Quote by paddyoneil
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.

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 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.