Results 1 to 2 of 2

Thread: Abstract Algebra: Suppose that G is a group with |G|=p3. Prove that |Z(G)| != p2.

  1. February 16th 2011, 03:38 PM #1
    joestevens
    joestevens is offline
    Junior Member
    Joined
    Sep 2010
    Posts
    42

    Question Abstract Algebra: Suppose that G is a group with |G|=p3. Prove that |Z(G)| != p2.

    Suppose that G is a group with |G|=p^3 where p is prime. Prove that |Z(G)|\ne p^2.
    Follow Math Help Forum on Facebook and Google+
  2. February 16th 2011, 03:58 PM #2
    Tinyboss
    Tinyboss is offline
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    If so, then |G/Z(G)|=p, and so G/Z(G) is cyclic. But G/Z(G) cyclic implies G is abelian, a contradiction.
    Follow Math Help Forum on Facebook and Google+
« Reduce the matrix [A|I] | If G is a simple group, then is the group algebra KG simple??(K is a field) »

Similar Math Help Forum Discussions

  1. [SOLVED] Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: March 2nd 2011, 12:36 AM
  2. [SOLVED] Need help getting started.. Abstract Algebra group problem
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 30th 2010, 10:51 AM
  3. abstract algebra (group)
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 10th 2010, 10:07 AM
  4. group theory abstract algebra
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: October 26th 2009, 04:54 PM
  5. Abstract Algebra- order of a group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: July 1st 2009, 06:23 PM

Search Tags


/mathhelpforum @mathhelpforum