2

This is my simple curiosity

If $H$ is normal in $G$ then $H\setminus G$ is a group : $$ Ha \cdot Hb = H ab $$ But if $H$ is not normal then the coset space $H\setminus G$ is not a group, but we guess that there exists a some property which can be satisfied in a group.

Let $H$ be a subgroup of a finite group $G$. Then we have $H\setminus G = \{ Hg_i \}_{i=1}^m$ where $g_i$ is representative and $m=[G:H]$. Here

Question : $$ \{ Hg_i \}_{i=1}^m = \{ Hg_i^{-1} \}_{i=1}^m$$

That is, $Hg_i=Hg_j^{-1}$ for some $j$. This holds ?

If not, $Hg_i^{-1}=Hg_j^{-1}$ for some $i\neq j$ so that $g_i^{-1} g_j\in H$. I cannot prove or disprove.

Specific Case : $$G=N\times_\phi H,\ N=\langle a\rangle={\bf Z}_p,\ H=\langle b \rangle={\bf Z}_q\subset {\rm Aut}\ (N)\ (q|(p-1))$$ where $p$ and $q$ are primes.

Then there exists $x\in G$ s.t. $H'=\langle x\rangle $ has order $q$ and $ H'\setminus G = \{ H'c^k \}_{k=0}^{p-1}$ where $c\in G$ has order $p$. In this case the above equality holds.

Hanul Jeon
  • 27,376
HK Lee
  • 19,964

2 Answers2

4

Not every right transversal to $H$ in $G$ has that property, but it is always possible to choose a particular right transversal with this property. There is a Theorem of P. Hall which asserts that whenever $H$ is a subgroup of a finite group $G,$ there is a right transversal $T$ to $H$ in $G$ which is also a left transversal. That is, if $T = \{t_{i}: 1 \leq i \leq n \},$ where $n = [G:H],$ then we have $G = \bigcup_{i=1}^{n} Ht_{i}$ and $G = \bigcup_{i=1}^{n}t_{i}H$. Then for $i \neq j,$ we have $t_{i}t_{j}^{-1} \not \in H$ and $t_{i}^{-1}t_{j} \not \in H.$ The second condition tells us that the right cosets $Ht_{i}^{-1}$ and $Ht_{j}^{-1}$ are distinct. Hence we do have the equality asked for in the question for this particular transversal.

Here is an explicit counterexample to the necessary disjointness for a general transversal. Take $G = S_{4}$ and $H$ to be the stabilizer of $4$ in $G,$ which is the natural copy of $S_{3}$ inside $G.$ Note that the cosets $H(134)$ and $H(234)$ are distinct, because $(134)(432) = (412) \not \in H.$ However, the cosets $H(431)$ and $H(432)$ are equal, because $(431)(234) = (123) \in H.$

Geoff Robinson
  • 24,035
2

If $G=S_3$ and $H=\{1,(12)\}$ take $g=(123)$ so $g^{-1}=(132).$

Then $Hg=\{(123),(13)\}$ while $Hg^{-1}=\{(132),(23)\}.$

[note multiplied left-to right here.]

Added: If the number of cosets is a finite $m$, then your list of cosets $Hg_i$ for $1 \le i \le m$ is the same as the number of cosets $Hg_i^{-1}$. The latter are also cosets, and are pairwise disjoint by the fact that the $Hg_i$ are pairwise disjoint. So both lists of cosets are complete lists of the cosets of $H$ in $G$. For example consider some fixed coset $Hg_j^{-1}.$ Then $g_j^{-1}$ must lie in one of the cosets $Hg_i$, and for that $i$ we have $Hg_i=Hg_j^{-1}.$

Note the following does not show disjointness of the cosets with representatives taken as the inverses. The question remains open.

A comment asked for why the $Hg_i^{-1}$ are disjoint. Suppose $Hg_1^{-1}=Hg_2^{-1}.$ Multiplying by $g_1$ on the right then gives $H=Hg_2^{-1}g_1,$ so that $g_2^{-1}g_1\in H$. But then $g_1,g_2$ are in the same coset of $H$, i.e. $Hg_1=Hg_2,$ against the choice of the coset reps for the cosets $Hg_i.$

coffeemath
  • 29,884
  • 2
  • 31
  • 52
  • Thank you for your answer. But $g^2=g^{-1}$ so that ${ H, Hg, Hg^2} ={H, Hg^{-1}, Hg^{-2}}$ So this is not a counter example. – HK Lee Mar 01 '14 at 07:47
  • Oh I thought you meant for each specific $g$ the equality should hold. I'll try for this other question. – coffeemath Mar 01 '14 at 07:54
  • @HeeKwonLee I added some argument to the effect that your switching the representatives to their inverses should work in a finite group, or even in an infinite group if the number of cosets is finite. – coffeemath Mar 01 '14 at 08:23
  • I am not convinced that the $Hg_{i}^{-1}$ are disjoint, at least for the reason you give, – Geoff Robinson Mar 01 '14 at 08:23
  • That is, you do have $g_{i}g_{j}^{-1} \not \in H$ when $i \neq j,$ but for the cosets $Hg_{i}^{-1}$ and $Hg_{j}^{-1}$ to be distinct (and hence disjoint) you need $g_{i}^{-1}g_{j} \not \in H.$ – Geoff Robinson Mar 01 '14 at 08:28
  • @GeoffRobinson I just added an argument on disjointness of the $Hg_i^{-1}.$ – coffeemath Mar 01 '14 at 08:29
  • 1
    I know but I am not convinced by your argument. – Geoff Robinson Mar 01 '14 at 08:32
  • @coffemath Thank you. I can understand the disjointness. – HK Lee Mar 01 '14 at 08:32
  • @GeoffRobinson I agree with you there is something wrong with the disjointness argument as it stands. Will try to fix... – coffeemath Mar 01 '14 at 08:33
  • 1
    @HeeKwonLee In the above comment Geoff Robinson has pointed out an error in the disjointness proof, so I'd suggest for now that you retract the "accept" i.e. uncheck the green checkmark, since the answer may not be right. In fact maybe there is an example where your reversal to inverses of coset representatives doesn't give a complete list of disjoint cosets. – coffeemath Mar 01 '14 at 08:49