Group Theory Question involving nonabelian simple groups and cyclic groups

[paddyoneil]

Homework Statement

Let A be a normal subgroup of a group G, with A cyclic and G/A nonabelian simple. Prove that Z(G)= A

Homework 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)?"

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.

 

[Deveno]

Homework Statement

Let A be a normal subgroup of a group G, with A cyclic and G/A nonabelian simple. Prove that Z(G)= A

Homework 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)?"

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 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.