Classification of Groups of Order 12.

[quantumdude]

Homework Statement

Classify the groups of order 12.

Homework Equations

None.

The Attempt at a Solution

The professor has worked this out up to a point. He proved a corollary that states:

"Let G be a group of order 12 whose 3-Sylow subgroups are not normal. Then G is isomorphic to A_4."

After the proof he states:

"Thus, the classification of groups of order 12 depends only on classifying the split extensions of Z_3 by groups of order 4."

OK, fine. So I know that split extensions are semidirect products, and that there are only 2 groups of order 4. So I need to compute the following:

D_4 \times_{\alpha} \mathbb{Z}_3
\mathbb{Z}_4 \times_{\beta} \mathbb{Z}_3

(sorry, don't know how to make the symbol for semidirect products)

Here's where the confusion begins. If I compare the semidirect products above with the definition of the same, then I see that I have to find the homomorphisms \alpha: \mathbb{Z}_3 \rightarrow Aut(D_4) and \beta: \mathbb{Z}_3 \rightarrow Aut(\mathbb{Z}_4).

The second one isn't so bad, but I would really like to turn the first one around so that the homomorphism comes out of D_4. That's because I've already done a homework exercise that gives me all of the homomorphisms out of D_{2n}.

So, first question: Is D_4 \times_{\alpha} \mathbb{Z}_3 for some \alpha isomorphic to \mathbb{Z}_3 \times_{\gamma} D_4 for some \gamma? In other words, can I arrange it so that I'm looking for homomorphisms from D_4 to Aut(\mathbb{Z}_3)?

Hope the question is clear.

 

[StatusX]

In general, no. In a non-trivial semi-direct product one of the groups in the product is normal and the other isn't (where we think of the groups making up the product as embedded in semi-direct product in the natural way). So it isn't symmetric. But Aut(D_4) is a pretty simple group, and the homomorphisms from Z_3 into it are really easy to classify (ie, find the automorphisms of order dividing 3).

 

[quantumdude]

In general, no.

That's what I suspected.

In a non-trivial semi-direct product one of the groups in the product is normal and the other isn't (where we think of the groups making up the product as embedded in semi-direct product in the natural way). So it isn't symmetric. But Aut(D_4) is a pretty simple group, and the homomorphisms from Z_3 into it are really easy to classify (ie, find the automorphisms of order dividing 3).

OK thanks, I'll try it.

Here's a stupid question. The definition of semidirect product of H and K with respect to \alpha is as follows.

"Let \alpha: K \rightarrow Aut(H) be a homomorphism. By the semidirect product of H and K with respect to \alpha, written H \times_{\alpha} K, we mean the set H \times K with the binary operation given by setting

(h_1,k_1) \cdot (h_2,k_2) = (h_1 \cdot \alpha(k_1)(h_2),k_1k_2)"

I'm a little unsure of what the right hand side of that last equation means. Since here H=D_4 and K=\mathbb{Z}_3, I suppose that in the first coordinate of the ordered pair I'll be multiplying in D_4, and in the second coordinate "k_1k_2" means "addition of k_1 and k_2 mod 3". Is that right? Also, I am supposing that the object "\alpha(k_1)(h_2)" is to be read as the product of \alpha(k_1) and h_2 in D_4. Is that also right?

 

[StatusX]

\alpha is a homomorphism from K to Aut(H), so \alpha(k_1) is an automorphism, and \alpha(k_1)(h_2) is the element in H that this automorphism sends h_2 to.