Solve Symmetric Group Homework: Find Subgroups of S6, S4 & S3 x S3

[Ted123]

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?

 

[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.

 

[Ted123]

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]

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}.

 

[Ted123]

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 $S_3 \times S_3$

 

[Dick]

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.

 

[Ted123]

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]

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.

 

[Ted123]

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]

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.

 

[Ted123]

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]

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 ___?

 

[Ted123]

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?

 

[Dick]

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]

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.

 

[Ted123]

$A=\{ e, (12), (23) \}$ 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]

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.

 

[Ted123]

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]

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?

 

[Ted123]

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]

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

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

 

[Ted123]

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

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

 

[Dick]

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.

 

[Ted123]

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]

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?

 

[Deveno]

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 $S_6$?

that's a good point.

this is a specific example of a more general theorem:

suppose A,B are subgroups of a finite group G such that:

ab = ba, for all a in A and b in B, and that:

A∩B = {e}.

prove that:

a) AB = {ab: a in A, b in B} is a subgroup of G
b) AxB is isomorphic to AB

note that since G is finite, you only need to show closure to prove (a). to prove (b), it suffices to show that if ab = e, then a = b = e (why?).

 

[Ted123]

that's a good point.

this is a specific example of a more general theorem:

suppose A,B are subgroups of a finite group G such that:

ab = ba, for all a in A and b in B, and that:

A∩B = {e}.

prove that:

a) AB = {ab: a in A, b in B} is a subgroup of G
b) AxB is isomorphic to AB

note that since G is finite, you only need to show closure to prove (a). to prove (b), it suffices to show that if ab = e, then a = b = e (why?).

I remember a 'direct product theorem' from my Groups and Rings course that said if $A,B<G$ satisfy

(i) $A,B \triangleleft G$
(ii) $G=AB$
(iii) $A\cap B = \{ e\}$

then $G \cong A\times B$

Now the 2 subgroups in my case are normal subgroups since they are abelian and certainly $A \cap B = \{e\}$. But you say that it is enough for $AB<G$ rather than $G=AB$.

 

[Deveno]

I remember a 'direct product theorem' from my Groups and Rings course that said if $A,B<G$ satisfy

(i) $A,B \triangleleft G$
(ii) $G=AB$
(iii) $A\cap B = \{ e\}$

then $G \cong A\times B$

Now the 2 subgroups in my case are normal subgroups since they are abelian and certainly $A \cap B = \{e\}$. But you say that it is enough for $AB<G$ rather than $G=AB$.

in the case at hand S₃xS₃ couldn't possibly be isomorphic to all of S₆, the first has only 36 elements, while the latter has 720 (yes, that many).

it is NOT true that "abelian subgroup = normal subgroup".

what IS true is "subgroup of abelian group = normal subgroup".

for example, {e,(1 2)} is an abelian subgroup of S₃, but it is NOT normal.

i was careful (at least i HOPE i was) NOT to mention normality at all in my statement. neither A NOR B need be normal subgroups of G in order for AB to be a subgroup. in fact, S₃ ISN'T a normal subgroup of S₆:

(1 4)(1 2 3)(1 4)^-1 = (1 4)(1 2 3)(1 4) = (2 3 4):

1→4→4→1
2→2→3→3
3→3→1→4
4→1→2→2

so the subgroup of S₆ that fixes {4,5,6} isn't normal, because i just exhibited a conjugate of (1 2 3) that moves 4.

if G is FINITE (i DID mention this), that the only condition for AB to be a subgroup is closure:

(ab)(a'b') must be in AB.

the trouble is, (ab)(a'b') = a(ba')b' and it's the ba' part that gives us the trouble. we want to get "all the a's on the left, and all the b's on the right".

now, if G was abelian, that would be no trouble at all. but, S₆ isn't abelian. but if elements of A and B commute (this is not the same as saying everything in G commutes), then...?

 

[Ted123]

in the case at hand S₃xS₃ couldn't possibly be isomorphic to all of S₆, the first has only 36 elements, while the latter has 720 (yes, that many).

it is NOT true that "abelian subgroup = normal subgroup".

what IS true is "subgroup of abelian group = normal subgroup".

for example, {e,(1 2)} is an abelian subgroup of S₃, but it is NOT normal.

i was careful (at least i HOPE i was) NOT to mention normality at all in my statement. neither A NOR B need be normal subgroups of G in order for AB to be a subgroup. in fact, S₃ ISN'T a normal subgroup of S₆:

(1 4)(1 2 3)(1 4)^-1 = (1 4)(1 2 3)(1 4) = (2 3 4):

1→4→4→1
2→2→3→3
3→3→1→4
4→1→2→2

so the subgroup of S₆ that fixes {4,5,6} isn't normal, because i just exhibited a conjugate of (1 2 3) that moves 4.

if G is FINITE (i DID mention this), that the only condition for AB to be a subgroup is closure:

(ab)(a'b') must be in AB.

the trouble is, (ab)(a'b') = a(ba')b' and it's the ba' part that gives us the trouble. we want to get "all the a's on the left, and all the b's on the right".

now, if G was abelian, that would be no trouble at all. but, S₆ isn't abelian. but if elements of A and B commute (this is not the same as saying everything in G commutes), then...?

$(ab)(a'b') = a(ba')b' = a(a'b)b' = (aa')(bb') \in AB$