Homework Help: Subgroups of S4

1. Dec 25, 2009

3029298

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?

2. Dec 25, 2009

Dick

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?

Last edited: Dec 25, 2009
3. Dec 25, 2009

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?

4. Dec 26, 2009

Dick

Cauchy's theorem says so. 2 and 3 are prime divisors of 6.

5. Dec 26, 2009

3029298

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 :)

Last edited: Dec 26, 2009
6. Dec 26, 2009

Dick

Right. There are 4 subgroups of order 3. You also have to show there is only one group of order 6 that contains each one.

7. Dec 26, 2009

3029298

Is there an easier way to show that there are no products of disjoint 2-cycles in the subgroup, other than writing all te products with all 3-cycles? (and thus showing that you generate more than 6 elements)

8. Dec 26, 2009

Dick

Sure, you don't have to do ALL of them. Your order 3 subgroups are all conjugate. You really just have to show that one of them is contained in a unique order 6 subgroup, say the one generated by (123). The only 'interesting' 2 cycle to multiply by is (14) (the others will just generate an S3). (123)(14)=(1423). That's not good. It has order 4. No subgroup of order 6 can contain a subgroup of order 4. Stop there. Stuff like that.

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