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    I want to find the normal subgroups of  S_4 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  S_4 . 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  S_3 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  S_4 . e.g  4! element, however apparently there are  \frac{4!}{4} 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.
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    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).
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    (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
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    (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.
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    In simple terms: cycles are not the same as elements.
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    (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
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    (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
    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
    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.
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    (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
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    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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    (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.).
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    (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.
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    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??
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    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...
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    (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

    (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)
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    (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??
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    (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??
    No it doesn't make them anything special as far as I'm aware.

    Also a subgroup of index 2 is always normal.
 
 
 
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