# Symmetric Group

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?

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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.

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 $S_3 \times S_3$ ?

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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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 $S_3 \times S_3$

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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.

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

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.

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?

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.

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 $S_3 \times S_3$?

Gold Member
MHB
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?

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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.

Gold Member
MHB
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?

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?

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!

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?

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

Homework Helper
(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?

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.

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 $S_6$?

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?

Gold Member
MHB
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 $G=AB$.

Gold Member
MHB
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....?

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