Subnormal implies Lagrange-like

Statement
Suppose $$L$$ is a finite loop and $$S$$ is a subnormal subloop of $$L$$. Then, $$S$$ is a Lagrange-like subloop of $$L$$, i.e., the order of $$S$$ divides the order of $$L$$.

Related facts

 * Normal implies Lagrange-like
 * Characteristic not implies Lagrange-like

Facts used

 * 1) uses::Normal implies Lagrange-like
 * 2) uses::Lagrange-like is transitive