The Group A_4 as a Split Extension.

[quantumdude]

Homework Statement

Let H be all the elements of the alternating group A_4 of exponent 2. Show that A_4 is a split extension of H by \mathbb{Z}_3.

Homework Equations

None.

The Attempt at a Solution

I have already shown that H is normal in A_4. I also have the proof of the following proposition:

Let H and K be groups whose orders are relatively prime and let f:G \rightarrow K be an extension of H by K. Then the extension splits iff G has a subgroup of order |K|.

Let H=H (as defined in the problem statement), let K=\mathbb{Z}_3, and let G=A_4. Then certainly the orders of H and K are relatively prime. Also, A_4 certainly has a subgroup of order 3. Just take the identity, a product of disjoint transpositions, and its inverse and you have a subgroup. So referring back to the definition of an extension, I've reduced the problem to finding a surjective homomorphism f: A_4 \rightarrow \mathbb{Z}_3 with \ker(f)=H.

Now I'm stuck. I know I need the homomorphism to map all 4 elements of H to 0 in \mathbb{Z}_3. But there doesn't seem to be enough variety in the remaining elements of A_4 for me to decide which of them should be mapped to 1 and which should be mapped to 2.

Little help?

 

[StatusX]

It doesn't matter, there's an automorphism of Z_3 taking 1 to 2. Remember the map is constant on cosets of H.

 

[quantumdude]

Good grief, I can't believe I didn't notice that.

Thank you.

 

[quantumdude]

I finished the problem. In case anyone's interested, here's the rest of the solution. Not being a mathematical genius, I had to write out the multiplication table for the elements of A_4 that are not in H. I noticed that there were two distinct equivalence classes of elements, that when multiplied by each other give an element of H and when multiplied by an element of the other class given an element of A_4 not in H. So I arbitrarily chose to send one class to 1 and the other to 2 (in \mathbb{Z}_3, that is).

Elements going to 1:
(1 2 3)
(1 4 2)
(1 3 4)
(2 4 3)

Elements going to 2:
(1 3 2)
(1 2 4)
(1 4 3)
(2 3 4)

Clearly, f defined in this way is a surjective homomorphism from A_4 to \mathbb{Z}_3 whose kernel is H. And now that I'm done, I see that I have an even slicker proof that H is normal in A_4, since it is the kernel of a homomorphism out of A_4. I rule.

 

[quantumdude]

I caught a mistake in the opening post.

Also, A_4 certainly has a subgroup of order 3. Just take the identity, a product of disjoint transpositions, and its inverse and you have a subgroup.

That doesn't give a subgroup at all. To get the desired subgroup of order 3 you have to take the identity, a 3 cycle, and its inverse.

OK, I think I'm finally done with this. :zzz: