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Math Help - Group Theory - Basic question on existence of sub-groups

  1. #1
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    Group Theory - Basic question on existence of sub-groups

    Let G be a group, \mid G \mid = p^n, where p is prime.
    Prove G has a sub-group for each of the order  p^\alpha, where \alpha \in [0,n]


    Can someone provide me hints/sketches to attempt this? Have been stuck for a long time now.

    Thanks
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  2. #2
    ynj
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    Quote Originally Posted by aman_cc View Post
    Let G be a group, \mid G \mid = p^n, where p is prime.
    Prove G has a sub-group for each of the order  p^\alpha, where \alpha \in [0,n]


    Can someone provide me hints/sketches to attempt this? Have been stuck for a long time now.

    Thanks
    use induction on n
    for any k<n, if p^k<=|Z(G)|, then Z(G) have a subgroup of order p^k since it is abelian. if p^k>|Z(G)|,then by induction,G/Z(G) must have a subgroup of order p^k/|Z(G)|,by the corresponding theorem, we can write it as H/Z(G), then H will be a subgroup of order p^k
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    Sorry but a little elaboration will help, not able to follow the argument. Maybe you can mention the relevant theorm for e.g. I am not able to follow why "if p^k<=|Z(G)|, then Z(G) have a subgroup of order p^k since it is abelian"
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    Quote Originally Posted by ynj View Post
    use induction on n
    for any k<n, if p^k<=|Z(G)|, then Z(G) have a subgroup of order p^k since it is abelian. if p^k>|Z(G)|,then by induction,G/Z(G) must have a subgroup of order p^k/|Z(G)|,by the corresponding theorem, we can write it as H/Z(G), then H will be a subgroup of order p^k
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  4. #4
    ynj
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    Quote Originally Posted by aman_cc View Post
    Sorry but a little elaboration will help, not able to follow the argument. Maybe you can mention the relevant theorm for e.g. I am not able to follow why "if p^k<=|Z(G)|, then Z(G) have a subgroup of order p^k since it is abelian"
    Thanks
    There is the theorem: |G|=n ,G is an abelian, m|n, then G has a subgroup of order m.
    Proof:
    G is isomorphic to H=Z(p1^k1)*...Z(pt^kt), where pi is prime.
    let n=p1^k1*p2^k2...*pt^kt,m=p1^l1....*pt^lt, where li<=ki; f be the isomorphism of H to G
    then the group generated by element f(p1^(k1-l1),0,0..0),f(0,p2^(k2-l2),0...0).....f(0..0,pt^(kt-lt))will have order p1^l1*...pt^lt=m
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  5. #5
    ynj
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    Actually, you can assert that for every p^k||G|,where p is a prime, then G has a subgroup of order p^k according to the First Sylow Theorem
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    Quote Originally Posted by aman_cc View Post
    Let G be a group, \mid G \mid = p^n, where p is prime.
    Prove G has a sub-group for each of the order  p^\alpha, where \alpha \in [0,n]


    Can someone provide me hints/sketches to attempt this? Have been stuck for a long time now.

    Thanks
    proof by induction over n: there's nothing to prove if n=0 (or \alpha = n). now consider two cases:

    1) G is abelian: by Cauchy's theorem G has an element x of oreder p. now apply induction to the abelian p group \frac{G}{<x>}.

    2) General case: let |Z(G)|=p^m, where 0 < m \leq n. if \alpha \leq m, we're done by 1). if \alpha > m, let G_1=\frac{G}{Z(G)}. we have |G_1|=p^{n-m} < p^n. so, by induction, G_1 has a subgroup H_1=\frac{H}{Z(G)}

    of order p^{\alpha-m}, which gives us |H|=p^{\alpha}. done again!
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    Thanks. I get the idea.

    Quote Originally Posted by NonCommAlg View Post
    proof by induction over n: there's nothing to prove if n=0 (or \alpha = n). now consider two cases:

    1) G is abelian: by Cauchy's theorem G has an element x of oreder p. now apply induction to the abelian p group \frac{G}{<x>}.

    2) General case: let |Z(G)|=p^m, where 0 < m \leq n. if \alpha \leq m, we're done by 1). if \alpha > m, let G_1=\frac{G}{Z(G)}. we have |G_1|=p^{n-m} < p^n. so, by induction, G_1 has a subgroup H_1=\frac{H}{Z(G)}

    of order p^{\alpha-m}, which gives us |H|=p^{\alpha}. done again!
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