# Find an example of sets of cosets of different cardinality

Mathematics Asked on January 7, 2022

$$G$$ is a finite group. Let $$H$$ be a subgroup of $$G$$. Is there an example of $$G$$ and $$H$$ such that
$${rm Card}({Hxhmid hin H})neq{rm Card}({Hyhmid hin H}),$$
where $$x,yin Gsetminus H$$? 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 $${Hxhmid hin H}$$ and $${Hyhmid hin H}$$ such that the number of cosets contained in one set is different from the other one.

Could you give me some help? Thank you!

$$G:=S_5$$ 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 $${Hxhmid hin H}$$ 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 $${Hyhmid hin H}$$ 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.

Answered by user792898 on January 7, 2022

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 $$gin D_8$$.

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 $${C_2ah|hin C_2}$$ is two cosets, whilst $${C_2a^2h|hin C_2}$$ is one.

Answered by tkf on January 7, 2022

First consider $$G=S_3$$ and $$H={e,(1 2)}$$, with $$x=e$$ and $$y=(1 2 3)$$. Then $${Hxhmid hin H} = {H} quadtext{while}quad {Hyhmid hin H} = {H(1 2 3), H(1 2)}.$$

But wait, you say, we're not allowed to take $$xin H$$? This isn't actually that serious a restriction, since for any nontrivial group $$K$$ and any $$kin Ksetminus{e}$$, we can now replace $$G$$ by $$Gtimes K$$ and $$H$$ by $$Htimes{e}$$, and $$x$$ and $$y$$ by $$xtimes k$$ and $$ytimes k$$.

Answered by Greg Martin on January 7, 2022