# generators for the handlebody group of genus two

Is the handlebody group of genus two surface generated by Dehn twists along properly embedded disks and annuli?

Are there alternative ways to describe a set of generators that are conceptually simple (not necessarily finite but conceptually simple)? For this part, really I'm asking the question for the handlebody subgroup of the mapping class group so I am considering only the restriction of the homeomorphisms to the boundary of the handlebody. Therefore a description in terms of Dehn twists that are tractable is appreciated.

Context: Define a handlebody $B$ as a regular neighbourhood of a graph embedded in $\mathbb{R}^3$. The genus of a handlebody is defined as the rank of its fundamental group. The boundary of a genus $g$ handlebody is a surface of genus $g$ then. Let the handlebody group of genus $g$ be the group of homeomorphisms of the handlebody up to isotopy. For any properly embedded disk $D$ or annulus $A$ in $B$, one can define a positive Dehn twist along $D$ or $A$ in the standard way by either twisting once counter-clockwise around the center of a regular neighborhood of $D$ or twisting along the $S^1$ direction in a regular neighborhood of $A$.

As stated in Ian Agol's answer, the mapping class group of a handlebody $B$ maps onto $Out(\pi_1B)$. This is easy to show by lifting known generators for the automorphism group of a free group. Twists along disks lie in the kernel of the surjection to $Out(\pi_1B)$, and it is a theorem of Luft (Math. Ann. 234 (1978) 279-292) that this kernel is generated by twists along disks. (McCullough showed this kernel is not finitely generated when the genus of $B$ is at least two.)
Twists along annuli need not lie in the kernel of the map to $Out(\pi_1B)$, and in fact the subgroup they generate maps onto a subgroup of $Out(\pi_1B)$ of index at most two. For example, if one considers $B$ as obtained by attaching a 1-handle $H$ to a handlebody $B'$ of genus one less than the genus of $B$, then the homeomorphism of $B$ obtained by dragging one end of $H$ around an embedded loop in $\partial B'$ disjoint from the disk where the other end of $H$ attaches can be realized by a twist along an annulus. The annulus lies in $B'$ and is isotopic to an annulus in $\partial B'$ which is a neighborhood of the loop used for the dragging operation, with one end of $H$ lying in the interior of the annulus in $\partial B'$ and the other end lying outside this annulus. The induced automorphism of $\pi_1B$ fixes $\pi_1B'$ and takes the generator corresponding to $H$ to the product of itself with the element of $\pi_1B'$ represented by the embedded loop. Such automorphisms generate an index two subgroup of $Out(\pi_1B)$, the subgroup mapping to automorphisms of $H_1(B)$ of positive determinant. To get the rest of $Out(\pi_1B)$ one can take a half-twist along a disk in $B$ splitting off a solid torus.
Those two operations give trivial outer automorphisms of $\pi_1(B)$, whereas the handlebody group surjects $Out(\pi_1(B))$. So Dehn twists about disks and annuli don't suffice (in fact, I think it's known that these generate the kernel of the map to $Out(\pi_1(B))$).