1. ## homomorphism from A4

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 ?

2. Let's see some work for the first one.

Originally Posted by xmcestmoi

(2) there is a homomorphism from A4 onto a group of order 3.

Can anyone please give me some hints ?
How much do you know about $A_4$? A (moderately) common fact is that $A_4$ has only one subgroup of order four (namely $\{e, (12)(34), (13)(24), (14)(23)\}=N$). But, it is fairly easy to prove that since $N$ is the only subgroup of order four and conjugation by any element of that subgroup (i.e. $gNg^{-1}$) is another subgroup of order four that $N$ must be invariant under conjugation. Thus, $N\unlhd A_4$. So, the canonical homomorphism is $\theta:G\to G/N$ given by $g\mapsto gN$. This is a surjective homomorphism and $|G/N|=[G:N]=\frac{12}{4}=3$

3. Thank you !!

4. Now I'm curious. How do you know there's no homorphism onto groups with order 4 or 6?

Now I'm curious. How do you know there's no homorphism onto groups with order 4 or 6?
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.

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# no.of homomorphism between A4 to A4

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