Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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

1 Answer 1

up vote 5 down vote accepted

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 $m \ge 1.$

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.