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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 (?).

Facts used

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