Twisted Alexander polynomials and surjectivity of a group homomorphism

If phi: G-->G' is a surjective homomorphism, we prove that the twisted Alexander polynomial of G is divisible by the twisted Alexander polynomial of G'. As an application, we show non-existence of surjective homomorphism between certain knot groups.Comment: Published by Algebraic and Geometric Topology at http://www.maths.warwick.ac.uk/agt/AGTVol5/agt-5-51.abs.htm