Consequences of the Sylow Theorems
نویسندگان
چکیده
Proof. Any element of odd order in a symmetric group is an even permutation, so the 3-Sylow and 5-Sylow subgroups of S5 lie in A5. Therefore it suffices to focus on A5. Since |A5| = 60 = 22 · 3 · 5, the 3-Sylow subgroups have size 3 and the 5-Sylows have size 5. Call the numbers n3 and n5. By Sylow III, n3 | 20 and n3 ≡ 1 mod 3, so n3 = 1, 4, or 10. The number of 3-cycles (abc) in A5 is 20, and these come in inverse pairs, giving us 10 subgroups of size 3. So n3 = 10. Turning to the 5-Sylows, n5 | 12 and n5 ≡ 1 mod 5, so n5 is 1 or 6. Since A5 has at least two subgroups of size 5 (the subgroups generated by (12345) and by (21345) are different), n5 > 1 and therefore n5 = 6. Theorem 2.2. In Aff(Z/(5)), n2 = 5 and n5 = 1.
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