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Thread: induction on p-groups

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    induction on p-groups

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

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

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

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

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

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

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

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