2D Groups State and prove Lagrange's Theorem. Lagrange's Theorem: If G is a finite group and H a subgroup then | H| divides |G|. Proof: Define a relation ~ on G by g1 ~ g2 iff gg2 E H. This is reflexive (since H contains the identity), symmetric (since g2 g1 = (g71g2)-1 and H is closed under inverses) and transitive (as g, 93 = (91 92)(g2 93) and H is closed under products). Hence it is an equivalence relation, and G is partitioned into equivalence classes. We now claim that for each g E G its equivalence class [g] has size | H\, so that (G| = |H| x #equivalence classes is divisible by |H|. Well, define a map f : [g] > H by x + g-1x. f well-defined: If x E [g] then g 1x E H so f(x) E H. f bijective: Its inverse is x => gx (this is similarly well-defined H > [g]). Thus [g]| = |H| and we're done. [6] Show that the dihedral group of order 2n has a subgroup of order k for every k dividing 2n. The dihedral group D2n is the symmetry group of a regular n-gon in the plane. If m > 1 divides n then one can form a regular m-gon with vertices every (n/m)th vertex of the n-gon. The symmetry group of this m-gon (of order 2m) and its rotational subgroup (of order m) form subgroups of the symmetry group of the n-gon. So we have found subgroups of Den of order m and 2m for every m > 1 dividing n, and hence of order k for every k | 2n (for k = 1 just take the subgroup {e}, and for k = 2 the subgroup generated by a reflection). [4] [10 5D Groups (a) Let G be a finite group, and let g E G. Define the order of g and show it is finite. Show that if g is conjugate to h, then g and h have the same order. The order o(g) of g is the smallest positive integer n such that g" = e. Claim: The order exists (is finite). Proof: We need to show that the set {n > 0 : g" = e} is non-empty. Let [G] = N and consider the N +1 elements of G given by e, g, g2, . . . , gN.
By the pigeonhole principle, there exist distinct i, j € {0, 1, ... , N + 1} with g' = g . WLOG i < j. Then g]- = e so j - i E {n > 0 : gn = e} and we're done. Suppose g = khk-1. Then for n E Z we have g" = kh"k-1 so g" = e => kh"k-1 = e => h"(= k-1k) = e. Thus {n E Z : g" = e} = {n E Z : h" = e} so o(g) = o(h). [5] (b) Show that every g E Sn can