Group of parallelizations of $M^3$ finitely generated? Let $M^3$ be a compact orientable 3-manifold.  Then $TM$ is trivial and let's go ahead and fix a trivialization $\tau : M \times \mathbb{R}^3 \to TM$.  Then given a map $g : (M, \partial M) \to (SO(3), 1)$ we can consider the new trivialization $g \cdot \tau$ that is given by $g \cdot \tau (p,v) = \tau(p,g(p)(v))$.
The group of homotopy classes $[(M,\partial M), (SO(3),1)]$ then parameterizes all of the homotopy classes of trivializations of $M$ that agree with $\tau$ on the boundary of $M$.  I know that this group is abelian, that the degree map is a homomorphism to $\mathbb{Z}$, and that the degree map is rationally an isomorphism.  
Is this group finitely generated?  
As more of an aside, for $M$ instead an $n$-manifold is the group $[(M,\partial M), (SO(n), 1)]$ abelian?
 A: I believe that for any finite $n$-complex $X$, the group $$[X, SO(n)]_*$$ is finitely generated. I will follow Igor Belegradek's approach. In fact I only think $H^*(X;\Bbb Z)$ finitely generated and maybe $\pi_1$ finitely generated is necessary.

There is a fibration $B\text{Spin}(n) \to BSO(n) \to B^2(\Bbb Z/2)$ coming from the fact that the extension is central, where the last map induces an isomorphism on $\pi_2$; this loops to the usual fibration (but now where every map is a loop map) $$\text{Spin}(n) \to SO(n) \to B(\Bbb Z/2).$$
In general this implies that if $X$ is a CW complex we have $$0 \to [X, \text{Spin}(n)] \to [X, SO(n)] \to H^1(X;\Bbb Z/2)$$ is an exact sequence of groups, which gives the desired result as soon as you know that $[X, \text{Spin}(n)]$ is finitely-generated. 
Now we have the short exact sequences $$[X, \Omega S^n] \to [X, \text{Spin}(n)] \to [X, \text{Spin}(n+1)] \to [X, S^n].$$ 
A priori the last map is exact at the level of pointed sets, but in fact for $X$ a $(\leq n)$-complex this is the same as the map $[X, \text{Spin}(n+1)] \to [X, K(\Bbb Z, n)] = H^n(X;\Bbb Z)$, and this is indeed a group map, given by looping the map $B\text{Spin}(n+1) \to K(\Bbb Z, n+1)$ given by the Euler class $e$. 
So this is a short exact sequence of groups when $X$ is a $(\leq n)$-complex, and $[X, S^n] = H^n(X;\Bbb Z)$ is a finitely-generated group. It follows from applying the Postnikov tower carefully that $[X, \Omega S^n]$ is also finitely-generated. So we have $$A \to [X, \text{Spin}(n)] \to [X, \text{Spin}(n+1)] \to B$$ for finitely generated abelian groups $A, B$ so long as $X$ is a $(\leq n)$-complex; passing to a quotient $A'$ of $A$ and a subgroup $B'$ of $B$ we have an exact sequence $$0 \to A' \to [X, \text{Spin}(n)] \to [X, \text{Spin}(n+1)] \to B' \to 0.$$ 
It follows that $[X, \text{Spin}(n)]$ is finitely generated iff $[X, \text{Spin}(n+1)]$ is so long as $X$ is a $(\leq n)$-complex. For us, we conclude that since $X$ is an $n$-complex, we have $[X, \text{Spin}(n)]$ f.g. iff $[X, \text{Spin}]$ f.g. 
To conclude observe that from the fibration $\text{Spin} \to SO \to B(\Bbb Z/2)$ (where all maps are loop maps) we conclude there is an exact sequence of groups $$0 \to [X, \text{Spin}] \to [X, SO] \to H^1(X;\Bbb Z/2),$$ and in particular $[X, \text{Spin}]$ is a subgroup of $[X, SO] = K^1(X)$. Now it follows from the Atiyah-Hirzebruch spectral sequence that $H^*(X;\Bbb Z)$ finitely generated implies that $K^1(X)$ is finitely generated.
