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Math Help - Soluble Groups....

  1. #1
    AAM
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    Soluble Groups....

    Hi guys,

    How can you prove that S3, S4 & A4 are soluble, & that A5 & S5 are not?

    I know that every group of order less than 60 is soluble, but how would you proove it without using that fact in these cases? :-)

    I have absolutely no idea where to start! :-s (our lecturer did not explain this bit very well AT ALL! :-( )

    Any of the proofs, or even some explanation about what Soluble Groups actually mean & their significance, would be VERY MUCH appreciated! :-)

    Many, many thanks. x
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  2. #2
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    Quote Originally Posted by AAM View Post
    Hi guys,

    How can you prove that S3, S4 & A4 are soluble, & that A5 & S5 are not?

    I know that every group of order less than 60 is soluble, but how would you proove it without using that fact in these cases? :-)

    I have absolutely no idea where to start! :-s (our lecturer did not explain this bit very well AT ALL! :-( )

    Any of the proofs, or even some explanation about what Soluble Groups actually mean & their significance, would be VERY MUCH appreciated! :-)

    Many, many thanks. x
    A_3 and S_3/A_3 are both solvable because they are abelian. therefore S_3 is solvable. V, the klein-4 group, and A_4/V are both abelian and hence solvable. therefore A_4

    is solvable. also obviously S_4/A_4 is abelian and hence solvable. thus S_4 is solvable. A_5', the commutator subgroup of A_5, is a normal subgroup of A_5. so either A_5'=\{1\}

    or A_5'=A_5 because A_5 is simple. but A_5' \neq \{1 \} because A_5 is not abelian. thus A_5'=A_5 and hence A_5^{(k)}=A_5 \neq \{1 \}, for all k. so A_5 is not solvable which obviously

    implies that S_5 is not solvable because every subgroup of a solvable group is solvable.
    Last edited by NonCommAlg; April 5th 2009 at 01:19 PM. Reason: typo!
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  3. #3
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    Quote Originally Posted by AAM View Post
    Hi guys,

    How can you prove that S3, S4 & A4 are soluble, & that A5 & S5 are not?

    I know that every group of order less than 60 is soluble, but how would you proove it without using that fact in these cases? :-)

    I have absolutely no idea where to start! :-s (our lecturer did not explain this bit very well AT ALL! :-( )

    Any of the proofs, or even some explanation about what Soluble Groups actually mean & their significance, would be VERY MUCH appreciated! :-)

    Many, many thanks. x
    As NonCommAlg said you need to know that A_5 is a simple group - which means it has no proper non-trivial normal subgroup (so if N\triangleleft A_5 \implies N = \{ e \}\text{ or }N=A_5). If A_5 is simple then it certainly cannot be solvable. This is easy to see since you cannot write a subnormal series because you will not be able to form normal subgroups of A_5 other than the trivial and itself, hence  \{ e \}\triangleleft A_5 is the only subnormal series for A_5. Since the factor group A_5/\{e\}\simeq A_5 is not abelian it means the group cannot be sovable. This means S_5 (in fact S_n) cannot be solvable because A_5 is a subgroup, and a subgroup of a solvable group is solvable. Thus, all we need to show is that A_5 is simple.
    ---
    Here is one method with orbits and stablizers. We will derive the class equation for A_5. First, we will work with S_5 to help figure out the conjugacy classes for A_5. For S_5 there are 7 conjugacy classes. These consist of: 1-cycle, 2-cycles, 3-cycles, 4-cycles, 5-cycles, (3-cycles)(2-cycles),(2-cycles)(2-cycles). The number of 1-cycles is just 1. The number of 2-cycles is {5\choose 2} = 10. The number of 3-cycles is 2!\cdot {5\choose 3} = 20. The number of 4-cycles is 3!\cdot {5\choose 4} = 30 and the number of 5-cycles is 4! = 24. The number of (3-cycles)(2-cycles) is {5\choose 3}=20. The number of (2-cycles)(2-cycles) is 3{5\choose 4} = 15. These are the conjugacy classes for S_5 and the number of elements in each class. What we will be interested in is the class equation for A_5.

    Two elements that are conjugate in A_5 must be conjugate in S_5, obviously. However, elements that are conjugate in S_5 are not necessarily conjugate in A_5. Remember a result from group theory that the number of elements conjugate to x\in A_5 would be equal to the index of C(x) (centralizer of x in A_5) in A_5. The conjugacy classes for A_5 consist of: 5-cycles, 3-cycles, 1-cycle, (2-cycles)(2-cycles). We need to count the number of elements in each conjugacy class, we will use our work above to help answer this question. Let us start easy, 1-cycle, is just the identity, so there is just one element in that conjugacy class. Let x=(12)(34), the number of conjugates to this in S_5 is 15, but necessary in A_5. We will prove that it is 15 for A_5 as well. Let [(12)(34)]_{S_5} be the order of the conjugacy class of (12)(34) in S_5, therefore 15=[(12)(34)]_{S_5} = 120/|C_{S_5}(x)| \implies |C_{S_5}(x)|=8. We notice that (if you do the computation) \left< (1324),(13)(24)\right> \subseteq C_{S_5}(x), but since |\left<(1324),(13)(24)\right>| = 8 it means C_{S_5}(x) = \left<(1324),(13)(24)\right>. Thus, C_{A_5}(x) = A_5\cap C_{S_5}(x) = \{ e, (12)(34),(13)(24),(14)(23)\}. Thus, |C_{A_5}(x)| = 4 which means [(x)]_{A_5} = [A_5:C_{A_5}(x)] = 15 which means the conjugacy class of x=(12)(34) in A_5 is the entire conjugacy class for that element in S_5. Let y=(123) then (all these arguments are similar) |(123)|_{S_5} = 20 so |C_{S_5}(y)| = 6, it can now be shown C_{S_5}(y) = \left< (123),(45)\right> thus C_{A_5}(y) = A_5\cap C_{S_5}(y) = \left< (123)\right>. This means [y]_{A_5} = 20 and so the conjugacy class of (123) is also all of conjugates in S_5. Now we get to z=(12345) (the last remaining case to consider). Again (similar argument, steps excluded), C_{S_5}(z) = \left< (12345)\right> and so C_{A_5}(z) = \left< (12345)\right>. However, this means [(12345)]_{A_5} = 12. Thus, conjugacy class for (12345) in A_5 is not all of the ones in S_5 (which have 24 of them). Repeating the same argument with the remaining 12 5-cycles we see (by the same argument as just now) that they consist of 12 conjugate elements too. Thus, the conjugacy class equation for A_5 is 60 = 1 + 12 + 12 + 15 + 20 . It now follows that A_5 is simple since we cannot write a non-trivial proper divisor of 60 as a summand of those numbers.
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