Register to reply 
Group Theory Question involving nonabelian simple groups and cyclic groups 
Share this thread: 
#1
Sep1610, 08:21 PM

P: 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 <=> C_{G}(G) = A = {a in G: ag = ga for all g in G} My professor's hint was "what is G/C_{G}(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
Mar3011, 09:10 PM

Sci Advisor
P: 906

now if A is cyclic, show that C_{G}(A) is normal in G. now every element of A certainly commutes with every other (A is abelian). thus C_{G}(A) is a normal sbgroup of A, containing A, so C_{G}(A) = G. now show that this implies A = Z(G) (containment of A in Z(G) is easysee above). use the fact that G/A is nonabelian to show that if g is not in A, g does not commute with some member of G. 


Register to reply 
Related Discussions  
Groups of permutations and cyclic groups  Linear & Abstract Algebra  5  
Help clarifying a question regarding (i think) cyclic groups  Calculus & Beyond Homework  1  
Questions involving simple groups  Calculus & Beyond Homework  9  
Simple proof involving groups  Calculus & Beyond Homework  3  
Group homomorphisms between cyclic groups  Linear & Abstract Algebra  2 