Place the base point c halfway around the tube from this hole, so that c is on the inside.
Draw two loops f and g, based at c, wrapping around the tube, so that the hole is between f and g. If the hole was not there, we could slide f and g together, with c acting as a hinge. But the hole is there, so this is not an option. Still, f and g are homotopic. Slide f all the way around the torus until it coincides with g. Note that this homotopy moves the base point. In fact we cannot slide f around onto g without moving the base point. The curves are homotopic, but do not represent the same elements in the fundamental group. Since homotopyc loops are conjugate in π1, write hf = gh for some loop h based at c. If π1 were abelian, we would have f = g, which is a contradiction. Therefore the fundamental group of the punctured torus is nonabelian.
In this case, h is the loop that runs all the way around the inside of the tire and returns to c. This is the trace of the base point as f slides around the torus and on to g.
To be rigorous, we should prove there is no homotopy from f to g that fixes c. I'm not sure how to do this, based on the theorems we have developed thus far, but if you try to slide f around with c fixed, I think you'll see that it can't be done.
So where exactly does the group become nonabelian? At hg, which is not the same as gh. If it were, then hg/h would equal g; yet hg/h = f, and g and f are distinct elements in π1.
This is an interesting result, because higher homotopy groups, i.e. the homotopy classes of spheres into S, are always abelian. We'll see this later on. Only the circle, i.e. the one dimensional sphere, can produce a nonabelian homotopy group.