This may be very either very simple or very unknown, but here goes: Let $F_n$ be the free group on $n$ generators and $Out(F_n)$ its outer automorphism group.
Can embeddings $Out(F_n) \hookrightarrow Out(F_m)$ be characterized for $m > n$?
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$\begingroup$ This sort of question has been looked at from the perspective of mapping class groups, see for example arxiv.org/abs/1011.1855 $\endgroup$– ndkrempelCommented Dec 15, 2010 at 9:51
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1$\begingroup$ Martin Bridson has done some work on this question. arxiv.org/abs/1007.2598 ; he is also giving a talk on it this Friday at the CUNY Graduate Center. $\endgroup$– Steve DCommented Dec 15, 2010 at 15:20
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$\begingroup$ Dr. Shello: This doesn't have to do with your question, but a couple of months ago you asked me about my write-up of a result of Frobenius on simple groups of order p(p-1)(p+1)/2. I don't have an address for you, so I'll inform you here that I put the result up on arXiv. $\endgroup$– paul MonskyCommented Aug 7, 2011 at 0:29
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2 Answers
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Check out
Bogopolski O.V. and Puga D.V., On embeddings of Out(F_n), the outer automorphism group of the free group of rank n, into Out(F_m) for m>n, Algebra and Logic, v. 41, no. 2 (2002), 69-73.
answered Dec 16, 2010 at 0:48
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$\begingroup$ Oh, that sounds promising! Thanks, will take a look. $\endgroup$ Commented Dec 16, 2010 at 4:46
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There is a bunch of negative results here:
1) Khramtsov has shown that there are no embeddings for $m = n+1$ (when $n>1$)
2) Bridson and Vogtmann have shown that any homomorphism $Out(F_n) \to Out(F_m)$ factors through $\mathbb{Z} / 2 \mathbb{Z}$ if
a) $ m < n, n>2$
b) $n > 8, n< m < 2n-2$ for $n$ odd, and $n < m< 2n$ for $n $ even. This is a slightly stronger version of a result of Potapchik--Rapinchuk.
3) I have a result showing that in fact all such homomorphisms factor through $\mathbb{Z} / 2 \mathbb{Z}$ if $n > 5$ and $ n < m < {n \choose 2}$.
There are two positive results as well, one by Aramayona--Leininger--Souto and one by Bogopol'skii--Puga and Bridson--Vogtmann; in both cases $m$ grows at least exponentially with $n$.
As far as I am aware, this is all that is known. I'd be happy to give you further details if you are interested.
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$\begingroup$ Nice. My naive idea was that one should first expect an embedding when m=2n-2, because that's the first time you can make sense of such a thing for a complex computing the rational homology of $Out(F_n)$. It's interesting to see that you are getting a quadratic negative range. $\endgroup$ Commented Dec 16, 2010 at 15:26