# Homework Help: Symmetric Group

1. Feb 15, 2012

### Ted123

1. The problem statement, all variables and given/known data

Let $G=S_6$ acting in the natural way on the set $X = \{1,2,3,4,5,6\}$.

(a)(i) By fixing 2 points in $X$, or otherwise, identify a copy of $S_4$ inside $G$.

(ii) Using the fact that $S_4$ contains a subgroup of order 8, find a subgroup of order 16 in $G$.

(b) Find a copy of $S_3 \times S_3$ inside $S_6$. Hence, or otherwise, show that $G$ contains a subgroup of order 9.

(c) Find a subgroup of order 5 in $G$.

3. The attempt at a solution

How do I find a subgroup of S6 which is isomorphic to S4 by fixing 2 elements of X?

For (a)(i) I know:

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

As written, this can be regarded as a subgroup of S6.

Now, if I want to fix 5 and 6, what permutations in S6 does that leave me with? And conversely, if I write S4 as I've done above, what is the action of each element of S4 on 5 and 6?

For (a)(ii) I know that the following is a subgroup of $S_4$ of order 8: $$\{ e, (24), (13), (12)(34), (14)(23), (1432), (1234), (1324) \}$$

How do I find the subgroup of order 16 by building it up from this subgroup of order 8?

Last edited: Feb 15, 2012
2. Feb 15, 2012

### Dick

If you regard the elements of your S4 as acting on the elements {1,2,3,4} of S, then the don't act on {5,6} at all. 5 and 6 are fixed. You could take your subgroup of order 8 and add a permutation to it that doesn't fix {5,6} to make a larger group.

3. Feb 15, 2012

### Ted123

Yeah I see how to do it.

For (b) what are the elements of the direct product $S_3 \times S_3$ ?

4. Feb 15, 2012

### Dick

You found a copy of S4 inside of S6. Try and find a copy of S3 inside of S6 in the same way. Now try and find two copies of S3 in S6 that act on disjoint subsets of {1,2,3,4,5,6}.

Last edited: Feb 15, 2012
5. Feb 15, 2012

### Ted123

Well a copy of $S_3$ in $S_6$ is $$\{ e, (12), (23), (13), (123), (132) \}$$

$S_3 \times S_3 = \{ (a,b) : a,b \in S_3 \}$ with operation $(a,b)(c,d)=(ac,bd)$

I'm not getting how to express cyclic permutations of $S_3 \times S_3$

Last edited: Feb 15, 2012
6. Feb 15, 2012

### Dick

That's one S3. Try using {4,5,6} as elements to permute to get the second S3 in the cross product.

7. Feb 15, 2012

### Ted123

OK so $$S_3 \times S_3 = \{ e, (12), (23), (13), (123), (132) \} \times \{ e, (45), (56), (46), (456), (465) \}$$

8. Feb 15, 2012

### Dick

Yes, both of your S3 groups act on {1,2,3,4,5,6} and it's a direct product because they don't interfere with each other. They commute.

9. Feb 15, 2012

### Ted123

So is this a copy of $S_3 \times S_3$ in $S_6$?

How do I find a subgroup of order 9 from this?

10. Feb 15, 2012

### Dick

Yes, it is. Use the same sort of idea you used to get a subgroup of order 16 from a subgroup of order 8. Think about it.

11. Feb 15, 2012

### Ted123

Well what is the order of $S_3 \times S_3$?

12. Feb 15, 2012

### Deveno

the order of S3 x S3 is 36. perhaps if you picked a subgroup of order 3 from each "factor", you might have a subgroup of order ___?

13. Feb 15, 2012

### Ted123

So can I just say that $A=\{ e, (12), (23) \}$ is a subgroup of $S_3$ and so is $B=\{ e, (45), (56)\}$.

Then $A \times B$ is a subgroup of $S_3 \times S_3$ which in turn is a subgroup of $S_6$ of order 9?

Last edited: Feb 15, 2012
14. Feb 15, 2012

### Dick

$A=\{ e, (12), (23) \}$ is not a subgroup.

15. Feb 15, 2012

### Deveno

A is not a subgroup of S3, neither is B.

neither one of these 3-element sets is closed under multiplication. for example:

(1 2)(2 3) = (1 2 3) (applying (2 3) first), which is not in A.

16. Feb 15, 2012

### Ted123

The alternating group $A_3$ is a subgroup:

$A=\{ e, (123), (132) \}$

$B=\{ e, (456), (465) \}$

So $A \times B < S_3 \times S_3 < S_6$

So $A \times B < S_6$

How do I go about finding a subgroup of order 5?

17. Feb 15, 2012

### Dick

18. Feb 15, 2012

### Ted123

Well can I do something similar to the previous 2 parts? i.e. identify a copy of a certain group inside S6 and deduce a subgroup of order 5? I can't think of any groups to use though!

19. Feb 15, 2012

### Dick

It's not very similar to the other parts. Does S6 have an element of order 5?

20. Feb 15, 2012

### Ted123

(12345)(6) is an element of order 5