Square-free implies solvability-forcing

Statement

Suppose $n=p_{1}p_{2}\dots p_{r}$ 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 (?).

Facts used

1. Every Sylow subgroup is cyclic implies metacyclic
2. Metacyclic implies solvable

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.

References

Textbook references

• The Theory of Groups by Marshall Hall, Jr., Page 148, Corollary 9.4.1, ^{More info}