Task:
Find all homomorphisms $A_4 \rightarrow \mathbb{Z}/6\mathbb{Z}$
$ A_4 = \{e; (12)(34); (13)(24); (14)(23); (123); (132); (124); (142); (134); (143); (234); (243) \} $
$\mathbb{Z}/6\mathbb{Z} = \{0,1,2,3,4,5,6\} $
My solution:
We take two elements of order 3 from $A_4$ and check the order of their multiplication: $$(1 2 3) (1 2 4) = (1 3) (2 4)$$ It means that from two elements of order 3, we can got an element of order 2
In $\mathbb{Z}/6\mathbb{Z}$ we cannot do this, in $\mathbb{Z}/6\mathbb{Z}$ we get either an element of order 0 or an element of order 3 from two elements of order 3, because we have only two elements of order 3: 2 and 4.
2 + 4 = 4 + 2 = 0 (order 0)
2 + 2 = 4 (order 3)
4 + 4 = 2 (order 3)
And with homomorphism, the order of the image must divide the order of the original element, so we cannot transfer an element of order 3 to an element of order 2. Therefore, we have only one trivial homomorphism
Am I right or not? thanks for all tips
$A_4=\langle(123),(124)\rangle$ and both $(123)$ and $(124)$ have order $3$; now let $f:A_4\to\Bbb{Z}_6$ be an homomorphism, we know that given a set of generators an homomorphism is uniquely determined by the images of the generators; since $o(f(g))|o(g)$ and elements in $\Bbb{Z}_6$ have either order $1, 2, 3, 6$ you can send $(123)$ and $(124)$ either in $[0]$, $[2]$ or $[4]$. Now consider that $(123)^3=(124)^3=1$ and $\left((123)(124)\right)^2=\left((13)(24)\right)^2=1$; therefore we must send $(123)$ in $[2]$ and $(124)$ in $[4]$ or vice versa, or both of them in $[0]$ so the possible homomorphisms are 3 in total.