2-Groups which contain exactly three involutions - download pdf or read online

By Konvisser M.W.

Show description

Read Online or Download 2-Groups which contain exactly three involutions PDF

Similar symmetry and group books

Linear and projective representations of symmetric groups by Alexander Kleshchev PDF

The illustration idea of symmetric teams is among the most pretty, renowned, and critical elements of algebra with many deep family to different parts of arithmetic, equivalent to combinatorics, Lie thought, and algebraic geometry. Kleshchev describes a brand new method of the topic, in keeping with the new paintings of Lascoux, Leclerc, Thibon, Ariki, Grojnowski, Brundan, and the writer.

Extra resources for 2-Groups which contain exactly three involutions

Sample text

With 1 = 1H, as necessary. Rule (iii) tells us simply that c1 = c. We now apply the first two rules. Since we don’t what a1 is, let’s denote it 2: a1 = 2. Similarly, let a2 = 3. Now a3 = a31, which according to (ii) must be 1. Thus, we have introduce three (potential) cosets 1,2,3, permuted by a as follows: a a a 1 → 2 → 3 → 1. What is b1? We don’t know, and so it is prudent to introduce another coset 4 = b1. Now b4 = 1, and so we have b b 1 → 4 → 1. We still have the relation cba. We know a1 = 2, but we don’t know what b2 is, and so set b2 = 5.

Let G be a finite group, and let (G : 1) = pr m with m not divisible by p. (a) Any two Sylow p-subgroups are conjugate. (b) Let sp be the number of Sylow p-subgroups in G; then sp|m, and sp ≡ 1 mod p. (c) Any p-subgroup of G is contained in a Sylow p-subgroup. GROUP THEORY 41 Let H be a subgroup of G. Recall (p27) that the normalizer of H in G is NG (H) = {g ∈ G | gHg −1 = H}, and that the number of conjugates of H in G is (G : NG (H)). For a Sylow p-subgroup P , the number of conjugates of P is (G : NG (P )) = (G : 1) (G : 1) m = = .

24) that conjugating an element preserves the type of its disjoint cycle decomposition. 1234... 26. (ijk) = (1234... ijk4... )(123)(ijk4... ) . S. 27. For 1 < k ≤ n, there are needed so that we don’t count n(n−1)···(n−k+1) k distinct k-cycles in Sn . The 1 k is (i1i2 . . ik ) = (ik i1 . . ik−1 ) = . . k times. Similarly, it is possible to compute the number of elements in any conjugacy class in Sn , but a little care is needed when the partition of n has several terms equal. For example, the number of permutations in S4 of type (ab)(cd) is 1 4×3 2×1 = 3.

Download PDF sample

Rated 4.28 of 5 – based on 43 votes