Question: Let $M$ and $N$ be $3$-manifolds obtained by zero-surgery on (left-handed) trefoil and figure-eight knot respectively.

What is the easy way to prove that $M$ and $N$ are not homeomorphic?

Note: When they are knot homology spheres (they are both homology $S^1 \times S^2$'s), I cannot use the classical invariants.

Answer: 
If you're happy bringing in heavy machinery then you could compute some sort of Floer homology, like the 'hat' version of Heegaard Floer homology: this has rank 2 for $S^3\_0(3\_1)$ and rank 4 for $S^3\_0(4\_1)$, so they're different.

On the other hand, in cases like this where you have a very specific pair of 3-manifolds in mind, it's often enough to distinguish their fundamental groups by counting covers of some finite order. Here you could work this out by hand starting from Wirtinger presentations of the respective knot groups, or you could just ask SnapPy to do it:

```
In[1]: len(Manifold('3_1(0,1)').covers(9))
Out[1]: 4

In[2]: len(Manifold('4_1(0,1)').covers(9))
Out[2]: 2

```

This counts 9-fold covers of each 0-surgery, and since the numbers are different, they must have different fundamental groups.
This is a topological counterpart to Josh's geometric argument.

Since $H\_1(M) \cong H\_1(N) \cong \mathbb{Z}$, there is only one 6-fold cyclic cover of either of them; call them $M'$, $N'$. Both $M'$ and $N'$ are both torus bundles over the circle, and their monodromies are the sixth powers of the monodromies of $M$ and $N$. Since the monodromy of $M$ is a sixth root of the identity but the monodromy of $N$ isn't, $M'$ is $T^3$, while $N'$ is not.