$.

• I don't think that "any group $G$ with a subgroup $H$ that is properly contained in some conjugate of $H$" will do. I tried this first and got nowhere. Perhaps it is possible to obtain an example from any such situation, it is only I that cannot see how$\ldots$ Sep 2, 2014 at 13:58
• @chizhek If $H$ is a proper subgroup of $yHy^{-1}$, then $yHy^{-1}$ is a union of (more than one) right cosets of $H$, $yHy^{-1}=H\cup Hx\cup\dots$. So $yH=Hy\cup Hxy\cup\dots$. Sep 2, 2014 at 14:21
• But of course. Sometimes I am quite unable to see the obvious. Sep 2, 2014 at 14:45

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}} $$\newcommand{\set}{{\{#1\}}}$$\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}{{\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.

$\newcommand{\card}{{\left|#1\right|}}$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 $Hx^ky$, $0\leq k<m$.