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Math Help - solvable group

  1. #1
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    solvable group

    How i can show that group of order 250 is a solvable group ?

    |G| = 2*(5^3)

    But i count show that (5^3)/{e} is abelian group

    Thanks
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  2. #2
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    Quote Originally Posted by s.lateralus View Post
    How i can show that group of order 250 is a solvable group ?

    |G| = 2*(5^3)

    But i count show that (5^3)/{e} is abelian group

    Thanks
    P, the Sylow 5-subgroup of G, is normal because [G:P]=2 and it's solvable because it's nilpotent. also G/P is abelian and hence solvable. thus G is solvable.

    more generally and with the same argument we see that if p < q are primes, then every group of order pq^n, \ n \geq 0, is solvable.
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  3. #3
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    a more direct method: again let P be the Sylow 5-subgroup of G. we know that P contains subgroups P_1 \subset P_2 such that |P_1|=5, \ |P_2|=5^2. now look at this subnormal series:

    (1) \lhd P_1 \lhd P_2 \lhd P \lhd G.
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  4. #4
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    How do you know that 5^2 is normal in 5^3 ?
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  5. #5
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    Code:
    How do you know that 5^2 is normal in 5^3 ?
    Let  |P_1| = 5^2, |P_2| = 5^3 then [P_2:P_1] = 5 thus P_2/P_1 is abelian. A group of p-order is cyclic

    There's a little theorem that states that any sub-group with index p is normal
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  6. #6
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    Quote Originally Posted by Dinkydoe View Post
    Code:
    How do you know that 5^2 is normal in 5^3 ?
    Let  |P_1| = 5^2, |P_2| = 5^3 then [P_2:P_1] = 5 thus P_2/P_1 is abelian. A group of p-order is cyclic

    There's a little theorem that states that any sub-group with index p is normal

    Not exactly: the theorem states that if p is the least prime dividing the order of a finite group G, then any sbgp. of index p is normal in G.

    Tonio
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