Why is this sentence true?
For every not nullhomologous loop $f$ without selfintersections on orientable compact surface $M$ there exists a surjection $\phi: H_1(M) \to \mathbb{Z}/8\mathbb{Z}$ such that $f \notin \ker(\phi)$. Where $H_1$ is abelianization map.
I know that we can present $\phi(f)=a_1^{k_1} b_1^{k_2} \cdots a_g^{k_{2g-1}} b_g^{k_{2g}} x_1^{k_{2g+1}} \cdots x_{b-1}^{k_{2g+b-1}}$
where fundamental group of $M$ (with genus $g$ and $b$ boundary components) is: $$ \pi_1(M, *) = \langle a_1, b_1, \ldots , a_g, b_g, x_1, \ldots , x_b \mid [a_1, b_1]\cdots [a_g, b_g]= x_1\cdots x_b \rangle . $$
so I think the point is to show why always there exists at least one $k_i$ such that $k_i \not \equiv 0\pmod{8}$
Thanks for any help.
Draw a smooth simple (oriented) loop $a$ on your surface. Since $a$ is not homologycally trivial, it does not separate the surface. Therefore there exists a smooth transversal oriented loop $b$ on the surface which intersects $a$ transversally and in a single point. By choosing orientation on $b$ correctly, you can assume that the oriented intersection number is positive, $+1$. Now, you define a homonorphism $h: H_1(M)\to Z$ by computing oriented intersection number with $b$. The rest you should be able to do yourself. A good reference for such constructions is Guillemin and Pollack "Differential Topology".