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    Group Theory Question involving nonabelian simple groups and cyclic groups

    Homework Statement Let A be a normal subgroup of a group G, with A cyclic and G/A nonabelian simple. Prove that Z(G)= AHomework 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)?" The Attempt at a Solution A is cyclic => A is...
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