  Let S be a minimal complex surface of general type with $q(S)=0$. We prove
the following statements concerning the algebraic fundamental group:
  I) Assume that K^2_S\leq 3\chi(S). Then S has an irregular etale cover if and
only if S has a free pencil of hyperelliptic curves of genus 3 with at least 4
double fibres. II) If K^2_S=3 and \chi(S)=1, then S has no irregular etale
cover.
  III) If K^2_S<3\chi(S) and S does not have any irregular etale cover, then
the order of the algebraic fundamental group is lesser or equal to 9, and if
equality occurs then K^2_S=2,
  \chi(S)=1.
