Can all terms of the Johnson filtration be hom-mapped onto the same nontrival group?

Question

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.

Answer 1

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 $m \ge 1.$