Groups of Order 6
Every group of order is isomorphic to or .
Lemma 6.6 (Groups of Order 6)
If , then or .
Proof
Cauchy’s theorem gives elements and of orders and . The subgroup generated by has index , so the other coset is both and . Thus . The cases and give and respectively; is impossible.
Related
Stated in
- Lemma 6.6 (Groups of Order 6)§6 Finite Groups