## Subgroups of S4

My question:
I was asked to give a subgroup of order 6 of the permutation group S4. That part is not so difficult, for example S3 has order 6 and is a subgroup of S4. But now I have to show how many subgroups of order 6 are in S4. Intuitively thinking, there are four of them, each of them leaving 1, 2, 3 or 4 fixed. But how can you prove this?

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 Recognitions: Homework Help Science Advisor A subgroup of order 6 must contain an element of order 3 and an element of order 2. That means it contains a 3 cycle and a 2 cycle or a product of two disjoint 2 cycles. Can you show it's a 2 cycle? Let's pick the 3 cycle to be (1,2,3) (how many other choices are there that lead to different order 3 subgroups?). Now you need to add a 2 cycle. If you add a 2 cycle chosen from the elements 1, 2 and 3, like (1,2), you generate S3. If you add an element like (1,4) can you show you get all of S4? Do you see how this proves your conclusion about the 4 different S3's?
 Hmm, why does the subgroup need to have an element of order 2 and an element of order 3? Why not only elements of order 2 or 3?

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## Subgroups of S4

 Quote by 3029298 Hmm, why does the subgroup need to have an element of order 2 and an element of order 3? Why not only elements of order 2 or 3?
Cauchy's theorem says so. 2 and 3 are prime divisors of 6.

 Ah, I see! It cannot be a product of two disjoint 2-cycles because then you again generate S4. And there are four choices of 3-cycles, which gives you 4 subgroups. Is this correct? Thanks a lot :)

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