# Symmetric Group

## Homework Statement

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 $G$.

## 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?

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## Answers and Replies

Dick
Science Advisor
Homework Helper
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.

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.

Yeah I see how to do it.

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

Dick
Science Advisor
Homework Helper
Yeah I see how to do it.

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

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}.

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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}.

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 $S_3 \times S_3$

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Dick
Science Advisor
Homework Helper
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 $S_3 \times S_3$

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

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

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

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

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.

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.

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?

Dick
Science Advisor
Homework Helper
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?

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.

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.

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

Deveno
Science Advisor
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 ___?

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 ___?

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?

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Dick
Science Advisor
Homework Helper
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?

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

Deveno
Science Advisor
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.

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

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?

Dick
Science Advisor
Homework Helper
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?

Are you thinking about these before you ask the question? Give me your thoughts on that.

Are you thinking about these before you ask the question? Give me your thoughts on that.

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!

Dick
Science Advisor
Homework Helper
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!

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

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

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

Dick
Science Advisor
Homework Helper
(12345)(6) is an element of order 5

Ok, so what's the order of the subgroup it generates?

Ok, so what's the order of the subgroup it generates?

Also 5. But how do I find the subgroup it generates?

Dick
Science Advisor
Homework Helper
Also 5. But how do I find the subgroup it generates?

The group is generated by the powers of a=(12345). Find the powers of a! a^2, a^3, etc.

The group is generated by the powers of a=(12345). Find the powers of a! a^2, a^3, etc.

I've just thinking about the previous part.

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

is not a copy of $S_3 \times S_3$ inside $S_6$ because it is not a subgroup of $S_6$ (subgroups of $S_6$ consist of elements of $S_6$).

How can I convert the pairs of elements into elements of $S_6$?

Dick
Science Advisor
Homework Helper
I've just thinking about the previous part.

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

is not a copy of $S_3 \times S_3$ inside $S_6$ because it is not a subgroup of $S_6$ (subgroups of $S_6$ consist of elements of $S_6$).

How can I convert the pairs of elements into elements of $S_6$?

Multiply the two elements of the pair to get an element of S6. The product of the two subgroups is isomorphic to the direct product. Why?