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. Normal implies Lagrange-like
2. Lagrange-like is transitive