1. The problem statement, all variables and given/known data The problem is: Let G be a group of order 12 ( o(G)=12). Let's assume that G has a normal sub-group of order 3 and let a be her generator ( <a>=G ). In the previous parts of the questions i've proved that: 1. a has 2 different conjucates in G and o(N(a))=6 or o(N(a))=12 (N(a) is the normalizer of a) 2. there is a b in G of order 2 who is commutative with a: ab=ba 3. if o(N(a))=6 then N(a) is a group of order 6 that is generated by ab ( <ab>=N(a) ) I have no clue in the next 2 parts of the question: 4. Prove that b is in the center of the group G (C(G)) . There's a clue: notice that b is in N(a). 5. prove that if 2 doesn't divide o(C(G)) then G is isomorphic to A4 (A4 is the group of all even permutations)... 2. Relevant equations None.... 3. The attempt at a solution In question 5 , I can't realy understand how the two groups can be isomorphic if A4 has order 13 and G has order 12... There's a clue to question 5 too: you can make use of Cayle's theorem... I will be delighted if someone will help me...I have no clue in 4 and 5.... detailed answers will be received with happines... TNX!