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Thread: finite group (prove)

  1. #1
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    finite group (prove)

    Let G be a finite group, $\displaystyle \pi$ a set of primes, $\displaystyle \Omega$ the set of normal $\displaystyle \pi$-subgroups of G, and $\displaystyle \Gamma$ the set of normal subgroups X of G with G/X a $\displaystyle \pi$-group. Prove
    (1) If H, K $\displaystyle \in \Omega$ then HK $\displaystyle \in \Omega$. Hence $\displaystyle \langle \Omega \rangle$ is the unique maximal member of $\displaystyle \Omega$.
    (2) If H, K $\displaystyle \in \Gamma$ then $\displaystyle H \cap K \in \Gamma$. Hence $\displaystyle \bigcap_{H\in\Gamma} H$ is the unique minimal member of $\displaystyle \Gamma$

    I would really appreciate your help.
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  2. #2
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    Re: finite group (prove)

    In other words we have to prove that $\displaystyle O_\pi (G)$ and $\displaystyle O^\pi (G)$ are well-defined, where $\displaystyle O^\pi (G)$ denotes the largest normal pi-subgroup of G and $\displaystyle O_\pi (G)$ denotes the smallest normal subgroup H of G such that G/H is a pi-group.
    Thank you very much for your help in advance!
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  3. #3
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    Re: finite group (prove)

    Quote Originally Posted by doug View Post
    Let G be a finite group, $\displaystyle \pi$ a set of primes, $\displaystyle \Omega$ the set of normal $\displaystyle \pi$-subgroups of G, and $\displaystyle \Gamma$ the set of normal subgroups X of G with G/X a $\displaystyle \pi$-group. Prove
    (1) If H, K $\displaystyle \in \Omega$ then HK $\displaystyle \in \Omega$. Hence $\displaystyle \langle \Omega \rangle$ is the unique maximal member of $\displaystyle \Omega$.
    (2) If H, K $\displaystyle \in \Gamma$ then $\displaystyle H \cap K \in \Gamma$. Hence $\displaystyle \bigcap_{H\in\Gamma} H$ is the unique minimal member of $\displaystyle \Gamma$

    I would really appreciate your help.
    1) clearly $\displaystyle HK$ is a normal subgroup. now, since $\displaystyle |HK|=\frac{|H| \cdot |K|}{|H \cap K|},$ every prime divisor of $\displaystyle |HK|$ is either a prime divisor of $\displaystyle |H|$ or $\displaystyle |K|$. thus, since $\displaystyle H$ and $\displaystyle K$ are $\displaystyle \pi$-groups, $\displaystyle HK$ is a $\displaystyle \pi$-group too.

    2) define a group homomorphism $\displaystyle \varphi: G \longrightarrow G/H \times G/K$ by $\displaystyle \varphi(g)=(gH, gK).$ then $\displaystyle \ker \varphi =H \cap K$ and so $\displaystyle G/(H \cap K)$ is isomorphic to a subgroup of $\displaystyle G/H \times G/K.$ thus $\displaystyle |G/(H \cap K)|$ divides $\displaystyle |G/H| \cdot |G/K|$ and you're done.
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