Mapping Class Group (MCG) of connected sum of 3-torus and $S^2\times S^1$ I know that the MCG (isotopy-classes of orientation preserving homeomorphisms) of 3-torus $(S^1\times S^1 \times S^1)$ is $SL(3,Z)$, since it is an Eilenberg–MacLane space, giving $ MCG(T^3)=Out(\pi_1(T^3))=Out(Z^3)=SL(3,Z)$.
But  $S^2\times S^1$ is not an Eilenberg–MacLane space. I want to know what $MCG(T^3\# S^2\times S^1)$ is? And, what relation does it have to $\pi_1((T^3\# S^2\times S^1)) $? Is there a general way of finding MCG of connected sum of 3-manifolds? 
Thanks
 A: The mapping class groups of all compact orientable 3-manifolds are essentially known.  A fairly detailed summary of the results, focusing on the nonprime case and with references to proofs in the literature, can be found in Section 2 of a paper of mine with Nathalie Wahl: "Stabilization for mapping class groups of 3-manifolds", Duke Math. J. 155 (2010), 205-269, arXiv:0709.2173. (Section 2 can be read independently of the rest of the paper.)  The results are stated in terms of the map $\Phi:MCG(M)\to Out(\pi_1M)$. For $M$ closed and orientable the kernel of $\Phi$ is a direct sum of finitely many copies of ${\mathbb Z}_2$ generated by Dehn twists along 2-spheres. Also, $\Phi$ is surjective if this is true for each prime connected summand of $M$ and a certain other natural condition holds involving permuting different summands with isomorphic fundamental groups.
In the case of $T^3 \# (S^1\times S^2)$ the kernel of $\Phi$ is a single ${\mathbb Z}_2$ generated by a Dehn twist along the $S^2$ factor of $S^1\times S^2$. (Interestingly, the twist along the $S^2$ separating the two connected summands is isotopic to the identity.) The image of $\Phi$ is an index two subgroup of $Out({\mathbb Z}^3*{\mathbb Z})$ since one can only realize $SL(3,{\mathbb Z})$, not all of $GL(3,{\mathbb Z})$, by orientation-preserving homeomorphisms of $T^3$. 
The structure of the automorphism group of a free product seems to be well understood in terms of the automorphism groups of the factors. For $M=T^3 \# (S^1\times S^2)$ this means its mapping class group is generated by the MCG's  $SL(3,{\mathbb Z})$ and ${\mathbb Z}_2 \times {\mathbb Z}_2$ of the two summands, along with homeomorphisms obtained by regarding $M$ as obtained from $T^3$ by attaching a "1-handle" and dragging either end of this handle around loops in $T^3$.
$\bf Added\  later\colon$ Generators for the mapping class group of $T^3 \# (S^1\times S^2)$ can be described explicitly, using the algebraic fact that the automorphism group of a free product of groups $G_1,\cdots,G_n$ is generated by four types of automorphisms:
(1) Automorphisms of the individual factors $G_i$,
(2) Permutations of isomorphic factors.
(3) Partial conjugations, in which one $G_i$ is conjugated by an element $x_j\in G_j$ and the other factors $G_k$ are fixed. (If $i=j$ this gives an automorphism of type (1), an inner automorphism of $G_i$.) 
(4) If some $G_i$ is infinite cyclic generated by $g_i$, say, and $x_j$ is an arbitrary element of some $G_j$ with $j\neq i$, there is an automorphism sending $g_i$ to $g_ix_j$ or $x_jg_i$ and fixing the other factors $G_k$, $k\neq i$.
For $T^3 \# (S^1\times S^2)$ the type (1) automorphisms for the ${\mathbb Z}^3$ factor form a copy of $SL(3,{\mathbb Z})$ which is the mapping class group of $T^3$ (a result originally due to Waldhausen I think).  Since one can isotope an orientation-preserving homeomorphism of $T^3$ to fix a ball pointwise, this homeomorphism of $T^3$ gives rise to a homeomorphism of the connected sum which is the identity on the other summand.  For the $S^1\times S^2$ summand the automorphism group of $\pi_1$ is ${\mathbb Z}_2$ with a generator realized by a homeomorphism of $S^1\times S^2$ that reflects both factors simultaneously.  This too can be isotoped to extend to the connected sum via the identity on the complement.
Note that $SL(3,{\mathbb Z})$ is generated by elementary matrices that differ from the identity matrix in a single off-diagonal entry, which is 1, and these generating automorphisms of ${\mathbb Z}^3$ are realized by Dehn twists along tori in $T^3$.
There are no type (2) automorphisms of ${\mathbb Z}^3 * {\mathbb Z}$. For type (3) automorphisms, conjugation of the ${\mathbb Z}^3$ factor by an element of the ${\mathbb Z}$ factor is actually an inner automorphism of the whole group since the ${\mathbb Z}$ factor is abelian. We are interested in $Out({\mathbb Z}^3 * {\mathbb Z})$ so inner automorphisms are factored out. On the other hand, conjugating the ${\mathbb Z}$ factor by an element of ${\mathbb Z}^3$ is realizable by composing two type (4) automorphisms.
Type (4) automorphisms are realized by the homeomorphisms that I described earlier as dragging one end of the "1-handle" $S^1\times S^2$ around a loop $\gamma$ in the $T^3$ summand. Concretely, this homeomorphism is a Dehn twist along a certain torus $T_\gamma\subset T^3 \# (S^1\times S^2)$ defined as follows. First take a solid torus neighborhood of the loop $\gamma$. We can view connected sum with $S^1\times S^2$ as removing two disjoint open balls in $T^3$ and identifying the resulting two boundary spheres. Choose one of these two balls in the interior of the solid torus neighborhood of $\gamma$ and choose the other ball in the exterior of this solid torus.  The boundary torus of the solid torus then gives the $T_\gamma$ we want. The homeomorphism $h_\gamma$ we are looking for is a twist in the longitudinal direction along $T_\gamma$. The automorphism of $\pi_1={\mathbb Z}^3*{\mathbb Z}$ induced by $h_\gamma$ is the identity on the factor $\pi_1(T^3)$ since generators for this ${\mathbb Z}^3$ can be chosen disjoint from the solid torus. If $x\in \pi_1(T^3)$ is the element represented by $\gamma$ and $g$ is a generator of $\pi_1(S^1\times S^2)$, then it's not hard to check that the automorphism induced by $h_\gamma$ sends $g$ to $xg$ or $gx$. (The two automorphisms given by $xg$ and $gx$ differ by an inner automorphism in the special case we are considering.) Thus a type (4) generator is realized by a twist along a torus. 
In summary, the mapping class group of $T^3\#(S^1\times S^2)$ is generated by twists along tori and 2-spheres, and also the homeomorphism reflecting the two factors of $S^1\times S^2$. 
