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Math Help - induction on p-groups

  1. #1
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    induction on p-groups

    Let p be a prime and let G be a group of order p^\alpha. Prove that G has a subgroup of order p^\beta, for every \beta with 0<=\beta<=\alpha.

    Hint: induction on \alpha and use the theorem: If p is a prime and G is a group of prime power order p^\alpha for some \alpha>=1, then G has a nontrivial center.
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    Quote Originally Posted by dori1123 View Post
    Let p be a prime and let G be a group of order p^\alpha. Prove that G has a subgroup of order p^\beta, for every \beta with 0<=\beta<=\alpha.

    Hint: induction on \alpha and use the theorem: If p is a prime and G is a group of prime power order p^\alpha for some \alpha>=1, then G has a nontrivial center.
    if \alpha = 0 \ \text{or} \ 1, then there's nothing to prove. so let \alpha > 1 and suppose that the claim is true for all groups of order p^{\beta} with 0 \leq \beta < \alpha. let G be a group of order p^{\alpha} and let 0 \leq \gamma \leq \alpha. we want

    to show that G has a subgroup of order p^{\gamma}. it's trivial for \gamma = 0. so we'll assume that 1 \leq \gamma \leq \alpha. we have to consider two cases:

    Case 1: G is abelian. in this case choose x \in G with o(x)=p. since G is abelian, < x > is normal in G. now G/< x > is a group of order p^{\alpha - 1}. so by induction hypothesis it has a subgroup H/<x>

    of order p^{\gamma -1}. therefore: |H|=p^{\gamma} and we're done in this case.

    Case 2: G is non-abelian. let Z(G) be the center of G and |Z(G)|=p^{\delta}. then: 1 \leq \delta < \alpha. if \gamma \leq \delta, then by induction hypothesis Z(G) has a subgroup of order p^{\gamma} and we're done. if \gamma > \delta, then

    consider G/Z(G), which is a group of order p^{\alpha - \delta}. by induction G/Z(G) has a subgroup K/Z(G) of order p^{\gamma - \delta}. hence: |K|=p^{\gamma}. \ \ \Box
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