Originally Posted by

**WannaBe** 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:

1a. a has 2 different conjucates in G and o(N(a))=6 or o(N(a))=12 (N(a) is the normalizer of a)

1b. there is a b in G of order 2 who is commutative with a: ab=ba

1c. if o(N(a))=6 then N(a) is a group of order 6 that is generated by ab ( <ab>=N(a) )

1d. Prove that b is in the center of the group G (C(G)) . There's a clue: notice that b is in

N(a).

I don't under stand how to prove this one:

2. prove that if 2 doesn't divide o(C(G)) then G is isomorphic to A4 (A4 is the group of all even permutations)...

We know that G has no normal sub groups of order 3... We know that G is isomorphic to a subgroup of S4...But how can we know that the image is A4?

Help is needed!

TNX