It is well-known that the braid group $B_{n}$ injects into the group of automorphisms of the free group $F_{n}$. However, there is certainly a kernel when mapping to the outer automorphism group $Out(F_{n})$. Namely, the kernel contains the generator of the center of $B_{n}$. Could someone please explain or give a reference to the fact (?) that the whole kernel of $B_{n} \rightarrow Out(F_{n})$ is the center of $B_{n}$?
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1$\begingroup$ $B_n$ naturally injects into $Aut(F_n)$. In your question you seem to have it injecting into $Aut(F_{n+1})$ so perhaps you're thinking of a different injection? The injection I'm thinking of comes from thinking of $B_n$ as the mapping class group of a disc with $n$ marked points, and the injection comes from considering the action on the fundamental group of the $n$-times punctured disc, base-point on the boundary. $\endgroup$– Ryan BudneyCommented Jun 9, 2011 at 21:39
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Assuming we're talking about the same map and there's just a $n \longmapsto n+1$ mix-up, this question reduces to studying $Inn(F_n)$ intersected with the image of $B_n \to Aut(F_n)$.
An inner automorphism of $F_n$ is a conjugation automorphism, so we're looking at braids that act on $F_n$ by conjugation. Let the generators of $F_n$ be denoted $x_1, \cdots, x_n$ and consider the product $x_1x_2\cdots x_n$. The image of the braid group fixes this element (the image of $B_n \to Aut(F_n)$ is precisely the subgroup of $Aut(F_n)$ that fixes this element). But if $c x_1x_2 \cdots x_n = x_1x_2 \cdots x_n c$, $c$ must be a power of $x_1x_2\cdots x_n$ since otherwise $F_n$ wouldn't be a free group.
answered Jun 9, 2011 at 21:47
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$\begingroup$ Yes, it was a typo on my part, confusing $n$ and $n+1$. Thank you for the explanation. $\endgroup$ Commented Jun 9, 2011 at 22:16