If $U,V \subset S_n$ are subgroups with $S_n//U = \{id,g_2,...,g_e\}$ and $\alpha_j$ is $\frac{1}{j}$ times the number of $i\in [e]$ s.t $[V:V \cap g_i U g_i^{-1}]=j$ then $(\alpha_1,...,\alpha_e)$ is a partition of $e$.

Now let $U=V \in \{V_4,D_4\}$ and $n=4$ ($V_4$ is the group with only the double transpositions). Then the number of $V_4$-orbits of $S_4/V_4$ of size $1$ is $6$, and there are none of greater size. One $D_4$-orbit of $S_4/D_4 = \{D_4,(12)D_4,(14)D_4 \}$ is $\{D_4\}$; for the other two, we have $|orb_{D_4}((ab)D_4)|=[D_4 : D_4 \cap (ab)D_4(ab)]=[D_4:V]=2$, so there is one orbit of size $1$ and two of size $2$.

I don't understand what I've done wrong here. I got a partition of $[S_4 :V_4]$, $1^6$, but apparently something went wrong with $D_4$, but I have no idea why? In the article which all this stuff pertains to it is asserted that there is one orbit of size one and one of size two, when $D_4$ acts on $S_4/D_4$. It cannot be that the two orbits should be considered as one, for if so then there would only be one orbit of size one wrt $S_4/V_4$.

(This is one of those questions that's difficult to pose because there's too much background, but I think this info should do, hopefully. Otherwise, if you're feeling more helpful than usual, check out beginning of section 3 and Table 2 here.)

Edit: Some calculations. $orb_V(V)={V}$, trivial ($V=V_4$). $orb_V((12)V)=\{(12)V,(34)V,(1324)V,(1423)V \}$ and $orb_V((13)V)=\{(13)V, (1234)V, (24)V, (1432)V\}$. To see that each of these orbits is really just $\{V\}$, use $\sigma V = \tau V \iff \tau^{-1} \sigma \in V$. Thus we have three orbits of size one so far.

With $D_4$, we have $orb_{D_4}((12)D_4)=\{ (12)D_4, (134)D_4, (243)D_4, (123)D_4, (142)D_4, (34)D_4, (1423)D_4, (1324)D_4\}$ and $orb_{D_4}((14)D_4) = \{ (14)D_4, (234)D_4, (132)D_4, (143)D_4, (124)D_4, (1342)D_4, (1243)D_4, (23)D_4 \}$. Some manipulations show that these two orbits have size two, yet only one of them counts?

  • $\begingroup$ I think you are $n$-counting orbits of size $n$. If the orbit of $x$ is $\{x,y\}$ then the orbit of $y$ is $\{x,y\}$ as well, so we only count that as one orbit of size two, namely $\{x,y\}$, rather than two orbits of size two, $\{x,y\}$ from $x$ and $\{x,y\}$ from $y$. $\endgroup$ – Jack Schmidt Jul 15 '13 at 17:58
  • $\begingroup$ @JackSchmidt What do you mean by $n$-counting orbits of size $n$? But, if I'm supposed to count orbits like you said, then there wouldn't be six $V_4$-orbits of $S_4/V_4$ of size one. I'll make an edit to the original post to show you some computations. $\endgroup$ – Erik Vesterlund Jul 15 '13 at 18:10
  • $\begingroup$ When $n=2$, I mean you are “double counting” orbits of size 2. For orbits of size $n=1$, you are single counting, which is good. $\{x\}, \{y\}, \{z\}, \{a\}, \{b\}, \{c\}$ are six orbits of size 1. $\endgroup$ – Jack Schmidt Jul 15 '13 at 18:12
  • $\begingroup$ @Jack The article is a mess to be honest. Anyway, wrt double counting, I'm afraid I still don't understand what that means (indicative of me not entirely knowing what I'm doing, yes). As you see in the calculations I added, nothing different is done with regards to counting or anything, singleton orbits become so by the same manipulations. $\endgroup$ – Erik Vesterlund Jul 15 '13 at 18:24
  • $\begingroup$ Ok, I'll just write up the calculation and the counts as an answer. Hopefully it'll be clear. $\endgroup$ – Jack Schmidt Jul 15 '13 at 18:26

There are a few unstated conventions. Which $D_4$? How do we multiply permutations? I'll go with GAP conventions for both as they are likely to agree with the paper and most of the world of computational group theory and finite group theory.

Take $U=D_4 = \{ (), (3,4), (1,2), (1,2)(3,4), (1,3)(2,4), (1,3,2,4), (1,4,2,3), (1,4)(2,3) \}$ and multiply permutations in reading order such that $(1,2)(2,3)=(1,3,2)$.

$G//U=S_4//D_4 = \{ ()D_4, (1,2,3)D_4, (1,3,2)D_4 \}$ so $e=3$.

If $V=D_4$ as well, then

$V \cap g_1Ug_1^{-1} = V \cap ()U()^{-1} = V \cap V = V$ has index 1. Note that the orbit of $D_4$ on the seed $()D_4$ is $\{ D_4 \}$ of size 1.

$V \cap g_2 U g_2^{-1}$ has size 4 (it is $V_4$, the Klein viergruppe of double transpositions). That means it has index 2. Note that the orbit of $D_4$ on the seed $(1,2,3)D_4$ is $\{ (1,2,3)D_4, (1,3,2)D_4 \}$.

$V \cap g_3 U g_3^{-1}$ has size 4 (it is $V_4$, the Klein viergruppe of double transpositions). That means it has index 2. Note that the orbit of $D_4$ on the seed $(1,3,2)D_4$ is $\{ (1,2,3)D_4, (1,3,2)D_4 \}$.

Now we get $\alpha_1 = \tfrac{1}{1} 1 = 1$ since only $g_1$ gave index 1.

We get $\alpha_2 = \tfrac{1}{2} 2 = 1$ since $g_2$ and $g_3$ gave index 2.

We get $\alpha_3=\tfrac{1}{3} 0 = 0$ since nothing gave index 3.

Sure enough $1 \alpha_1 + 2\alpha_2 + 3 \alpha_3 = 1 + 2 + 0 = 3$ is a partition.

In terms of orbits, this is because the three cosets $\{ ()D_4, (1,2,3)D_4, (1,3,2)D_4 \}$ split into two orbits under the left action of $D_4$, namely, $\{ ()D_4 \}$ and $\{ (1,2,3)D_4, (1,3,2)D_4 \}$.

  • $\begingroup$ Looks like my crucical mistake was, as you pointed out, not dividing $\alpha_j$ by $j$. I'll look over this a few more times and try other examples before I accept it as the answer ;) Thank you! $\endgroup$ – Erik Vesterlund Jul 15 '13 at 18:44
  • $\begingroup$ No problem. A keyword for this is "double coset". Hall's theory of groups section 1.7 explains the relation between the index formula and the size of orbits, but I don't think it directly addresses you (and certainly doesn't say anything about galois theory). $\endgroup$ – Jack Schmidt Jul 15 '13 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.