Take a compact 3-manifold $M$ with $b_1(M)\geq 2$. Then there are many homomorphisms $\pi_1(M)\to \mathbb{Z}$, since $\mathbb{Z}^{b_1(M)}\leq H_1(M)$. Further, if the manifold fibers over $S^1$ corresponding to a map $\phi:M\to \mathbb{Z}$, then $ker(\phi)$ is finitely generated. If $\phi:M\to \mathbb{Z}$ is not fibered, then a theorem of Stallings implies that the cohomology class is not dual to a fiber. For example, consider the link L4a1:

The complement is a compact manifold $M$ with $H_1(M)=\mathbb{Z}^2$. Orienting the two circles of the link in two different ways (up to negation) gives two different homomorphisms to $\mathbb{Z}$ (via the linking number). One orientation corresponds to a fibering, while the other does not (there is an annulus running between the two components). Also, the intersection number with the meridian is the same (up to sign) for each choice of orientation, so the cyclic subgroup condition is satisfied. So the kernel of one map is finitely generated (in fact free), while the other is infinitely generated.

a lotmore information than having an extension of groups. $\endgroup$ – Theo Johnson-Freyd Oct 21 '10 at 5:54