homomorphism from $C_6$ to $A_3$ Is it possible to construct a nontrivial homomorphism from $C_6$ to $A_3$? I have tried to construct one but failed. Is there a good way to see when there will be a homomorphism?
 A: Obviously you can't construct an isomorphism $C_6 \to A_3$, since $C_6$ has order $6$ and $A_3$ has order $3$. So you'll need to construct a homomorphism $\theta : C_6 \to A_3$ which is not injective. 
A useful fact to note is that $A_3 \cong C_3$, so write $C_6 = \{e, a, \dots, a^5 \}$ and $A_3 = \{ e, b, b^2 \}$.
A good way to do this is to consider the orders of the elements of the two groups you're working with. In $C_6$, you have an element of order $1$ (the identity), one element of order $2$, two elements of order $3$ and two elements of order $6$. In $A_3$ you have one element of order $1$ and two elements of order $2$.
Since $C_6$ is generated by either of its elements of order $6$, namely $a$ or $a^5$, it makes sense to choose a sensible element of $A_3$ to map such an element (say $a$) to so that the homomorphism is nontrivial. Then since any homomorphism $\theta$ satisfies $\theta(x^n) = \theta(x)^n$, this must determine the images of the rest of the elements of your group.
In this case, any non-identity choice of $\theta(a)$ will work; but this philosophy applies more generally when constructing homomorphisms.
A: I have the following recipe in mind:
Proposition:
The number of homomorphisms from $\mathbb Z_m$ to $\mathbb Z_n$ is $(m,n)$.
Proof. 
Observe that a homomorphism from a cyclic group is fixed by fixing the image of a generator, $1$. 
$$\varphi(1)=a \implies \varphi (2)=\varphi(1)+\varphi(1)=a+a~~~\text{and so on} \cdots$$
And, the order of the element divides the order of the image of that element. These facts when cooked appropriately should prove the fact. $\blacksquare$
Explicit Maps:
Set $(m,n)=d$.
$$\phi: \mathbb Z_m \to \mathbb Z_n$$ Every map defined by  $$\phi([x]_m)=[k\frac{n}{d}x]_n~~~~\text{$k=0,1,2,3,\cdots d-1$}$$ is a homomorphism and conversely, any homomorphism is of this form.

Note that $A_3 \cong C_3$. This means that the number of homomorphisms is $GCD(6,3)=3$.

Another way of looking at this problem will be, to consider the universal property of the quotients. This is a very useful thing and is a often recurring theme in Algebra.
$$\begin{array}{ccccccccc}
\mathbb Z \\

\downarrow & \searrow \\

\mathbb Z_n & \xrightarrow{\varphi} & \mathbb Z_m
\end{array}$$
Call the map from $\mathbb Z$ to $\mathbb Z_m$ by $\Pi_m$. In keeping with the standard language, $\Pi_n$ factors through $\operatorname{Ker} \Pi_m=m\mathbb Z$ and $\varphi$ is the induced homomorphism on $\mathbb Z_m$.
Let's recall that, generally, when our maps depended on coset representatives, we require the fact that the map didn't change when the representative of the coset changes. (i.e.) maps must be well defined. We can prove in a more gneral setup, this is always the case, when $\operatorname{Ker} \Pi_m \subseteq \operatorname{Ker} \Pi_n$. In this case, it would just mean, $m \mathbb Z \subseteq n\mathbb Z \implies n|m$   
This is a useful criteria for figuring out if a non-trivial homomorphism exists. 
One Should note that the set of all homomorphisms between two groups is never empty, thanks to the identity map that always exists. 
Further, between finite groups, it helps to think of the first isomorphism theorem. $$\dfrac{|G|}{|\operatorname{Ker} \theta|}=|Im \theta|$$
How does this help? This helps when you consider the fact that $Im \theta$ is a subgroup of the co-domain and by asking for those cardinalities of the $\operatorname{Ker} \theta$ that permit such values. 
One common mistake I have seen people make is the assumption a homomorphism is onto! No, first isomorphism theorem says if the map is onto, then going modulo the kernel, you have the isomorphic copy of the range! 
I hope this helps!
