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Quick review of examples of group actions

  • Dihedral group D2n acts on vertices of regular polygon with n sides. The stabilizer of a vertex is of order two; cosets are in bijection with n vertices. This group also acts on itself by conjugation. What are the orbits?

  • GL2(R) acts transitively on lines through the origin in R2. The stabilizer of the x-axis are the matrices of shape (abf0d) The cosets are in bijection with points of the real projective line, with representatives [u1] and [10]. The action on a coset representative is by linear fractional transformations in u.

  • The symmetric group acts on the set {1,,}n through its natural realization as bijections of this set to itself.

  • The symmetric group acts on its conjugacy classes. What can you say about S4 acting on its conjugacy classes?

  • Let F be a finite graph. The automorphisms of F are the maps from F to itself that preserve the edges (so connected vertices stay connected). Examples?

Proposition: Suppose that G has order n and p is the smallest prime dividing n. Then any subgroup of G of index p is normal.

Proof: Consider the action of G on the p cosets of H. The kernel of this action is the set of gG such that gg1H=g1H, which means that gg1=g1h for some hH. This in turn means that gg1Hg11. Since g1 is arbitrary, we must have g in the intersection of all conjugates g1Hg11 as g1 ranges over G. This is the largest normal subgroup of G contained in H – call it K.

The group G/K acts faithfully on the cosets G/H, so it is isomorphic to a subgroup of Sp. Therefore [G:K] divides p!. But [G:K]=[G:H][H:K]=p[H:K] and if this is a divisor of p! then [H:K] can only have prime divisors less than p. Since p is the smallest prime divisor of the order of G, this means [H:K]=1 so K=H and H is normal.

Remark: In the course of this we proved that, in general, the kernel of the action of G on G/H is the largest normal subgroup of G contained in H.

Class equation and applications

Lemma: The stabilizer of an element gG under conjugation is the centralizer of {g}.

Theorem: Let G be a finite group. Let G act on itself by conjugation, yielding a partition of G into disjoint conjugacy classes K1,,Kg. Choose a representative gi for each class. Then

G∣=i=1gKi∣=i=1g[G:CG(gi)].

Proposition: Let G be a group of prime power order. Then G has nontrivial center.

Corollary: If G has order p2 for some prime p, then G is abelian.

Proof: If xZ generates the quotient group, then every element of G is of the form xizj with zZ. This forces G to be abelian.

Conjugacy in Sn

The conjugacy classes in Sn correspond to the cycle decompositions, and there is one class for each partition of n as a sum of positive integers.

The centralizer of a cycle are the permutations which fix the integers appearing in the cycle.

The normalizer of a cycle was computed in the homework, at least in one case.