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

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.

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Gilman proved that $\mathrm{Aut}(F_n)$ is residually finite alternating. So there is an example unless $\mathrm{Aut}(F_n/\gamma_k(F_n))$ map onto arbitrarily large alternating groups as $k\to\infty$. I've no idea whether the latter is true or not. –  HJRW Aug 12 '10 at 21:05
Thank you very much indeed. What is required of Aut(Fnk(Fn)) to guarantee existence of an example in question seems to be probable. –  VladAr Aug 12 '10 at 21:28
The paper in question is: Gilman, Robert, Finite quotients of the automorphism group of a free group, Canad. J. Math. 29 (1977), no. 3, 541--551. –  HJRW Aug 12 '10 at 21:31

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 $m \ge 1.$

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Nice argument!. –  HJRW Aug 21 '10 at 5:11
Thanks. $\phantom{a}$ –  VladAr Aug 21 '10 at 5:50