Group Theory: Math30038, Sheet 5

Solutions GCS

1. Let G = S_{n} be the symmetric group on {1, 2, …, n}, so |G| = n! and the elements of G are the permutations of {1, 2, …, n}.

(a) Suppose that (a_{1}, a_{2}, …, a_{t}) ∈ G is a cycle, and that g ∈ G. Show that g^{−1}(a_{1}, a_{2}, …, a_{t})g = (a_{1}g, a_{2}g, …, a_{t}g). Solution: If x ≠ a_{i}g for any i, then xg^{−1} ≠ a_{i} for any i, so xg^{−1}(a_{1}, a_{2}, …, a_{t})g = xg^{−1}g = x. However, if x = a_{i}g for some i, then xg^{−1}(a_{1}, a_{2}, …, a_{t})g = a_{i}gg^{−1}(a_{1}, a_{2}, …, a_{t})g = a_{i}(a_{1}, a_{2}, …, a_{t})g = a_{i+1}g. Thus we can describe g^{−1}(a_{1}, a_{2}, …, a_{t})g by the notation (a_{1}g, a_{2}g, …, a_{t}g).

(b) Each element of G can be expressed as a product of disjoint cycles (elements of G are disjoint if their supports are disjoint). Show that the number of conjugacy classes in S_{n} is the number of ways of writing n as an ascending sum a_{1} + a_{2} + ⋯ + a_{t} of positive integers a_{1} ≤ a_{2} ≤ ⋯ ≤ a_{t}. Thus there are 3 conjugacy classes in S_{3} because 3 can be written as an ascending sum in three ways: 3, 1+2, 1+1+1. Also in S_{4} there are 5 conjugacy classes be cause 4 is 4, 1 + 3, 1 + 1 + 2, 2 + 2 and 1 + 1 + 1 + 1. Solution: Conjugation sends cycles to cycles (of the same length). Morover, the calculation performed for Question 1 shows how g must be chosen to effect the required conjugation. In particular any two cycles of length r are conjugate in S_{n} by a suitably chosen g. Conjugation sends disjoint cycles to conjugate disjoint cycles so conjugacy classes in S_{n} consist of all elements with a given ‘cycle shape’. Thus the cycle shapes characterize the conjugacy classes, and these shapes can be specified by listing the lengths of the disjoint cycles in ascending order.

(c) Determine the number of conjugacy classes in S_{5}, and the size of each conjugacy class, and describe the centralizer in G of a chosen representative of each conjugacy class. Solution: Here is a transversal for the conjugacy classes: id, (1, 2), (1, 2, 3), (1, 2, 3, 4), (1, 2, 3, 4, 5), (1, 2)(3, 4, 5) and (1, 2)(3, 4). The corresponding conjugacy classes have sizes 1, 10, 20, 30, 24, 20 and 15 and happily these numbers sum to 120 = 5!. The corresponding centralizers must have orders 120, 12, 6, 4, 5, 6 and 8. This enables us to identify them as G = S_{5}, ⟨(1, 2), (3, 4, 5), (3, 4)⟩, ⟨(1, 2, 3), (4, 5)⟩, ⟨(1, 2, 3, 4)⟩, ⟨(1, 2, 3, 4, 5)⟩, ⟨(1, 2), (3, 4, 5)⟩ and ⟨(1, 2, 3, 4), (1, 2)⟩.

2. Let G = S_{n}. Let x = (1, 2, …, n) ∈ G. Prove that C_{G}(x) = ⟨x⟩. Solution: The conjugacy class of x has size (n−1)! so C_{G}(x) has index (n − 1)! and therefore order n. However ⟨x⟩ ≤ C_{G}(x) and |⟨x⟩| = n so C_{G}(x) = ⟨x⟩.

3. Show that in S_{4} there is a non-identity element y such that C_{G}(y) ≠ ⟨y⟩. Solution: (1,2) centralizes (1,2)(3,4) (and vice versa).

4. Suppose that G is a finite group. Show that the number of elements in each conjugacy class of G must divide G. Solution: Suppose that x ∈ G and that C is the conjugacy class of x. We know that |C| = |G : C_{G}(x)| but |G| = |G : C_{G}(x)| · |C_{G}(x)| and we are done.

5. Let G be a group with a subgroup H such that g^{−1}Hg ⊆ H for every g ∈ G. Prove that g^{−1}Hg = H for every g ∈ G. Solution: This is a standard trick. If g^{−1}Hg ⊆ H for every g ∈ G, then replacing g by g^{−1} we obtain that gHg^{−1} ⊆ H for every g ∈ G. Therefore H ⊆ g^{−1}Hg for every g ∈ G and so H = g^{−1}Hg for every g ∈ G.

6. (Challenge) Does there exist a group G containing an element g and a subgroup H such that g^{−1}Hg ⊆ H but g^{−1}Hg ≠ H. Solution: Yes. Let G be the set of 2 by 2 invertible rational matrices. Let H be the set of upper unitriangular matrices with integer entry in the top right position. Let g be the diagonal matrix diag(1, 2).

7. Suppose that G is a group and that H ≤ G. Choose g ∈ G. Prove that g^{−1}Hg ≤ G. Solution: g^{−1}1g ∈ g^{−1}Hg ≠ ∅. If g^{−1}ag, g^{−1}bg ∈ g^{−1}Hg, then g^{−1}agg^{−1}bg = g^{−1}abg ∈ g^{−1}Hg, As for inversion, the inverse of g^{−1}ag is g^{−1}a^{−1}g ∈ g^{−1}Hg,

8. Let G be a finite group of order n which has t conjugacy classes. Elements x and y are each selected uniformly at random from G. What is the probability that x and y commute? Does this make sense for abelian group s? Solution: Let Γ = {(x, y)|x, y ∈ G, xy = yx}. Now the required probability is

p = \frac{1}{|G|^{2}} |Γ| = \frac{1}{|G|^{2}} ∑_{x∈G} |C_{G}(x)|

and by not Burnside’s counting lemma this is t/n.

9. Show that a finite group with exactly two conjugacy classes must have two elements. Solution: We have proved that the number of elements in a conjugacy class of a finite group must divide the order of the group. If the group order is n this forces n − 1 to divide n which forces n = 2.

10. Let G be a containing H a subgroup of finite index. Let S = {x^{−1}Hx | x ∈ G}. Let G act on S by conjugation so if K ∈ S then K ·g = g^{−1}Kg. Verify that this is a group action, and deduce that |S| = |G : N_{G}(H)| where N_{G}(H) = {g ∈ G | gH = Hg}. Deduce that |S| is finite and divides |G : H|. Solution: Suppose that A ∈ S so that A = x^{−1}Hx for some x ∈ G. Take elements a, b ∈ G. Now A · 1 = 1^{−1}A1 = A and A · ab = b^{−1}a^{−1}Aab = (A · a) · b. Thus we have a group action. There is a single orbit of this action, and the orbit is in bijective correspondence with the right cosets of the stabilizer of H. This is {g ∈ G | g^{−1}Hg = H} a group known as N_{G}(H) and called the normalizer of H in G (and of course H ≤ N_{G}(H) and in fact H ⊴ N_{G}(H)). Thus |S| = |G : N_{G}(H)| but of course |G : H| = |G : N_{G}(H)| · |N_{G}(H) : H| and we were given that |G : H| < ∞. Thus |S| is finite and a divisor of |G : H|.