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I have the following problem:

Find the number of subgroups of 4 elements in the direct product of permutation groups $S_2 \times S_4$

I started with writing down all the elements in $S_4$ group. Here we have $4!=24$ elements:

$$S_4=\{ (1)(2)(3)(4),\\ (1)(2)(34),(1)(3)(24), (12)(34), (13)(24), (1)(4)(23), (14)(23), (12)(3)(4), (13)(2)(4), (14)(2)(3), \\ (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(132),\\ (1243), (1234), (1342), (1324), (1423), (1432) \}$$

The group $S_2$ is simple: $\{(1)(2), (12)\}$

In our subgroup we must have the neutral element $\{ (1)(2), (1)(2)(3)(4)\}$. But how can I find out how another $3$ elements can look like?

I understand, that we have to fulfill here 2 conditions:

  1. The product of any two elements must be in the subgroup
  2. The inverse of each element must be in the subgroup as well
Shaun
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Anna Schmidt
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    Your title was much more general than what you really asked for. You can start with the subgroups $H$ of $1\times S_4\cong S_4$ of order $4$. They are listed here already. Then $S_2\times G$ for any subgroup $G$ of order $2$ in $S_4$, which are also listed there. – Dietrich Burde Sep 02 '22 at 10:59
  • You are talking about groups but did not explicit the operation involved, try to start there – Aseed Sep 02 '22 at 11:02
  • @Aseed I added the information about the operation involved. – Anna Schmidt Sep 02 '22 at 11:14
  • @DietrichBurde yes you are right, I mean here this operation. I already found the subgroups of $S_4$ of order 4, but what should I do futher? And what dio you mean by looking at $S_2 \times G$? – Anna Schmidt Sep 02 '22 at 11:17
  • Well, $S_2\times {(1),(12)}$ is also a subgroup of $S_2\times S_4$ with $2\cdot 2=4$ elements. You need to count them, too. So I meant $G={(1),(12)}$ and so on. – Dietrich Burde Sep 02 '22 at 11:19
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    What a horrible question! According to a computer calculation the answer should be $31$. – Derek Holt Sep 02 '22 at 11:26
  • @DerekHolt yes, the answer is 31, but I am trying here to understand how it can be counted by hand. – Anna Schmidt Sep 02 '22 at 11:32
  • @DietrichBurde for example we consider the group $S_2 \times G$ with G - subgroup of order 4. There exist 7 such subgroups for G. For each of this subgroups we have $2 \times 4 = 8$ possible combinations. That means together 7*8 = 56? But it is much more than 31. – Anna Schmidt Sep 02 '22 at 11:34
  • No, $G$ does not have order $4$, but order $2$. This gives only $9$ groups $S_2\times G$. Here $G$ can also have double transpositions, so ${(1),(12)(34)}$. – Dietrich Burde Sep 02 '22 at 11:41
  • See https://en.m.wikipedia.org/wiki/Goursat%27s_lemma – calc ll Sep 02 '22 at 12:12
  • @DietrichBurde But then we have 9+4+7=20, because we have 9 subgroups of order 2, 4 subgroups of order 3, 7 subgroups of order 4. – Anna Schmidt Sep 02 '22 at 13:20
  • @DietrichBurde okay, so for 4=14 we have 7+7=14 and for 4=22 we have 1*9=9. How can I find another 31-23=8? – Anna Schmidt Sep 02 '22 at 13:53
  • Indeed by Goursat. A subgroup of $S_2\times S_4$ need not be of the form $G\times H$ for $G\le S_2$ and $H\le S_4$. So we can have "mixed ones". See for example this post, or similar ones. – Dietrich Burde Sep 02 '22 at 13:57
  • @DietrichBurde can you please give an example for this case? I mean the example of such a subgroup, which is not of the form $G \times H$ for $G \leq S_2$ and $H \leq S_4$ – Anna Schmidt Sep 02 '22 at 14:03

1 Answers1

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Within the $S_4$ factor of $G$ there are three $C_4$ subgroups and four $C_2\times C_2$ subgroups.

The remaining subgroups of order $4$ all intersect the $S_4$ factor in a group of order $2$.

There are just three $C_4$ groups outside of $S_4$, generated by $(1,2)(3,4,5,6)$, $(1,2)(3,4,6,5)$ and $(1,2)(3,5,4,6)$.

There are three $C_2 \times C_2$ subgroup that intersect $S_4$ in $\langle(3,4)(5,6)\rangle$, namely $\langle (1,2),(3,4)(5,6) \rangle$, $\langle (1,2)(3,4),(3,4)(5,6) \rangle$, and $\langle (1,2)(3,5)(4,6),(3,4)(5,6) \rangle$, and similarly for the other two intersections of that type.

Finally, there are two $C_2 \times C_2$ subgroup that intersect $S_4$ in $\langle(3,4)\rangle$,namely $\langle (1,2),(3,4) \rangle$ and $\langle (1,2)(5,6),(3,4) \rangle$, and simialrly for the other $5$ intersections of that type.

So the total number of subgroups of order $4$ is $$3+4+3 + (3\times 3) + (2 \times 6) = 31.$$

Derek Holt
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