1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Group Theory Question involving nonabelian simple groups and cyclic groups

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A be a normal subgroup of a group G, with A cyclic and G/A nonabelian simple. Prove that Z(G)= A


    2. Relevant equations

    Z(G) = A <=> CG(G) = A = {a in G: ag = ga for all g in G}

    My professor's hint was "what is G/CG(A)?"

    3. The attempt at a solution
    A is cyclic => A is abelian
    A normal in G <=> gAg-1 = A
    So gA=Ag. Then gA is an element of G/A.

    I don't really know where to go. I have been working on this for several hours and am at a loss. Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 30, 2011 #2

    Deveno

    User Avatar
    Science Advisor

    the fact that G/A is simple means that there are no normal subgroups of G containing A except G itself.

    now if A is cyclic, show that CG(A) is normal in G. now every element of A certainly commutes with every other (A is abelian). thus CG(A) is a normal sbgroup of A, containing A, so CG(A) = G.

    now show that this implies A = Z(G) (containment of A in Z(G) is easy-see above). use the fact that G/A is non-abelian to show that if g is not in A, g does not commute with some member of G.
     
    Last edited: Mar 30, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Group Theory Question involving nonabelian simple groups and cyclic groups
Loading...