Results 1 to 3 of 3

Math Help - finite group (prove)

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    43

    finite group (prove)

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

    I would really appreciate your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2010
    Posts
    43

    Re: finite group (prove)

    In other words we have to prove that O_\pi (G) and O^\pi (G) are well-defined, where O^\pi (G) denotes the largest normal pi-subgroup of G and 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!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Re: finite group (prove)

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

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

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

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: October 13th 2011, 12:50 PM
  2. Prove that the finite cyclic group is isomorphic
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 19th 2011, 02:20 AM
  3. Prove that the finite cyclic group is isomorphic
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 18th 2011, 11:30 AM
  4. Replies: 4
    Last Post: April 13th 2010, 07:09 PM
  5. Replies: 1
    Last Post: November 5th 2009, 06:14 PM

Search Tags


/mathhelpforum @mathhelpforum