# Is a simple loop in a spine of a strongly irreducible Heegaard splitting primitive in the fundamental group?

Let $\gamma$ be a simple loop in a spine of a strongly irreducible Heegaard splitting of a closed 3-manifold $M$ with torsion-free fundamental group. Does $\gamma$ necessarily correspond to a primitive element of the fundamental group of $M$, or is it possible for $\gamma$ to be a power of some other element?

I suspect that $\gamma$ is not necessarily primitive in the fundamental group of $M$, but I do not know of any examples.

Edit: I added torsion-free fundamental group after Charlie's comment. In particular, I am most interested in this question for hyperbolic 3-manifolds.

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Take the genus one Heegaard splitting of $\mathbb{R}P(3)$. This is strongly reducible. As the fundamental group of $\mathbb{R}P(3)$ is $\mathbb{Z}_2$ it has roots of all odd orders so it isn't primitive. :) –  Charlie Frohman Jun 8 '10 at 22:54
Take a Seifert fibered space that doesn't have torsion in its fundamental group. Build a "vertical" Heegaard splitting starting with regular neighborhoods of singular fibers, and putting horizontal tubes in till the complement is a handlebody. The singular fibers will lie in the spine of one handlebody, and yet they are not primitive elements of the fundamental group, as they are powers of the generic fiber. You might be able to build a hyperbolic counterexample to order, but I don't know one of the cuff. –  Charlie Frohman Jun 9 '10 at 4:30
I guess I need to be careful that it is strongly reducible, so take a Seifert fibered space with three singular fibers over the sphere, so that its fundamental group is an extension of a hyperbolic triangle group, and its euler class is nontrivial. Build your Heegaard splitting from two fibers and one horizontal bar. That Heegaard splitting is strongly irreducible, as the manifold is small and irreducible and the splitting has minimal genus. –  Charlie Frohman Jun 9 '10 at 5:01
Its still wrong. As the singular fibers have the standard fibers as powers and not visa verse. –  Charlie Frohman Jun 9 '10 at 5:08
In fact, I think you can prove your statement for small Seifert fibered spaces because the vertical Heegaard splittings are the only strongly irreducible Heegaard splittings. Good question. –  Charlie Frohman Jun 9 '10 at 12:36

New Answer: Take a 2-bridge knot, and perform hyperbolic Dehn filling (so that the core of the Dehn filling is geodesic), and so that the filling slope has intersection number $>1$ with the meridian. Then the meridian will not be primitive, since it will be a multiple of the core of the Dehn filling. 2-bridge knots have a genus two Heegaard splitting, which has a spine for the handlebody which is a wedge of two meridians at the bottom. This remains a spine in the Dehn filling, so the loops represented by the meridians are not primitive. This also works for the (2,n) torus knots (which are 2-bridge), so I think Charlie's answer is right (at least for many small Seifert fibered-spaces).
Old (non)Answer: Here's almost an example. All punctured torus bundles have Heegaard genus $\leq 3$, and many have Heegaard genus 3 (I discussed this once in my defunct blog). One may find a genus 3 Heegaard splitting of any once punctured torus bundle by taking two copies of a fiber, tubing them together along the boundary on one side, and adding a handle to the other side (drill out discs from both fibers, and glue an annulus in). By a theorem of Moriah-Rubinstein, most Dehn fillings will also have Heegaard genus 3 if the punctured torus bundle does.