Proving No Homomorphism from A(4) Onto Groups of Order 2, 4, or 6

[wakko101]

Homework Statement

Show that there is no homomorphism from A(4) onto a group of order 2, 4 or 6, but that there is one onto a group of order 3.

Homework Equations

I have a feeling that I'm to use the First Isomorphism theorem

The Attempt at a Solution

I've been puzzling over this for quite some time, but I'm not really sure how to go about it. It seems that it might have something to do with the fact that the first group all have orders divisible by 2 whereas the latter has odd order, but I'm not even sure that's the correct approach. Any hints or suggestions would be appreciated.

Cheers,
W. =)

 

[Office_Shredder]

The question is heavily involved in the orders of the groups obviously. First observe the question has to be talking about non-trivial homomorphisms (since trivial homomorphisms always exist).

Start with mapping to the group of order 2. Since the homomorphism isn't trivial, it must be onto. So what can you conclude?

 

[wakko101]

If it is onto, we can use the first isomorphism th. to get the order of the kernel which would be 12. I know that a kernel must be a normal subgroup of A(4) and I've figured out (i.e. read somewhere) that the only normal subgroup of A(4) has order 3. So, using that, I can see how we might prove the hypothesis. Is this the only way...I mean, is knowing that A(4) has only one normal subgroup the only way to prove it?