TITLE: Can all terms of the Johnson filtration be hom-mapped onto the same nontrival group?
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: 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 $m \ge 1.$