1. ## Isomorphism Theorems

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

2. 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

I'm confused: at the beginning it is given that "Let's assume that G has a normal sub-group of order 3 and let a be her generator ( <a>=G ).", but this is absurd: I think they meant a is the generator of the normal sbgp. of order 3, not of G.
Also, in (2), whatever the conditions are, it can't be $\displaystyle G\cong A_4$ , since $\displaystyle A_4$ has no normal sbgp. of order 3, as given at the beginning...check this and write back with al the conditions clearluy stated.

Tonio

3. 1 and 2 have different assumptions!!!
The only thing we know in 2 is that 2 doesn't divide o(C(G)) and that o(C(G))=12 ...That's it! We use 1 only to get a contradiction when we assume G has a normal sbgrp of order 3... !!!

4. Originally Posted by WannaBe
1 and 2 have different assumptions!!!
The only thing we know in 2 is that 2 doesn't divide o(C(G)) and that o(C(G))=12 ...That's it! We use 1 only to get a contradiction when we assume G has a normal sbgrp of order 3... !!!

Ok, that could have been said at the beginning... , but still: are we seeing $\displaystyle G$ as embedded in some bigger group, or else what is $\displaystyle C(G)$?? Or is this a (non-standard) way to symbolize the center of the group G? Usually, $\displaystyle C(G)$ means the centralizer of the group G inside a bigger group...

Tonio

5. It's indeed the center of the group G...

6. Originally Posted by WannaBe
It's indeed the center of the group G...

Ok, but then I'm still a little baffled...let's see what we've here: G is a group of order 12 s.t. $\displaystyle 2 \nmid Z(G)$...and that's all we have, right? But then you added:
"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?"

How do "we know" G has no normal sbgp. of order 3 if we haven't received more information?? There are groups of order 12 with normal sbgps. of order 3 (all the abelian ones and one of the non-abelian ones), and if we already know then we're almost done, since $\displaystyle 12 = 2^2\cdot 3$, and a simple counting argument shows that either the Sylow 2-sbgp. or the Sylow 3-sbgp. MUST be normal, so if it isn't the 3-sbgp. then it must be the 2-sbgp. of order 4 ==> $\displaystyle G=A_4$ as it is the only non-abelian group of order 12 with normal sbgp. of order 4...

Tonio

7. I'll explain:
G has order 12... By the first part, we know that if there's a normal sub-group of order 3, there's an element in Z(G) of order 2, and that's a contradiction to lagrange's theorem... So we now know there's no normal sbgrps of order 3... that means that all the sbgrps are from order different from 3... We can defind a homo. that makes use of these facts...