Find an example of sets of cosets of different cardinality

Question

$G$ is a finite group. Let $H$ be a subgroup of $G$. Is there an example of $G$ and $H$ such that $${\rm Card}(\{Hxh\mid h\in H\})\neq{\rm Card}(\{Hyh\mid h\in H\}),$$ where $x,y\in G\setminus H$? Here ${\rm Card}$ means cardinality, namely the number of elements contained in a set, so I wonder if we can find two sets of cosets $\{Hxh\mid h\in H\} $ and $\{Hyh\mid h\in H\}$ such that the number of cosets contained in one set is different from the other one.

Could you give me some help? Thank you!

Answer 1

$G:=S_5$ and $H:=\{(1),(23),(24),(34),(234),(243)\}\cong S_3$. Set $x:=(35)$ and $y:=(13)(45)$. We have \begin{align} &Hx=\{(35),(253),(24)(35),(345),(2534),(2453)\},\\ &Hy=\{(13)(45),(132)(45),(13)(254),(1354),(13542),(13254)\}. \end{align} The set of cosets $\{Hxh\mid h\in H\}$ has $3$ elements and they are \begin{align} &\{(35),(253),(24)(35),(345),(2534),(2453)\},\\ &\{(235),(25),(2435),(2345),(25)(34),(245)\},\\ &\{(345),(2543),(2354),(45),(254),(23)(45)\}; \end{align} the set of cosets $\{Hyh\mid h\in H\}$ has $6$ elements and they are \begin{align} & \{(13)(45),(132)(45),(13)(254),(1354),(13542),(13254)\},\\ &\{(123)(45),(12)(45),(12543),(12354),(12)(354),(1254)\},\\ &\{(13)(245),(13452),(13)(25),(13524),(1352),(134)(25)\},\\ &\{(1453),(14532),(14253),(14)(35),(142)(35),(14)(253)\},\\ &\{(14523),(1452),(143)(25),(14)(235),(14352),(14)(25)\},\\ &\{(12453),(12)(345),(1253),(124)(35),(12)(35),(12534)\}. \end{align}

Hence we are done.

Answer 2

First consider $G=S_3$ and $H=\{e,(1\ 2)\}$, with $x=e$ and $y=(1\ 2\ 3)$. Then $$ \{Hxh\mid h\in H\} = \{H\} \quad\text{while}\quad \{Hyh\mid h\in H\} = \{H(1\ 2\ 3), H(1\ 2)\}. $$

But wait, you say, we're not allowed to take $x\in H$? This isn't actually that serious a restriction, since for any nontrivial group $K$ and any $k\in K\setminus\{e\}$, we can now replace $G$ by $G\times K$ and $H$ by $H\times\{e\}$, and $x$ and $y$ by $x\times k$ and $y\times k$.

Answer 3

Consider the symmetry group $D_8$ of a square. Let $C_2$ denote the subgroup fixing a corner $x$ (so $C_2$ consists of the identity, and reflection through the diagonal containing $x$). Then we can identify the $4$ corners of the square with the cosets of $C_2$. That is all the elements of $C_2g$ map $x$ to $xg$, so we may identify the coset $C_2g$ with the corner $xg$, for each $g\in D_8$.

The orbits of corners under $C_2$ have different sizes: one orbit is the two corners adjacent to $x$, another is the single corner opposite to $x$.

Thus if $a$ is a $90^\circ$ rotation, then $\{C_2ah|h\in C_2\}$ is two cosets, whilst $\{C_2a^2h|h\in C_2\}$ is one.