Manifold with $\pi_1(M)=F_n$ We may construct a 3-manifold $M_n$ with $\pi_1(M_n)\cong F_n$ (i.e. the free group on $n$ generators) as follows: consider the complement of $n$ pairs of open 3-balls in $\mathbb{R}^3$. For each pair, identify the corresponding boundary spheres. It is easy to see that this manifold has the free group on $n$ generators as a fundamental group. Lets identify with each pair of spheres a generating loop $x_i$. Clearly, $\pi_1(M_n)\cong <x_1,...,x_n>$. My question is as follows: if possible, how can we modify $M_n$ such that $x_ix_j=x_jx_i$ for some $i$ and $j$. If this is not possible what is the obstruction?
 A: Your construction of the 3-manifold $M_n$ can be reworded by saying that it is a connected sum of the form
$$(*) \qquad M_n \, = \, \mathbb{R}^3 \, \# \, \underbrace{(S^2 \times S^1) \, \# \, \cdots \, \# \, (S^2 \times S^1)}_{n \,\,\text{times}}
$$
The group you want to obtain instead is simply the free product $F_{n-2} * \mathbb{Z}^2$. This is obtained by altering your construction, replacing two of the $S^2 \times S^1$ connected summands with a single $T^2 \times \mathbb{R}$ connected summand:
$$(**) \qquad \mathbb{R}^3 \, \# \, \underbrace{(S^2 \times S^1) \, \# \, \cdots \, \# \, (S^2 \times S^1)}_{n-2 \,\,\text{times}} \, \# \, (T^2 \times \mathbb{R})
$$
As noted in the comments, you can drop the $\mathbb{R}^3$ connected summand of $(*)$ without changing the fundamental group and you get a compact manifold. But although you can also drop the $\mathbb{R}^3$ connected summand of $(**)$ without changing the fundamental group, that will not result in a compact manifold because the $T^2 \times \mathbb{R}$ connected summand remains noncompact.
A: Just for the sake of completeness: If $M^3$ is a closed connected 3-dimensional manifold (closed means compact and without boundary), its fundamental group cannot be isomorphic to the free product $F_n \star Z^2$ for any $n$. Otherwise, by Kneser's theorem, this free product decomposition would correspond to a connected sum decomposition of $M$ into pieces with fundamental groups isomorphic to $Z$ and to $Z^2$. However, a closed 3-manifold cannot have the fundamental group $Z^2$. 
