Math 306 Topics in Algebra, Spring 203 Homework 7 Solutions () (5 pts) Let G be a finite group. Show that the function defines an inner product on C[G]. We have Also Lastly, we have C[G] C[G] C c f + c 2 f 2, f 3 = (f, f 2 ) f, f 2 = = = f (g)f 2 (g) (c f + c 2 f 2 )(g)f 3 (g) ( c f (g)f 3 (g) + c 2 f 2 (g)f 3 (g) ) = c c f (g)f 3 (g) + c 2 f 2 (g)f 3 (g) f (g)f 3 (g) + c 2 = c f, f 3 + c 2 f 2, f 3 f, f 2 = f, f = f (g)f 2 (g) f (g)f 2 (g) f (g) f 2 (g) f (g) f 2 (g) = f 2, f f(g)f(g) = f 2 (g)f 3 (g) f(g) 2. This sum is greater than or equal to zero and is zero if and only if f(g) = 0 for all g G. (2) Instead of taking the trace of φ g to define the character, one might try to do the same by taking the determinant of φ g instead. This problem shows that this is not as useful since such a character would tell us nothing about non-abelian simple groups (and these are important). For φ a representation of a finite group G, define a function det φ: G C g (det φ)(g) = det(φ g ) (a) (4 pts) Show det φ is a representatation (and hence it s a character since characters are same as representations for one-dimensional representations).
(b) (5 pts) Show that if G is a non-abelian simple group, then det φ is the trivial character. (a) Follows from the multiplicativity of the determinant. (b) Since det φ is a homomorphism, its kernel is a normal subgroup of G. Since G is simple, it must be that ker(det φ) = e} or ker(det φ) = G. If the former is true, then det φ is injective, and its image is a subgroup of C, which is abelian. This G would be isomorphic to an abelian group, but this cannot by by assumption. So it must be that ker(det φ) = G, in which case det φ is the trivial homomorphism. (3) (4 pts) Show that χ φ ψ = χ φ + χ ψ. The matrix for each (φ ψ) g is a block matrix with blocks φ g and ψ g. The trace of a matrix like that is the sum of the traces of the blocks. Hence χ φ + χ ψ. (4) (5 pts/part) (a) Suppose A is a matrix over C of finite order, i.e. A n = I for some positive integer n. Show that the eigenvalues λ i of A are the nth roots of unity, namely they satisfy λ n i =. (Hint: Use the result that A is diagonalizable, which in turn follows from fact that for a representation of a finite group G, there exists a matrix T such that T φ g T is diagonal for all g G. Then look up what diagonalizability has to do with eigenvalues.) (b) Prove that for an irreducible representation of a finite group G, χ φ (g) = λ + + λ d, where λ i are the eigenvalues of φ g and d is the dimension of φ. (c) Show that, if a complex number ω is a root of unity, then ω = ω. (d) Prove that χ φ (g ) = χ φ (g). (a) If A is diagonalizable, then it is a basic result of linear algebra that its diagonal entries are its (distinct) eigenvalues λ i. Denote this diagonal matrix by D. Then D n = (T AT ) n = T A n T = T IT = T T = I But powers of a diagonal matrix are obtained by taking powers of its diagonal entries. Thus it follows that λ n i = as desired. (b) Since G is finite, φ g has finite order. Hence, by the previous part, there exists a matrix T such that T φ g T = D, where D is diagonal with eigenvalues λ i as the diagonal entries. Since characters take the same value on similar matrices, we have χ φ (g) = Tr(φ g ) = Tr(T φ g T ) = Tr(D) = λ + + λ d. (c) A conjugate of a complex number re iθ is re iθ. If ω is a root of unity, then it has the form e iθ/n for some n (but the important thing is that r = ; if r, then (re iθ ) n = r n e iθn, and this number could not have size (modulus) ). Then ωω = e iθ/n e iθ/n = e 0 =. So ω is the inverse of ω. (d) Consider χ φ (g ) = Tr(φ g ) = Tr(φ g ). Since trace is same on similar matrices, by part (a) we can replace φ g by the diagonal matrix with diagonal entries the eigenvalues λ i, and consequently we can replace φ g by the diagonal matrix whose entries are (since to obtain the inverse of a diagonal matrix, you take the inverse of the diagonal entries). Thus λ i Tr(φ g ) = i λ i.
Since λ i are roots of unity, by part (b) we have Tr(φ g ) = i λ i. But this sum is precisely Tr(φ g ) (or rather the trace of the conjugate of the diagonal matrix replacing φ g ). I.e. the sum is precisely χ φ. (5) (5 pts) Recall that, in the proof of the theorem that a representation φ is irreducible iff χ φ, χ φ =, we assumed φ m φ m s φ s and then claimed that Show that this equation indeed holds. χ φ, χ φ = m 2 + + m 2 s. The matrix for φ is a block matrix whose blocks are the matrices for m i φ i. It is then immediate that s s s χ φ = Tr(φ) = Tr(m i φ i ) = m i Tr(φ i ) = m i χ φi (we can pull out m i since it appears in each diagonal term of the ith block). So then we have s s χ φ, χ φ = m i χ φi, m j χ φj. However, inner product is linear by definition (and you verified this in an earlier problem for the inner product we re using here), so that the above can be computed by foiling the inner product. We thus have χ φ, χ φ = m i χ φi, m j χ φj. i,j n By orthogonality relations, the above inner products are only nonzero when i = j in which case they are and we get χ φ, χ φ = m i χ φi, m i χ φi = m 2 i χ φi, χ φi = i n j=i i n as desired (because of linearity, we were able to take each m i out as well). (6) (a) (4 pts) Show that a finite group G is abelian if and only if it has irreducible representations (over C). (b) (5 pts) Use part (a) to show Z/nZ is abelian. Do this without using that Z/nZ has n conjugacy classes. i n (a) A finite group is abelian iff each element is its own conjugacy class. Thus = Cl(G). But we know by a theorem from class that Cl(G) is the number of irreducible representations of G. (b) For 0 k n, define representations φ k : Z/nZ C m e 2πimk/n As we know from class and previous homework, each φ k is a well-defined irreducible representation, and all n of them are distinct. By the previous part, it follows that Z/nZ is abelian. (7) This problem explores the regular representation over C and the associated character. The formula you will in part in part (d) is an important application of representation theory to group theory. Recall from an earlier homework that the (left) regular representation of a finite group G is given by L: G GL(F [G]) g L g (v) = gv. m 2 i
(We also defined it in class using the dual vector space of F [G], but for this problem, we ll stick to the original definition above.) (a) (7 pts) Prove that the character of the regular representation is given by χ L : G C g χ L (g) =, g =, 0, g. (b) (5 pts) We observed in class that every representation φ of G breaks up as φ m φ m s φ s, where φ,..., φ s is the complete set of irreducible representations of G (some of the m i s might be zero). Prove that χ φ, χ φi = m i. (Hint: Use an earlier exercise and the linearity of the inner product.) (c) (7 pts) Suppose d i are the degrees of the irreducible representations φ i of G. Show that if φ = L g, the m i s from the previous part are precisely the degrees d i. In other words, show that the decomposition holds. (d) (4 pts) Prove that L d φ d s φ s = d 2 + + d 2 s. (a) Let G = g,..., g n }, where n =. Let v = g j (each v is a linear combination of g j s and in particular it suffices to define L on the g j since they are the basis). Then L g (g j ) = gg j. Thus if we regard L g as a matrix with respect to the basis G with the ordering g,..., g n, we have (L g ) ij =, g i = gg j ; 0, otherwise =, g = g i gj ; 0, otherwise. This is because g sends g j to g i, so, to represent this in matrix notation, we think of g j and g i as basis vectors with zeros except in the jth or ith slot; then the matrix that sends g j vector to g i vector is precisely (L g ) ij as defined above. In particular, when i = j, we get, g = ; (L g ) ii = 0, g. From this it then follows that (b) It follows from an earlier problem that Then χ L (g) = T r(l g ) =, g = ; 0, g. χ φ = m χ φ + + m s χ φs. χ φ, χ φi = m χ φ + + m s χ φs, χ φi = m χ φ, χ φi + + m i χ φi, χ φi + + m s χ φs, χ φi = m 0 + + m i + + m s 0 = m i The next to last equality is the orthogonality relations for characters.
(c) Since the result from the previous part in fact shows that the decomposition of a representation is unique and that a representation is determined up to equivalence to its character, it suffices to check the inner product of χ L with the χ φi : χ L, χ φi = χ L (g)χ φi (g) = χ i() = χ φi () = deg φ i = d i. Here we have used: Result about χ L (g) from part (a); In class we observed that χ φi () is the trace of the identity matrix, and it hence gives the dimension of the vector space, i.e. the degree of the representation φ i ; and consequently χ φi () is a real number, and so χ φi () = χ φi (). (d) From the previous part, we know L d φ d s φ s. Consequently, χ L = d χ φ + + d s χ φs. (we used this in part (b) already). Evaluating this equation at, we have χ L () = d χ φ () + + d s χ φs (). We know from part (a) that χ L () = and we recalled in part (c) that χ φi () = d i. So the above equation becomes = d d + + d s d s, as desired.