Suppose that $G$ is a group and $H$ and $K$ are its subgroups with $K\subset H$. Prove that $[G:H]\leq [G:K]$.
Intuitively i know that this is true. But I am trying to prove it in rigorous way: let's denote the set of all left cosets of $K$ in $G$ by $G/K$ and by $G/H$ the set of all left cosets of $H$ in $G$.
Consider the mapping $\varphi:G/K\to G/H$ defined by $\varphi(gK)=gH$ for any $g\in G$. It's easy to check that this map is well-defined and surjective. How to conclude rigorously the need inequality?
Would be very grateful for help!
EDIT: If $f:X\to Y$ the mapping between two sets and $f$ - surjective. Then two cases are possible:
1) If $Y$ is infinite then $X$ is also infinite.
2) If $Y$ is finite then $|X|\geq |Y|$.
In both cases we get that $|X|\geq |Y|$.
Is my reasoning correct?