Show that
(1) there is no homomorphism from A4 onto a group of order 2, 4, or 6
(2) there is a homomorphism from A4 onto a group of order 3.
Can anyone please give me some hints ?
Let's see some work for the first one.
How much do you know about ? A (moderately) common fact is that has only one subgroup of order four (namely ). But, it is fairly easy to prove that since is the only subgroup of order four and conjugation by any element of that subgroup (i.e. ) is another subgroup of order four that must be invariant under conjugation. Thus, . So, the canonical homomorphism is given by . This is a surjective homomorphism and
The kernel of group homomorphism is a normal subgroup. As the previous reply said, there is only one normal subgroup of order 4 in A_4. If there were group homomorphisms onto groups with order 4 and 6, by the first isomorphism theorem, there should, at least, exist normal subgroups of order 3 and 2 in A_4. However, this is not the case here.