I want to find the normal subgroups of and have looked at a previous thread on here:
http://www.thestudentroom.co.uk/showthread.php?t=884055
However I don't understand some of it and therefore still need some help.
What I have done so far is look for group with the same cycle structure as that of . Therefore you need cycle structures of:
1,1,1,1
1,1,2
2,2
1,3
4.
I then looked for the number of elements in the groups of this cycle structure. This is where I came across a problem first.
For cycle structure 1,1,1,1. There is of course only 1 element.
For cycle structure 1,1,2. There are 6 elements. found these by writing them down but is can been seen easily by intuition.
For cycle structure 2,2. There are of course 3 elements. Again intuition/ writing them down.
For cycle structure 1,3. This is where I get stuck as I seem to have found 24 elements, however in my lecture notes there are only supposed to be 8. This leads me to believe that some are the same and therefore are not supposed to be counted more than once. However I do not know how to show this.
The elements I have are:
(1)(2,3,4), (1)(2,4,3), (1)(3,2,4), (1)(3,4,2), (1)(4,2,3), (1)(4,3,2), (2)(1,3,4), (2)(1,4,3) etc, giving 6x4 (basically kind of like but with 4 different numbers in front of it). Hence 24 elements.
For cycle structure 4. There are supposed to be 6 elements, however this again confuses me as I would have said there would be 24 again due to just writing out . e.g element, however apparently there are and I don't get why you must divide by 4.
Therefore I am supposed to have 24 = 1+6+3+8+6, (which clearly works) and this (according to the other thread) comes from the CLASS EQUATION. However I have looked this up and don't fully understand that either and therefore can't use it.
I think I need running through the whole process basically, as I do not understand it and have honestly looked at several sources before coming here. Also for me it is the Easter break so I can't visit my tutors. (Note I have emailed them, and am still waiting on a response).
Rep will be given for this as I know it is a big ask. Also if it is easier, reply in the other thread and I shall watch it so as to have certain things written out already if you prefer.
Thank you so much in advance.

adie_raz
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 27032011 00:17

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 27032011 03:02
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 rightmost number to the leftmost position, you will get the same cycle. The same applies for a leftshift.
It appears you are making the same mistake with the 4cycles 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 14 in an arbitrary order, there are three equivalent cycles due to the left or rightshifting I described above, which explains the .
Another way of thinking about the (1,3) cycle type is as follows. There is one fixed number and a 3cycle. In there are only four numbers you can fix. For the remaining three numbers, regardless of what they are, there are only two possible 3cycles. 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 . The only two distinct possibilities are (2 3 4) and (2 4 3).Last edited by Dragon; 27032011 at 03:09. 
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 27032011 12:22
(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 rightmost number to the leftmost position, you will get the same cycle. The same applies for a leftshift.
It appears you are making the same mistake with the 4cycles 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 14 in an arbitrary order, there are three equivalent cycles due to the left or rightshifting I described above, which explains the .
Another way of thinking about the (1,3) cycle type is as follows. There is one fixed number and a 3cycle. In there are only four numbers you can fix. For the remaining three numbers, regardless of what they are, there are only two possible 3cycles. 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 . The only two distinct possibilities are (2 3 4) and (2 4 3).
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 
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 27032011 13:12
(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).
By this I mean, in (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. 
DFranklin
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 27032011 13:52
In simple terms: cycles are not the same as elements.

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 27032011 14:06
(Original post by DFranklin)
In simple terms: cycles are not the same as elements. 
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 27032011 14:31
(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
such that , 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 nonnormal subgroups.
Also are you aware that and 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
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 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 1cycle and a 3cycle.Last edited by Dragon; 27032011 at 14:40. 
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 27032011 17:00
(Original post by Dragon)
Since you're only considering subgroups containing entire conjugacy classes
Thank you so much 
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 27032011 17:04
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. 
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 27032011 18:34
(Original post by Dragon)
Because of this:
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.).Last edited by adie_raz; 27032011 at 18:43. 
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 27032011 19:04
(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.
is a subgroup of 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.). 
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 27032011 19:39
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.Last edited by adie_raz; 27032011 at 20:21. 
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 27032011 20:38
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
Therefore 1+3 is a subgroup, namely: . (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... 
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 27032011 20:55
(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??
(Original post by adie_raz)
With regard this statement how can I "not worry" about "normality" as you stated before??
(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.
(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
Therefore 1+3 is a subgroup, namely: . (How do I know this is normal??)
Same for 1+3+8 (which is the alternating group )Last edited by Dragon; 27032011 at 21:06. 
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 27032011 21:00
(Original post by Dragon)
Such subgroups will satisfy the "normality condition", but it won't necessarily be a subgroup. 
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 27032011 21:08
(Original post by adie_raz)
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??
Also a subgroup of index 2 is always normal.Last edited by Dragon; 27032011 at 21:10.
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