Can there exist a nontrivial homomorphism from an abelian group to a non-abelian group? Just some notations I make to shorten the text:
$(AG,+)$ denotes an abelian group, $(NG,\cdot)$ denotes a non-abelian group
I've seen examples of homomorphisms from a non-abelian group to an abelian group (and I can see why they are possible: In some particular cases one can find some $f:NG\rightarrow AG$ with $\forall n_1,n_2 \in NG: f(n_1\cdot n_2) = f(n_1)+f(n_2) = f(n_2\cdot n_1)$, which is not injective, but does the job).
But what about homomorphisms from an abelian group to a non-abelian group? My intuition tells me they aren't possible and I came up with the following proof:
Suppose there exists $f:AG\rightarrow NG$ a homomorphism. Then, by definition: $\forall a_1,a_2 \in AG: f(a_1 + a_2) = f(a_1)\cdot f(a_2)$, but $a_1+a_2 = a_2 +a_1$, so we have $\forall a_1,a_2 \in AG: f(a_1 + a_2) = f(a_1)\cdot f(a_2)=f(a_2 + a_1) = f(a_2)\cdot f(a_1)$, or more concisely, $\forall a_1,a_2 \in AG: f(a_1)\cdot f(a_2)=f(a_2)\cdot f(a_1)$. Due to the lack of commutativity in $NG$, this last proposition can only be true if $f$ is the trivial map $f:AG\rightarrow NG,\forall a\in AG:f(a):=e_{NG}$, where $e_{NG}$ is the neutral element in $NG$.
Is this correct or am I missing something?
 A: You're making a mistake when you say

Due to the lack of commutativity in $NG$, this last proposition can only be true if $f$ is the trivial map $f:AG\rightarrow NG,\forall a\in AG:f(a):=e_{NG}$, where $e_{NG}$ is the neutral element in $NG$.

A group being non-abelian doesn't necessarily imply that every pair of two non-identity elements do not commute. It implies that there exists at least one pair of (non-identity) elements that do not commute.
The condition
$$\forall a_1,a_2 \in AG: f(a_1)\cdot f(a_2)=f(a_2)\cdot f(a_1)$$
means that the image of $f$ is abelian, so $NG$ has an abelian subgroup. It doesn't mean that all of $NG$ has to be abelian - unless $f$ is also assumed to be surjective, as remarked by @DavidA.Craven in the comments.
In fact, for every non-abelian group $NG$, there exists a non-trivial abelian group $AG$ and a non-trivial homomorphism $AG \to NG$. Let $g \in NG$ be any non-trivial element, and let $AG = \langle g \rangle$, the cyclic (hence abelian) subgroup of $NG$ generated by $g$. Then the inclusion map $AG \to NG$ is a non-trivial homomorphism.
A: The group of rotations in $\Bbb R^2$ is abelian, and the group of rotations in $\Bbb R^3$ is not. But there is an obvious homomorphism from the first to the second; its range is the subgroup of rotations about the $z$-axis (which is, of necessity, itself abelian).
As Stef points out in a comment, the first group can be regarded as a subgroup of the second, so this is called an inclusion homomorphism.
A: Let $G$ be an abelian group and $H$ be a nonabelian group.  Then, the mapping $f: G\rightarrow H$, where $f(g)=1$ for all $g\in G$ is a homomorphism
