Let $F\_n$ where $n \ge 3$ be a free group and let $(\mathcal A\_n(k))$ where $k \ge 1$ be the
kernel of the homomorphism $Aut(F\_n) \to Aut(F\_n/\gamma\_{k+1}(F))$
determined by the natural homomorphism $F\_n \to F\_n/\gamma\_{k+1}(F).$


($(\mathcal A\_n(k) : k \ge 1)$ is called the Johnson filtration of $Aut(F\_n);$


$\gamma\_k(G)$ denotes the $k$-th terms of the lower central series of a group $G,$
$\gamma\_1(G)$ being equal to $G$).


I do not know an example of a group homomorphism
$f : Aut(F\_n) \to G$ which
takes ***all*** terms of the Johnson filtration
$(\mathcal A\_n(k))$ to the same ***nontrivial***
subgroup:
$$
1 \ne f(\mathcal A\_n(1))=f(\mathcal A\_n(2)) = \ldots = f(\mathcal A\_n(k)) = \ldots
$$
I would be very grateful for such an example, or for an argument
that homomorphisms like that do not exist.



Can all terms of the Johnson filtration be hom-mapped onto the same nontrival group?
The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A\_q$, a *simple* one (see his comment above). Let $K=\mathcal A(1).$ Then


$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad (m\quad times) \qquad (\*)$


Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F\_n) \to A\_q$ which doesn't vanish at $\alpha$.


Then 
$$ 
A\_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K). 
$$  

It follows that 
$$ 
\Delta( [K,K,\ldots,K])=A\_q 
$$ 
and by $(\*)$ $\Delta( \mathcal A(m))=A\_q $ for every $m \ge 1.$


