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Group Theory Question involving nonabelian simple groups and cyclic groups

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paddyoneil
#1
Sep16-10, 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 <=> 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.
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Deveno
#2
Mar30-11, 09:10 PM
Sci Advisor
P: 906
Quote Quote by paddyoneil View Post
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.
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.


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