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: