Symmetric Group

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Homework Statement



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

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

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

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

(c) Find a subgroup of order 5 in [itex]G[/itex].

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:

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

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 [itex]S_4[/itex] of order 8: [tex] \{ e, (24), (13), (12)(34), (14)(23), (1432), (1234), (1324) \}[/tex]

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

  • #2
Dick
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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
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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 [itex]S_3 \times S_3[/itex] ?
 
  • #4
Dick
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Yeah I see how to do it.

For (b) what are the elements of the direct product [itex]S_3 \times S_3[/itex] ?

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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  • #5
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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 [itex]S_3[/itex] in [itex]S_6[/itex] is [tex] \{ e, (12), (23), (13), (123), (132) \}[/tex]

[itex]S_3 \times S_3 = \{ (a,b) : a,b \in S_3 \}[/itex] with operation [itex](a,b)(c,d)=(ac,bd)[/itex]

I'm not getting how to express cyclic permutations of [itex]S_3 \times S_3[/itex]
 
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  • #6
Dick
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Well a copy of [itex]S_3[/itex] in [itex]s_6[/itex] is [tex] \{ e, (12), (23), (13), (123), (132) \}[/tex]

[itex]S_3 \times S_3 = \{ (a,b) : a,b \in S_3 \}[/itex] with operation [itex](a,b)(c,d)=(ac,bd)[/itex]

I'm not getting how to express cyclic permutations of [itex]S_3 \times S_3[/itex]

That's one S3. Try using {4,5,6} as elements to permute to get the second S3 in the cross product.
 
  • #7
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That's one S3. Try using {4,5,6} as elements to permute to get the second S3 in the cross product.

OK so [tex]S_3 \times S_3 = \{ e, (12), (23), (13), (123), (132) \} \times \{ e, (45), (56), (46), (456), (465) \}[/tex]
 
  • #8
Dick
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OK so [tex]S_3 \times S_3 = \{ e, (12), (23), (13), (123), (132) \} \times \{ e, (45), (56), (46), (456), (465) \}[/tex]

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
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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 [itex]S_3 \times S_3[/itex] in [itex]S_6[/itex]?

How do I find a subgroup of order 9 from this?
 
  • #10
Dick
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So is this a copy of [itex]S_3 \times S_3[/itex] in [itex]S_6[/itex]?

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.
 
  • #11
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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 [itex]S_3 \times S_3[/itex]?
 
  • #12
Deveno
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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
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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 [itex]A=\{ e, (12), (23) \}[/itex] is a subgroup of [itex]S_3[/itex] and so is [itex]B=\{ e, (45), (56)\}[/itex].

Then [itex]A \times B[/itex] is a subgroup of [itex]S_3 \times S_3[/itex] which in turn is a subgroup of [itex]S_6[/itex] of order 9?
 
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  • #14
Dick
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So can I just say that [itex]A=\{ e, (12), (23) \}[/itex] is a subgroup of [itex]S_3[/itex] and so is [itex]B=\{ e, (45), (56), \}[/itex].

Then [itex]A \times B[/itex] is a subgroup of [itex]S_3 \times S_3[/itex] which in turn is a subgroup of [itex]S_6[/itex] of order 9?

[itex]A=\{ e, (12), (23) \}[/itex] is not a subgroup.
 
  • #15
Deveno
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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
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[itex]A=\{ e, (12), (23) \}[/itex] is not a subgroup.

The alternating group [itex]A_3[/itex] is a subgroup:

[itex]A=\{ e, (123), (132) \}[/itex]

[itex]B=\{ e, (456), (465) \}[/itex]

So [itex]A \times B < S_3 \times S_3 < S_6[/itex]

So [itex]A \times B < S_6[/itex]

How do I go about finding a subgroup of order 5?
 
  • #17
Dick
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The alternating group [itex]A_3[/itex] is a subgroup:

[itex]A=\{ e, (123), (132) \}[/itex]

[itex]B=\{ e, (456), (465) \}[/itex]

So [itex]A \times B < S_3 \times S_3 < S_6[/itex]

So [itex]A \times B < S_6[/itex]

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.
 
  • #18
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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!
 
  • #19
Dick
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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?
 
  • #20
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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
 
  • #21
Dick
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(12345)(6) is an element of order 5

Ok, so what's the order of the subgroup it generates?
 
  • #22
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Ok, so what's the order of the subgroup it generates?

Also 5. But how do I find the subgroup it generates?
 
  • #23
Dick
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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.
 
  • #24
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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.

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

is not a copy of [itex]S_3 \times S_3[/itex] inside [itex]S_6[/itex] because it is not a subgroup of [itex]S_6[/itex] (subgroups of [itex]S_6[/itex] consist of elements of [itex]S_6[/itex]).

How can I convert the pairs of elements into elements of [itex]S_6[/itex]?
 
  • #25
Dick
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I've just thinking about the previous part.

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

is not a copy of [itex]S_3 \times S_3[/itex] inside [itex]S_6[/itex] because it is not a subgroup of [itex]S_6[/itex] (subgroups of [itex]S_6[/itex] consist of elements of [itex]S_6[/itex]).

How can I convert the pairs of elements into elements of [itex]S_6[/itex]?

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?
 

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