# Square-free implies solvability-forcing

## Statement

Suppose $n = p_1p_2 \dots p_r$ where the $p_i$ are pairwise distinct prime numbers. In other words, $n$ is a square-free number.

Then, $n$ is a solvability-forcing number: any Finite group (?) of order $n$ is a Solvable group (?), i.e., a Finite solvable group (?).

## Proof

The proof follows from facts (1) and (2), and the observation that in a group of square-free order, every nontrivial Sylow subgroup has prime order, and is hence cyclic.