$\newcommand{\ss}{\subset}\newcommand{\R}{\mathbb R}\newcommand{\cn}{\colon}\newcommand{\range}{\operatorname{range}}\newcommand{\d}{\partial}\newcommand{\ker}{\operatorname{ker}}\newcommand{\sr}[2]{\stackrel{#1}{#2}}\newcommand{\lra}{\longrightarrow}\newcommand{\Z}{\mathbb{Z}}$
Let $M$ be a closed orientable smooth surface and $L\ss M$ be a smooth $1$-submanifold diffeomorphic to the circle. Prove that the class $[L]\in H_1(M)$ is either zero or primitive.
Def. An element of an abelian group $A$ is primitive if $a\not\in kA$ for all $k>1$.
Def. Closed surface is a compact surface without boundary.
Def. Surface without boundary $M$ is a Hausdorff space in which every point has an open neighbourhood homeomorphic to some open subset of the Euclidean plane, but there exists some $z\in M$ which is not homeomorphic to any open subset of $\R^2_+=\{(x,y)\in\R^2\cn y\geq > 0\}$.
Def. Orientable surface $M$ is a surface such that $H_1(M)$ is free abelian.
Def. Smooth surface $M$ is a surface which has a tangent plane at every point.
Def. Differentiable map is called a diffeomorphism if it is a bijection and its inverse is also differentiable.
I don't even know where to start here so any hints would be appreciated.
Edit: 1. what is meant by class here? $H_1(M)=\ker(\d_1)/\range(\d_2)$, where the chain complex is $$\ldots \sr{\d_3}{\lra} C_2\sr{\d_2}{\lra}C_1\sr{\d_1}{\lra}C_0\lra 0.$$ How $L$ could be seen as a (formal linear combination of maps $\Delta^1\to M)\in\ker(\d_1)$?
- I've seen that $H_0(M)=\Z^C=H_2(M)$, where $C$ is the set of connected components of $M$. $H_1(M)=\Z^S$, for some set $S$, since $M$ is oriented. Also $H_i(M)=0$ for all $i>2$.