Solutions to Example Sheet 2 All groups are finite; all representations are over complex vector spaces. Group G acts on set X orbit of x X is xG := gx X g G ◦ 2 f 2 j 2 g stabilizer of x X is Gx := g G gx = x ◦ 2 f 2 j g fixed point set of g G is Xg := x X gx = x ◦ 2 f 2 j g Special case: G acts on itself via conjugation G 1 conjugacy class of x G is x := gxg− G g G ◦ 2 f 2 j 2 g x centralizer of x G is CG(x) := Gx = G = g G gx = xg ◦ 2 f 2 j g center of G is Z(G) := g G gx = xg for all x G = x G CG(x) ◦ f 2 j 2 g T 2 1 1 commutator of x and y is [x; y] := x− y− xy ◦ commutator subgroup of G is G0 := [x; y] x; y G ◦ h j 2 i G Recall: x = [G : Gx] j j 1. (a) Let ρ be a representation of G with character χ, and suppose dimension of ρ is d. Show ker ρ = g G χ(g) = d . f 2 j g (b) Show χ(g) d for all g G, and that if g = d, then ρ(g) = λId, where λ is a root of unity. j j ≤ 2 j j Solution. As in Problem 7, Example Sheet 1 (applied to the cyclic subgroup g G), we may h i E pick a basis so that ρ(g) = diag(λ1; : : : ; λd) where the λi's are nth roots of unity, n = ordG(g) = g . jh ij Note that χ(g) is basis independent and so χ(g) = λ1 + + λd. ··· (a) Clearly if g ker ρ, then χ(g) = tr ρ(g) = tr(Id) = d. Conversely, d = χ(g) = λ1 + + λd could 2 ··· only happen if λi = 1 for all i = 1; : : : ; d. Hence ρ(g) = Id. (b) By the triangle inequality χ(g) = λ1 + + λd λ1 + + λd = d. Equality occurs iff j j j ··· j ≤ j j ···iθ j j arg(λ1) = = arg(λd) = θ for some θ. So λ1 = = λd = e which must be a root of unity ··· iθ ··· since all λi's are and so we have ρ(g) = e Id. 2. Let χ be a character of G, and suppose g G has order 2, ie. g2 = 1. Show χ(g) Z, and χ(g) χ(1) (mod 2). 2 2 ≡ Solution. Let d be the dimension of χ. As in the solution of Problem 1, χ(g) is a sum of d square roots of unity, ie. 1. Since 1 1 (mod 2), χ(g) d = χ(1) (mod 2). ± − ≡ ≡ 2 2 2 3. (a) Let G = 1; i; j; k be the \quaternion group" Q8,(ij = k = ji, i = j = k = 1). Regarding{±G as± a subgroup± ± g of the quaternions H = C Cj, we get a two-dimensional− representation− of G. Show this is irreducible. ⊕ (b) Conclude the remaining irreducible representations of G are one-dimensional. Find them. Write the character table of G. 2 (c) Let G = D8 = symmetries of the square. Show G acts on C irreducibly, and determine the character table of G. (d) Compare the character tables of D8 and Q8. Comment? 1 1 1 Solution . First, a few easy facts about Q8. Clearly 1 are in Z(Q8); jij− = kik− = i ± − and so iQ8 = i . By the symmetry in the relations on i; j; k, we must also have jQ8 = j Q8 {± g 1 1 2 2 {± g and k = k and thus Z(Q8) = 1 . Since [i; j] = i− j− ij = ijij = i j = 1, we have {± g {± g − − [a; b] = 1 for every a; b i; j; k , a = b, again by symmetry and 1 Z(Q8). So Q0 = 1 − 2 {± ± ± g 6 ± 2 8 {± g and Q8=Q0 = 1; i; j; ij = Z=2 Z=2. The `header' of the character table is: 8 f g ∼ × 1The solution to this problem is worked out in full details. You are not required to write most of these steps down in your solutions (and please don't). The obvious steps will be dropped in future problems. 1 xQ8 1 1 i j k − {± g {± g {± g CQ (x) 8 8 4 4 4 j 8 j (a) If we regard H as a left C-vector space (ie. we write the scalars on the left), then right multipli- cation by elements in Q8 gives rise to a Q8-action on H, (ie. Q8 H H,(g; α) αg). Pick × ! 7! the basis 1; j and determine the Q8 action (to make things clearer, is used to denote group action andf isg used to denote scalar multiplication): ◦ · 1 ( i) = i 1 + 0 j 1 ( j) = 0 1 1 j 1 ( k) = 0 1 + ( i) j ◦ ± ± · · ◦ ± · ± · ◦ ± · ± · j ( i) = 0 1 + ( i) j j ( j) = 1 1 + 0 j j ( k) = i 1 + 0 j ◦ ± · ∓ · ◦ ± ∓ · · ◦ ± ± · · and this gives a matrix representation ρ : Q8 GL2(C) with ! 1 0 i 0 0 1 0 i ρ( 1) = ± ; ρ( i) = ± ; ρ( j) = ± ; ρ( k) = ± : ± 0 1 ± 0 i ± 1 0 ± i 0 ± ∓ ∓ ± Computing the trace gives the following entry in the character table: xQ8 1 1 i j k − {± g {± g {± g χρ 2 2 0 0 0 − ρ is irreducible since 1 χρ; χρ = χρ(x)χρ(x) h i Q8 X x Q8 j j 2 1 = χρ(x)χρ(x) X CQ8 (x) xQ8 j j 1 22 ( 2)2 0 0 0 = + − + + + = 1: 8 8 8 4 4 4 (b) Let d1; d2; d3 be the dimensions of the other non-trivial irreducible characters. Then since 2 2 2 2 2 1 + d + d + d + 2 = Q8 = 8; 1 2 3 j j we must have d1 = d2 = d3 = 1. Hence the remaining irreducible representations are one- Z dimensional. By Problem 7, the linear characters are exactly those lifted from Q8=Q80 ∼= =2 Z=2. Recall from the Problem 9, Example Sheet 1 that the irreducible (thus one-dimensional)× representations ( linear characters) of Z=2 Z=2 are: ≡ × Z=2 Z=2 (0; 0) (1; 0) (0; 1) (1; 1) × 1 1 1 1 χ1 1 1 1 1 − − χ2 1 1 1 1 − − χ3 1 1 1 1 − − Lifting the characters on Z=2 Z=2 to Q8 means to see where the map × π χ Q8 Q8=Q0 ' Z=2 Z=2 C× −! 8 −! × −! sends the conjugacy classes in Q8. For instance if we look at χ1, then 1 1 (0; 0) 1 i i (1; 0) 1 k ij (1; 1) 1 7! 7! 7! ± 7! 7! 7! − ± 7! 7! 7! − 1 1 (0; 0) 1 j 1 (0; 1) 1 − 7! 7! 7! ± 7! 7! 7! and we get (abusing notation, we use χ1 for the lifted character too): xQ8 1 1 i j k − {± g {± g {± g χ1 1 1 1 1 1 − − Repeating this for χ2 and χ3, we obtain the full character table of Q8: 2 xQ8 1 1 i j k − {± g {± g {± g 1 1 1 1 1 χ1 1 1 1 1 1 − − χ2 1 1 1 1 1 − − χ3 1 1 1 1 1 − − χρ 2 2 0 0 0 − (c) See: Some notes on Dn, An and Sn. 2 (d) D8 Q8 since D8 has two elements of order 2: a and b, while Q8 has only one element of order 2: 1. So the character table of a group does not determine the group up to isomorphism. − 4. Determine the character table for D12 = symmetries of the hexagon. Solution. See: Some notes on Dn, An and Sn. 5. Determine the character table for D10 = symmetries on the 5-gon. For every representation V of D10, we may regard it as a representation of Z=5 , D10. Determine, for each irreducible representation ! V of D10, how it decomposes into irreducible characters for Z=5. Solution. See: Some notes on Dn, An and Sn. 6. Determine the character table for the alternating group A4. Solution. See: Some notes on Dn, An and Sn. 7. Describe the commutator subgroup of a group in terms of the character table of G. Solution. We claim that the rows in the character table of G with 1 in the first column are precisely the characters lifted from G0 E G. If χ, a character of G, is lifted from χ0, a character of G=G0, then χ = χ0 π and so χ is one-dimensional, ie. χ(1) = 1. Conversely, if χ(1) = 1, then we have to find a ◦ χ0 : G=G0 C× such that χ = χ0 π. The obvious choice would of course be ! ◦ χ0 G=G0 C×; xG0 χ(x): −! 7! Since our definition of χ0 involves a choice of coset representative, we have to show that this is a well-defined map, ie. the image is independent of the coset representative we chose: χ(x) = χ(xg) for all g G0 or equivalently χ(g) = 1 for all g G0. The last condition is just G0 ker(χ0) and this is 2 2 ≤ obvious since G= ker(χ0) C× and so is abelian, thus G0 ker(χ0). Observe also the characters lifted ≤ ≤ from the irreducible character of G=G0 are all distinct (clear from how a lifted character is defined) and so the character table of G contains exactly G=G0 rows of dimension one. j j We further claim that G = ker(χ): (7.1) 0 \ χ(1)=1 In the earlier part, we have already shown that G0 ker(χ) for all one-dimensional character χ.