Normal Subgroups of S4 (symmetric group)

Indeed, some of the elements you listed are the same. (2 3 4) = (4 2 3) = (3 2 4) and (2 4 3) = (3 2 4) = (4 3 2). To see this, look at, say, (2 3 4). It takes 2 to 3, 3 to 4, and 4 to 2. Then observe that (3 4 2) and (4 2 3) do exactly the same thing. If you shift all the numbers in any cycle one place to the right, moving the right-most number to the left-most position, you will get the same cycle. The same applies for a left-shift.

It appears you are making the same mistake with the 4-cycles too. For example, (1 2 3 4) = (4 1 2 3) = (3 4 1 2) = (2 3 4 1). Hence if you write down the numbers 1-4 in an arbitrary order, there are three equivalent cycles due to the left or right-shifting I described above, which explains the

\frac{4!}{4}

.

Another way of thinking about the (1,3) cycle type is as follows. There is one fixed number and a 3-cycle. In

S_4

there are only four numbers you can fix. For the remaining three numbers, regardless of what they are, there are only two possible 3-cycles. That gives eight elements with cycle structure (1,3). For example, let's say you fix the number 1. Then you have elements of the form (1)(x y z) where

x,y,z \in \{2,3,4\}

. The only two distinct possibilities are (2 3 4) and (2 4 3).

(edited 13 years ago)

Reply 2

OP

Original post by Dragon

Indeed, some of the elements you listed are the same. (2 3 4) = (4 2 3) = (3 2 4) and (2 4 3) = (3 2 4) = (4 3 2). To see this, look at, say, (2 3 4). It takes 2 to 3, 3 to 4, and 4 to 2. Then observe that (3 4 2) and (4 2 3) do exactly the same thing. If you shift all the numbers in any cycle one place to the right, moving the right-most number to the left-most position, you will get the same cycle. The same applies for a left-shift.

It appears you are making the same mistake with the 4-cycles too. For example, (1 2 3 4) = (4 1 2 3) = (3 4 1 2) = (2 3 4 1). Hence if you write down the numbers 1-4 in an arbitrary order, there are three equivalent cycles due to the left or right-shifting I described above, which explains the

\frac{4!}{4}

.

Another way of thinking about the (1,3) cycle type is as follows. There is one fixed number and a 3-cycle. In

S_4

there are only four numbers you can fix. For the remaining three numbers, regardless of what they are, there are only two possible 3-cycles. That gives eight elements with cycle structure (1,3). For example, let's say you fix the number 1. Then you have elements of the form (1)(x y z) where

x,y,z \in \{2,3,4\}

. The only two distinct possibilities are (2 3 4) and (2 4 3).

I see, looking at my answer I actually did this for the cycles structure 1,1,2. However this is only because it is more obvious for a 2 cycle that if you shift right it stays the same. I did not think about it enough for the 3 and 4 cycles it seems. right now I know why we have 24 = 1+6+3+8+6, I then know:

|N| divides |S4|. Therefore we can have:

1+6+3+8+6,
6+6,
3+1,
3+8+1,
8,
1.

as the orders of N. Now how do I check which ones are normal subgroups and which ones aren't??

Thank you for your help :smile:

Reply 3

OP

Original post by Dragon

Indeed, some of the elements you listed are the same. (2 3 4) = (4 2 3) = (3 2 4) and (2 4 3) = (3 2 4) = (4 3 2).

May I ask why when looking at

S_4

do you look at the "results" under permutation, whereas when trying to find its subgroup you look at the "permuters" themselves.

By this I mean, in

S_4

(1,2,3,4) doesn't equal (2,3,4,1). They are two separate elements, giving 24 elements overall. Whereas in a normal subgroup of cycle structure 4, there are only 6 elements as (1,2,3,4) does in fact equal (2,3,4,1). Do you see what I mean??

I have clearly misunderstood something here, and I realise that I must be mixing something up in my head.

Reply 4

DFranklin

18

In simple terms: cycles are not the same as elements.

Reply 5

OP

Original post by DFranklin

In simple terms: cycles are not the same as elements.

I ma really sorry, but can you explain this a bit more clearly, as I do not get why, when all of the elements of S4 aren't cycles, when trying to find the subgroups (normal) the elements are cycles. Or am I looking at this in a ridiculous way?? If I am I do apologise, I can't seem to grasp the concept :frown:

Reply 6

Original post by adie_raz

I see, looking at my answer I actually did this for the cycles structure 1,1,2. However this is only because it is more obvious for a 2 cycle that if you shift right it stays the same. I did not think about it enough for the 3 and 4 cycles it seems. right now I know why we have 24 = 1+6+3+8+6, I then know:

|N| divides |S4|. Therefore we can have:

1+6+3+8+6,
6+6,
3+1,
3+8+1,
8,
1.

as the orders of N. Now how do I check which ones are normal subgroups and which ones aren't??

Thank you for your help :smile:

Since you're only considering subgroups containing entire conjugacy classes, you don't need to worry about the "normality" condition, that is:

\forall a \in K, \forall g \in G, \exists b \in K

such that

gag^{-1} = b

, where K is the subset in question.

But K has to be a subgroup in addition to satisfying the extra condition for normality, so you need to check the subgroup axioms for each of your subsets. I'd start with closure, I think that would be the quickest way to eliminate the non-normal subgroups.

Also are you aware that

S_4

and

\{e\}

are the trivial normal subgroups (where e is the identity)?

Original post by adie_raz

I ma really sorry, but can you explain this a bit more clearly, as I do not get why, when all of the elements of S4 aren't cycles, when trying to find the subgroups (normal) the elements are cycles. Or am I looking at this in a ridiculous way?? If I am I do apologise, I can't seem to grasp the concept :frown:

Not all elements of

S_4

are cycles. All elements of

S_4

are permutations, and every permutation (in any symmetric group) can be expressed as a product of disjoint cycles. This could be just one cycle, e.g. (1 2 3 4) or a product of two 2-cycles, e.g. (1 2)(3 4), or a number of other things.
A permutation by definition takes the numbers 1,2,3,4 and permutes them in some way. There are 24 different ways to permute those numbers. As I said before, (2 3 4) permutes the numbers in exactly the same way (4 2 3) does, so they are the same element, both in

S_4

and in any subgroups. By convention you write the numbers of the cycle in increasing order. Also note that (2 3 4) is not a cycle, it is a product of a 1-cycle and a 3-cycle.

(edited 13 years ago)

Reply 7

OP

Original post by Dragon

Since you're only considering subgroups containing entire conjugacy classes

Ok, I think I understand all you have said so far, except, how did you know I am only considering subgroups containing entire conjugacy classes??

Thank you so much :smile:

Reply 8

Because of this:

Original post by adie_raz

|N| divides |S4|. Therefore we can have:

1+6+3+8+6,
6+6,
3+1,
3+8+1,
8,
1.

Reply 9

OP

Original post by Dragon

Because of this:

Ahh, so I have already disregarded all of those subgroups which aren't normal by finding those which have an order which is a divisor of the order of S4.

Edit:

What are those sets with the correct cycle structure?? are they subsets or conjugacy classes?? All I know is that they have 1,3,6,6,8 as the number of elements in them. How do I know what they are?? (Sorry to be such a pain, but I want to be as concise as possible so as to leave no room for error. Even in the way I think of them.).

(edited 13 years ago)

Reply 10

Original post by adie_raz

Ahh, so I have already disregarded all of those subgroups which aren't normal by finding those which have an order which is a divisor of the order of S4.

The order of any subgroup divides the order of the group. Not just normal ones.

\{e, (1 \; 2)\}

is a subgroup of

S_4

but not a normal subgroup since (1 2) is conjugate to, say, (1 3), which is not in the subgroup. Hence if a subgroup has an element from a conjugacy class, it must have every other element from that conjugacy classes to be a normal subgroup. But that is not a sufficient condition for it to be a normal subgroup.

Original post by adie_raz

What are those sets with the correct cycle structure?? are they subsets or conjugacy classes?? All I know is that they have 1,3,6,6,8 as the number of elements in them. How do I know what they are?? (Sorry to be such a pain, but I want to be as concise as possible so as to leave no room for error. Even in the way I think of them.).

There is a theorem that states that two permutations are conjugate if and only if they have the same cycle structure. So S4 has 5 conjugacy classes: One of order 1, one of order 3, two of order 6 and one of order 8.

Reply 11

OP

Ok, so I am finding the conjugacy classes of S4.The normal subgroups of S4 must either be an entire conjugacy class of S4 or a combination of entire conjugacy classes.

Also for them to be subgroups (not necessarily normal) they must satisfy the condition |H| divides |G|, where G is S4 and therefore |G| = 24.

As the divisors of 24 are: 1,2,3,4,6,8,12,24, the order of the conjugacy classes I have found (or a combination of them) must add to one of these divisors.

Therefore my options of combinations of entire conjugacy classes are:

1, 3, 6, 8, 6+6, 1+3,1+3+8 or 1+3+6+6+8.

Now that I have found these, I just need to show which of these 7 combinations of entire conjugacy classes are subgroups of S4.

Is all I have said correct??

Original post by Dragon

Hence if a subgroup has an element from a conjugacy class, it must have every other element from that conjugacy classes to be a normal subgroup. But that is not a sufficient condition for it to be a normal subgroup.

With regard this statement how can I "not worry" about "normality" as you stated before??

(edited 13 years ago)

Reply 12

OP

Just continuing, now that I know to look at:

1,3,6,8,6+6,1+3,1+3+8 and 1+3+6+6+8 I show they are subgroups.

Well, 1 is the identity {e}, and 1+3+6+6+8 is the group itself {S4} so these are normal subgroups by triviality (as you stated above).

3, 6, 6 and 8 don't have identities (and therefore aren't groups) so 3, 6, 6+6 and 8 can't be subgroups.

Now we have 1+3 and 1+3+8 to look into.

In 1+3, each element is its own inverse. Also

h_1h_2 \epsilon H, \forall h_1, h_2 \epsilon H

Therefore 1+3 is a subgroup, namely:

\{ e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}

. (How do I know this is normal??)

Now I just need to check 1+3+8. I know the index of 1+3+8 is 2, however how does this show it is a normal subgroup. Surely I need to still check it is a subgroup, then the index shows me it is a normal subgroup...

Reply 13

Original post by adie_raz

Ok, so I am finding the conjugacy classes of S4.The normal subgroups of S4 must either be an entire conjugacy class of S4 or a combination of entire conjugacy classes.

Also for them to be subgroups (not necessarily normal) they must satisfy the condition |H| divides |G|, where G is S4 and therefore |G| = 24.

As the divisors of 24 are: 1,2,3,4,6,8,12,24, the order of the conjugacy classes I have found (or a combination of them) must add to one of these divisors.

Therefore my options of combinations of entire conjugacy classes are:

1, 3, 6, 8, 6+6, 1+3,1+3+8 or 1+3+6+6+8.

Now that I have found these, I just need to show which of these 7 combinations of entire conjugacy classes are subgroups of S4.

Is all I have said correct??

Yep that's all correct. I'm glad I explained it well enough for you to understand :smile:

Original post by adie_raz

With regard this statement how can I "not worry" about "normality" as you stated before??

Such subgroups will satisfy the "normality condition", but it won't necessarily be a subgroup. It may not be closed or it may not have an identity or not all the elements will have an inverse in the subgroup. Remember a normal subgroup has to be a subgroup as well as satisfying the normality condition.

Original post by adie_raz

Just continuing, now that I know to look at:

1,3,6,8,6+6,1+3,1+3+8 and 1+3+6+6+8 I show they are subgroups.

Well, 1 is the identity {e}, and 1+3+6+6+8 is the group itself {S4} so these are normal subgroups by triviality (as you stated above).

3, 6, 6 and 8 don't have identities (and therefore aren't groups) so 3, 6, 6+6 and 8 can't be subgroups.

That is all correct.

Original post by adie_raz

Now we have 1+3 and 1+3+8 to look into.

In 1+3, each element is its own inverse. Also

h_1h_2 \epsilon H, \forall h_1, h_2 \epsilon H

Therefore 1+3 is a subgroup, namely:

\{ e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}

. (How do I know this is normal??)

It is normal because it is a union of conjugacy classes, hence it satisfies

gKg^{-1} \subseteq K \; \forall g \in G

Same for 1+3+8 (which is the alternating group

A_4

)

(edited 13 years ago)

Reply 14

OP

Original post by Dragon

Such subgroups will satisfy the "normality condition", but it won't necessarily be a subgroup.

Does this make them anything in particular, eg a normal conjugacy class, or normal subset, or does this not make them anything until you have proved they are subgroups, in which case they then become normal subgroups??

Reply 15