If $X$ is a Riemann surface with boundary $\partial X$ and $\pi_1(X,p)$ is its fundamental group, $p \in X$, then we shall call class $[\gamma] \in \pi_1(X,p)$ primitive (or generator) if it can not be represented as a power of another class. It is stated that every primitive class has a non self-intersecting representative $\gamma$. But how to show this? I also think that every nonprimitive class doesn't have any non self-intersecting representative (intuitively), but is it true?
UPD. A 2-plane with 2 holes may be a counterexample. I should reformulate the question: is it always possible to choose generators of $\pi_1(X,p)$ in such a manner that the property from the question holds?
Non self-intersecting generators of $\pi_1(X)$ are easy enough to give. For boundaryless surfaces, if you puncture the surface and consider a deformation retract on to a finite wedge of circles, each of the embedded circles will generate $\pi_1(X)$. Now, if we then start cutting out disks, the boundary components will also contribute to the generators and should be included. This collection of boundary circles, and embedded wedge of circles are generating loops of the entire fundamental group by the classification of surfaces with boundary.