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.