REPRESENTATIONS AND CHARACTERS OF FINITE GROUPS

SUMMER PROJECT REPRESENTATIONS AND CHARACTERS OF FINITE GROUPS September 29, 2017 Miriam Norris School of Mathematics

Contents 0.1 Introduction........................................ 2 0.2 Representations of Groups............................... 2 0.3 FG-modules and Reducibility.............................. 6 0.3.1 Reducibility.................................... 7 0.3.2 Group Algebra................................... 7 0.3.3 FG-homomorphisms............................... 8 0.3.4 Maschke s Theorem............................... 9 0.3.5 Schur s Lemma.................................. 10 0.3.6 Irreducible Modules............................... 11 0.4 Conjugacy Classes.................................... 13 0.5 Characters......................................... 14 0.5.1 Inner Products and Orthogonality....................... 16 0.5.2 Character tables.................................. 16 0.5.3 Normal Subgroups and Lifted characters................... 18 0.6 Acknowledgements.................................... 21 References............................................ 22 1

0.1 INTRODUCTION This report covers the key elements of representation theory that I looked at during my summer project. I have constructed it so that it is hopefully clear how the earlier groundwork built up to allow me to find some interesting results. I wanted to center the report around two groups, the symmetries of a square, D 8 and the quaternions, Q 4 because of there interesting relationship that this report should demonstrate. 0.2 REPRESENTATIONS OF GROUPS Before we delve into representations recall the structure of a group. The following theorems and definitions are based on James and Leibeck Representations and Characters of Finite Groups [1]. Definition 1. A group, G, is a set of elements with some operation that combining elements in G to form another element in G such that it satisfies the following properties; 1. It must be associative such that for all g,h,k in G, i.e. (g h)k = g (hk); 2. There must exist an identity element e in G such that for all g in G, eg = g e = g ; 3. Every element g in G must have an inverse element g 1 in G such that g g 1 = g 1 g = e. Given this definition consider the set of invertible n n matrices over some field, F, along with standard matrix multiplication. If we combine any matrices, A and B, in this set we get another invertible matrix, AB, since (AB) 1 = A 1 B 1 and A and B are invertible. Since the three properties above are clearly satisfied this set is a group, we call it the general linear group, and write, GL(n, F ). A representation of a group over some field is a relationship between the group and the general linear group over this field. To define it properly we need to specify this relationship. Definition 2. Suppose G and H are groups. A group homomorphism is a function, f, from G to H, f : G H such that for all g,h G (g h)f = (g f )(h f ). Definition 3. A representation of a group, G, over a field, F, is a homomorphism, ρ, from G to GL(n,F ). Where the integern is the degree of ρ. Example 1. Consider the set of symmetries of a square, call it D 8, where a is the rotation by π/2 and b the reflection through the vertical line. All other symmetries can be generated by Page 2 of 22

combinations of powers of a and b. It is useful for describe the set of symmetries therefore using its generators, we write D 8 =< a,b : a 4 = b 2 = 1,b 1 ab = a 1 >. It is easy to see that this set satisfies the conditions above and is therefore a group. Example 2. Consider the quaternion group of order 8, we call it Q 8. We say Q 8 is the group generated by element a and b determine by the following conditions, a 4 = 1, a 2 = b 2,b 1 ab = a 1. We write this Q 8 =< a,b : a 4 = 1, a 2 = b 2,b 1 ab = a 1 >. Although this group is similar to the one above notice that there is a distinct difference in the relations as b 2 1. Consider Example 1., clearly if we construct a function that sends every element of D 8 to the 1 1 matrix (1) it satisfies Definition 3. This is called the trivial representation. In fact for any integer n we could construct a representation that sends every element in the group to the n n identity matrix I n. It therefore becomes useful to determine how many elements are sent to I n by the representation ρ, so we define the kernel of ρ as Kerρ = {g G : g ρ = I n }. This gives rise to the following definition. Definition 4. A representation ρ is said to be faithful if Kerρ = {1}, that is if the only element of G sent to I n by ρ is the identity element of G. It will also prove useful to define the relationships between representations themselves. Suppose we have some representation, ρ, of a group, G, and an invertible matrix T. Consider the function σ = T 1 ρt, for elements g,h in G we have, (g h)σ = T 1 ((g h)ρ)t = T 1 ((g ρ)(hρ))t = T 1 (g ρ)t T 1 (hρ)t = (g σ)(hσ) Therefore by Definition 3 σ is itself a representation of G. We define the relationship between these two representations as follows; Definition 5. Representations ρ : G GL(n,F ) and σ : G GL(n,F ) are said to be equivalent if there exists an nxn matrix T such that for all g G g σ = T 1 (g ρ)t. To sum up this section we can consider the following example. Page 3 of 22

Example 3. Let G = D 8 = a,b : a 4 = b 2 = 1,b 1 ab = a 1. Define the matrices A,B,C,D over C by ( ) ( ) i 0 0 1 A =, B =, 0 i 1 0 ( ) ( ) 0 1 1 0 C =, D =. 1 0 0 1 Prove that each of the functions ρ k : G GL(2,C) (k = 1,2,3), given by ρ 1 :a r b s A r B s, ρ 2 :a r b s A 2r ( B) s, ρ 3 :a r b s C r D s (0 r 5,0 s 1), is a representation of G. Which of these representations are faithful? Which are equivalent. Solution 1. To show each of these are representations we need to prove that for all i ρ i is a homomorphism. Take any pair of matrices A and B such that A 4 = B 2 = I n,b 1 AB = A 1. We know that for 0 r, t 3 0 s,k 1 a r b s a t b k = a i b j for some i and j detrmined by repeatedly using a 4 = b 2 = 1,b 1 ab = a 1.Since A and B also satisfy A 4 = B 2 = I n,b 1 AB = A 1 it follows that A r B s A t B k = A i B j. Therefore if ρ : a i b j A i B j then (a r b s a t b k )ρ = (a i b j )ρ = A i B j = A r B s A t B k = (a r b s )ρ (a t b k )ρ, hence ρ is a homomorphism and therefore a representation of G. It therefore remains only to show that for A,B,C,D above A 4 = B 2 = I n,b 1 AB = A 1, A 8 = B 2 = I n,b 1 A 2 B = A 2,C 4 = D 2 = I n,d 1 C D = C 1. It is clear that for 0 r 3,0 s 1 matrices A r B s and C r D s are all different, therefore ρ 1 and ρ 3 are faithful. However a 2 ρ 2 = I so ρ 2 is not faithful. Page 4 of 22

Lastly we can see that ρ 1 and ρ 3 are equivalent since for ( ) 1 1 T = i i T 1 (g ρ 3 )T = g ρ 1 for all g G. Page 5 of 22

0.3 FG-MODULES AND REDUCIBILITY It will be useful to now define an FG-module and its relationship to representations of G over F. Definition 6. A vector space, V, over F is an FG-module for some G if multiplication v g for some v V, g G is defined satisfying the following conditions for all v,u V, g,h G and λ F. 1. v g V 2. v(g h) = (v g )h 3. v1 = v 4. (λv)g = λ(v g ) 5. (u + v)g = ug + v g The permutation module is a particularly important example of an FG-module. If G is a subgroup of S n for some n, as all the examples above have been, then the permutation module, V, with basis v 1...v n has multiplication defined v i g = v i g fo all i and g G. These vector spaces tie in with representations through the following theorem Theorem 1. If ρ : G GL(n, F ) is a representation of G over F then V is a F G-module multiplication is defined v g = v(g ρ) and there is a basis β of V such that g ρ = [g ] β. Furthermore if V is an FG-module and β a basis the θ : [g ] β is a representation of G over F. The following example demonstrates this result. Example 4. Consider the representation ρ 1 in Example 3. For G = D 8 = a,b : a 4 = b 2 = 1,b 1 ab = a 1 we have, ( ) ( ) i 0 0 1 aρ =,bρ = 0 i 1 0 If we set V = F 2 then Theorem 1. gives us that V becomes an FG-module if we define v g = v(g ρ)(v V, g G). Therefore if v 1, v 2 is the basis, (1,0),(0,1) of V, then we have, v 1 a = i v 1, v 1 b = v 2, v 2 a = i v 2, v 2 b = v 1. Page 6 of 22

If β denotes the basis v 1, v 2 then, by Theorem 1. again, the representation g [g ] β (g G) is just the representation ρ 1 from Example 3. An important example of a FG-module for subgroups of the symmetric group S n is defined as follows and we will come back to later on in this report. Definition 7. Let G be a subgroup of S n. The FG-module V with basis v 1,..., v n such that v i g = v i g for all i and all g G is called the permutation module of G over F. The idea of equivalence we defined for representations corresponds to F G-modules through the following theorem. Theorem 2. Let V be an FG-module with basis β, let ρ be the representation of G over F defined by, ρ : g [g ] β. If β is another basis of V then the representation σ : g [g ] β of G is equivalent to ρ. If θ is a representation of G which is equivalent to ρ then there is a basis β of V such that, θ : g [g ] β. 0.3.1 Reducibility The concepts of reducibility and irreducibility will prove very important for representation theory, so we need the following definitions. Definition 8. A subset of the FG-module V, call it W, is an FG-submodule if it is closed under multiplication on the right by elements of G. Definition 9. If V is an FG-module with an FG-submodule W such that W is not equal to {0} or V then V is said to be reducible. If no such W exist then V is said to be irreducible A representation ρ : G GL(n,F ) is therefore said to be irreducible if is corresponding FGmodule with multiplication defined by v g = v(g ρ) is irreducible. 0.3.2 Group Algebra If F is R or C and G is a finite group with elements g 1,..., g n, then we can define a vector space, FG, over F with basis g 1,..., g n. All elements of FG are of the form λ 1 g 1 +... + λ n g n for all λ i F. Page 7 of 22

Definition 10. The vector space FG along with multiplication defined as is called the group algebra of G over F. ( λ g g )( µ h h) = λ g µ h (g h), g G h G g,h G We can use this definition to define an FG-module that will prove to be fundamental for the next few sections. Definition 11. Let G be a finite group and F be R or C. The vector space FG as defined above along with natural multiplication, v g is called the regular F G-module. Its corresponding representation formed by taking the basis β to be the natural basis of FG and sending g [g ] β is called the regular representation of G over F. 0.3.3 FG-homomorphisms Definition 12. An FG-homomorphism, θ, is a linear transformation from one FG-module, V to another, W such that (v g )θ = (vθ)g for all v V, g G. If θ is also invertible then it is an FG-isomorphism. Example 5. Consider the following elements in CD 8, the group algebra of D 8 over G, u 1 = 1 i a 1a 2 + i a 3, u 2 = b + i ba ba 2 i ba 3 v 1 = 1 + i a a 2 i a 3, v 2 = b i ba ba 2 + i ba 3 Let U = sp(u 1,u 2 ),V = sp(v 1, v 2 ), it is easy to check that with multiplication defined in the natural way these vector spaces satisfy Definition 6. and so are FG-modules. Prove that the function θ : u 1 v 2 u 2 v 1 is an FG-homomorphism. Solution 2. To see this we only need that (ua)θ = (uθ)a and (ub)θ = (uθ)b for all u U since these elements generate the group. Take u 1,u 2, v 1, v 2 to be the basis element respectively of U and V. On these basis elements we have u 1 a = i u 1, u 2 a = i u 2, u 1 b = u 2, u 2 b = u 1 v 1 a = i v 1, v 2 a = i v 2, v 1 b = v 2, v 2 b = v 1 Page 8 of 22

We can write any element u U as a product of the basis element, u = αu 1 +βu 2. Putting this together we get, (ua)θ = (αu 1 a + βu 2 a)θ = (αi u 1 βi u 2 )θ = αi v 2 βi v 1 (uθ)a = (αu 1 θ + βu 2 θ)a = (αv 2 + βv 1 )a = αi v 2 βi v 1 (ub)θ = (αu 1 b + βu 2 b)θ = (αu 2 + βu 1 )θ = αv 1 + βv 2 (uθ)b = (αu 1 θ + βu 2 θ)b = (αv 2 + βv 1 )b = αv 1 + βv 2. Therefore θ is an FG-homomorphism. In fact it is an FG-isomorphism as θ is clearly invertible. 0.3.4 Maschke s Theorem Maschke s Theorem is one of the fundamental building blocks of representation theory because of its consequences for understanding the structure of a representation. Theorem 3 (Maschke s Theorem). Let G be a finite group and let F be R or C and let V be an FG-module. If U is an FG-submodule of V, then there exists a FG-submodule w of V, such that V = U W. This result means that if ρ is a reducible representation of a finite group G over F of degree n, then ρ is equivalent to some representation of the form ( ) Ag 0 g 0 B g For our first important consequence we need to define the following term. Definition 13. An FG-module V is completely reducible if V = U 1... U r where each U i is an irreducible FG-submodule of V. Theorem 4. If G is a finite group and F = Cor R then every non-zero FG-module is completely reducible. Proof. For V a non-zero FG-module we will prove this by induction on dimv. If dimv = 1 then V is necessarily irreducible and if V is irreducible then the result holds. Therefore suppose V is reducible, meaning there exists an FG-submodule W, not equal to {0} or V, and that dimv = n. By Maschke s Theorem, there is another F G-submodule U such that Page 9 of 22

V = W U. Since dimu dimv and dimw dimv, by induction, W = W 1.. W r,u = U 1... U r, where each W i,u i are irreducible FG-modules. Combining this we get V = W 1.. W r U 1... U r a direct sum of irreducible FG-submodules. Example 6. Let G be a finite group and let ρ : G GL(2,C) be a representation of G. Suppose there are elements g, h in G such that the matrices g ρ and hρ do not commute. Prove that ρ is irreducible. Solution 3. Suppose that ρ is reducible and has a sub representation of degree 1. It therefore follows from Maschke s Theorem that ρ is equivalent to some representation σ where ( ) λg 0 g σ =. 0 µ g Since all diagonal matrices commute all matrices g σ and hσ commute and therefore so do g ρ and hρ. Therefore if there are elements g, h in G such that the matrices g ρ and hρ do not commute ρ must be irreducible. Example 7. Consider the CD 8 -module, U, from Example 5. prove that it is irreducible. Solution 4. Using Theorem 1. we see that the representation corresponding to U it ρ 1 from Example 4. Therefore we have ( ) ( ) 0 i 0 i (aρ)(bρ) = = (bρ)(aρ). i 0 i 0 So the result follows from Example 6. 0.3.5 Schur s Lemma Theorem 5. Let V and W be irreducible CG-modules. 1. If θ : V W is a CG-homomorphism, then wither θ is a CG-isomorphism, or vθ = 0 for all v V. 2. If θ : V V is a CG-isomorphism, the θ is a scalar multiple of the identity endomorphism 1 v. Let G be finite abelian group and V an irreducible CG-module. For x G we have v g x = v xg for all g G. Therefore the endomorphism v v x is CG-homomorphism, so by Schur s Page 10 of 22

ω = e πi /2 v 0 = 1 + a + a 2 + a 3 w 0 = bv 0 Miriam Norris Lemma this endomorphism is a scalar multiple of th identity, call it λ x 1 v. Therefore every subspace of V is CG-submodule but since V is irreducible this implies di mv = 1. we have therefore shown that for every finite group, G, every irreducible CG-module has dimensions 1. The following theorem follows from this. Theorem 6. Let G be the abelian group C n1... C nr with c i,1 i r the generators of C ni. If g i = (1,...,c i,...,1) with c i in the i th position we can rewrite G = g 1,..., g r with g n i i representations satisfying (g i 1 1...g i r r )ρ = (λ 1 i 1...λ i r r ) =. The where λ i is the n i th root of unity are irreducible and have degree 1. There are G of these representations, and every irreducible representation of G over C is equivalent to precisely one of them. 0.3.6 Irreducible Modules Theorem 7. Let CG be the regular CG-module, and write CG = U 1... U r, a direct sum of irreducible CG-submodules. Then every irreducible CG-module is isomorphic to one of the CG-modules U i. Given the theorem above it is extremely valuable to be able to find these irreducible CG-modules for any finite group, G and their corresponding representations. The following example illustrates how we can use the theory from this section to do exactly that. Example 8. Find all the irreducible representations of D 8 over C. Solution 5. To do this we want to decompose the group algebra CD 8 as a direct sum of irreducible CD 8 -submodules. We begin by defining the following elements of CD 8, where v 1 = 1 + ω 3 a + ω 2 a 2 + ωa 3 w 1 = bv 1 v 2 = 1 + ω 2 a + a 2 + ω 2 a 3 w 2 = bv 2 v 3 = 1 + ωa + ω 2 a 2 + ω 3 a 3 w 3 = bv 3. It is easy to see that v i a = ω i v i for i = 0,1,2,3 and likewise with w i, therefore we can deduce Page 11 of 22

that sp(v i ), sp(w i ) are C < a >-modules. We also have v 0 b = w 0 w 0 b = v 0 v 1 b = w 3 w 1 b = v 3 v 2 b = w 2 w 2 b = v 2 v 3 b = w 1 w 3 b = v 1 Therefore sp(v 0, w 0 ), sp(v 2, w 2 ), sp(v 1, w 3 ) and sp(v 3, w 1 ) are C < b >-modules and therefore CD 8 -submodules of CD 8. By Example 5. we have that U 4 = sp(v 1, w 3 ) and U 5 = sp(v 3, w 1 ) are isomorphic and from in Example 7. we know U 5 is irreducible. However sp(v 0, w 0 ), sp(v 2, w 2 ) are reducible as U 0 = sp(v 0 + w 0 ) and U 1 = sp(v 0 w 0 ) and U 2 = sp(v 2 + w 2 ) and U 3 = sp(v 2 w 2 ). Since v 0, v 1, v 2, v 3, w 0, w 1, w 2, w 3 form a basis of CD 8 we can write CD 8 = U 0 U 1 U 2 U 3 U 4 a direct sum of non-isomorphic irreducible CD 8 -submodules. Therefore by Theorem 6 there are exactly 5 non-isomorphic irreducible CD 8 -modules and every other irreducible CD 8 - module is isomorphic to one of them. Using Theorem 1. we can see that these modules correspond to the representations ρ 0,ρ 1,ρ 2,ρ 3 and ρ 4 as follows, aρ 0 = (1), bρ 0 = (1) aρ 1 = (1), bρ 1 = ( 1) aρ 2 = ( 1), bρ 2 = (1) aρ 3 = ( 1), bρ 3 = ( 1) ( ) ( ) ω 0 0 1 aρ 0 = 0 ω 3, bρ 0 =. 1 0 Before we end this section it will be useful to observe the following theorem that follows from Theorem 7. Theorem 8. Let V 1,...,V k form a complete set of non-isomorphic irreducible CG-modules. Then k (di mv i ) 2 = G. i=1 Page 12 of 22

0.4 CONJUGACY CLASSES Definition 14. For x, y G, we say x is conjugate to y in G if there exists some g G such that y = g 1 xg. The conjugacy class of x is the set of all element conjugate to x in G, we write it as x G = {g 1 xg : g G} Definition 15. For x G the centralizer of x in G, C G (x) is the set of elements that commute with x, we write it as C G (x) = {g G : xg = g x}. Example 9. Find the centralizer of 1, a, a 2,b, ab Q 8, the quaternion group of order 8. Solution 6. By the definition of the identity 1 commutes with everything therefore C Q8 (1) = Q 8. By the definition of the quaternion group in Example! a i and b do not commute and therefore neither do a i and a j b, if (a i ) 1 (a i ). This is the case for i = 1, therefore since a clearly commutes with all a i Q 8, C Q8 (a) =< a >. However (a 2 ) 1 = a 2 so it commutes with all of Q 8, C Q8 (a 2 ) = Q 8. Clearly b commutes with itself, We know b does not commute with a or a 3 and therefore does not commute with ab or a 3 b, but it does commute with a 2 and therefore a 2 b therefore C Q8 (b) = {1, a 2,b, a 2 b}. Lastly ab clearly commutes with itself and we have seen it commutes with a 2, it is easy to check it also commutes with a 3 b. But as we have already covered ab does not commute with a, a 3, b and it is easy to check {a 2 b}. Therefore C Q8 (ab) = {1, a 2, ab, a 3 b}. Theorem 9. For x G the size of x G is given by, x G = G : C G (x) = G / C G (x). We can use the theorem above to determine the conjugacy classes of all dihedral groups. To do so it is useful to distinguish between dihedral groups, D 2n where n is odd and where n is even. We will discuss the case where n is even so we can use it later on but the case when n is odd is very similar. When n is even, n = 2m, we have b 1 a m b = a m = a m so the centralizer of a m contains both a and b and is therefore G. by above we have the conjugacy class of a m has only one element and since conjugacy is an equivalence relation and so is reflexive it Page 13 of 22

must be {a m }. For a i,i m it is clear that C G (a i ) contains < a >, therefore, G : C G (a i ) G :< a > = 2 and on the other hand since b 1 a i b = a i we have {a i, a i } (a i ) G. Putting this together by the theorem above we have 2 G : C (a i ) = (a i ) G 2. So the equality holds and (a i ) G = {a i, a i } for1 i m 1. For every intger j, a j ba j = a 2j b, a j (ab)a j = a 2j +1 b. Therefore b G = {a 2j b : 0 j m 1},(ab) G = {a 2j +1 b : 0 j m 1}. Remark 1. The Dihedral group D 2n (n even) has precisely m + 3 conjugacy classes: {1},{a m },{a, a 1 }...,{a m 1, a m+1 },{a 2j b : 0 j m 1},{a 2j +1 b : 0 j m 1}. 0.5 CHARACTERS Now we have done all this groundwork we can begin to look at the Characters of groups and the interesting results that come from them. Lets start with some definitions and the basic theorems that help us use these characters. Definition 16. If V is a CG-module with basis β. Then the character of V is the function χ : G C defined by χ(g ) = tr [g ] β. This corresponds to representations in the obvious way. The character of the representation ρ : G GL(n,F ) is the character χ of the corresponding CG-module, χ(g ) = tr (g ρ),(g G). There are a couple of things to notice about this definition. Remark 2. Firstly it is clear that the chose of basis of a CG-module does not effect the character. If β and β are basis for V then there is an invertible matrix T such that [g ] β = T 1 [g ] β T, Page 14 of 22

but it is easy to see that tr (T 1 [g ] β T ) = tr ([g ] β ). Therefore tr ([g ] β ) = tr ([g ] β ). This result means that isomorphic CG-modules have the same character. Remark 3. Another thing that follows from this observation is that is x, y are conjugate elements in G then χ(x) = χ(y) for all characters χ of G. Clearly if x is conjugate to y, x = g 1 y g for some g G, therefore [x] β = [g 1 y g ] β = [g 1 ] β [y] β [g ] β. Therefore the result follows from above. Definition 17. We say χ is the character of G if it is the character of some CG-module. Therefore it makes sense that χ of G is irreducible if it is the character of an irreducible CG-module and χ of G is reducible if it is the character of a reducible CG-module. Consider the permutation module, V, for G over C, we introduced in section 0.3.0, such that if β is the basis v 1,..., v n, v i g = v i g. The i i -th entry in the matrix [g ] β is 1 if i g = i and 0 otherwise. Therefore the character π of V counts the number of i, 1 i n, such that i g = i. We write f i x(g ) = {i : (1 i n)(i g = i )} so that π(g ) = f i x(g ) (g G) and call π the permutation character. Example 10. Consider the group D 8 as a subgroup of S 4 where the elements are permutations of the corners of the square, such that a = (1,2,3,4) and b = (1,2)(3,4). Find the permutation module for D 8 over C. Solution 7. By Remark 1. we can see that the representatives of the 5 conjugacy classes of D 8 are e = (1), a = (1,2,3,4), a 2 = (1,3)(2,4),b = (1,2)(3,4), ab = (1,3), and by Remark 3. the permutation character is constant on the conjugacy classes it is enough to find the values just for these representatives. We have the following values for the permutation character, e a a 2 b ab π 4 0 0 0 2 Page 15 of 22

0.5.1 Inner Products and Orthogonality We need to cover a few more definitions and theorems to get to our finial results. Definition 18. the inner product of two functions, θ,φ from G to C is given by θ,φ = 1 G θ(g )φ(g ). Theorem 10. If χ 1...χ k are the irreducible characters of G, and ψ is any charcter, then g G ψ = d 1 χ 1 +... + d k χ k where d i = ψ,χ i. 0.5.2 Character tables Definition 19. Let χ 1...χ n be the irreducible characters of G and let g 1,...g k be the representatives of the conjugacy classes of G. The character table of G is the matrix whose i j -entry is χ i (g j ). One observation which we will use later is as follows. Theorem 11. The number of irreducible characters of G is equal to the number of conjugacy classes of G. Therefore a completed table is an n n matrix. Theorem 12. Let χ 1,..χ n be the irreducible characters of G and let g 1,...g k be the representatives of the conjugacy classes of G. 1. There exists the following row orthogonality relation: k χ r (g i )χ s (g i ) = δ r s. C G (g i ) i=1 2. There exists the following column orthogonality relation: k χ i (g r )χ i (g s ) = δ r s C G (g r ). i=1 Example 11. Find the character table of C 2 C 2. Page 16 of 22

Solution 8. Let us first label the elements of G = C 2 C 2 as follows G = {(1,1),(a,1),(1,b),(a,b)}, where a 2 = b 2 = 1. By Theorem 6 we have exactly 4 irreducible, non-isomorphic representations of G, ρ 0,ρ 1,ρ 2,ρ 3, such that for any element x = (a i,b j ) G xρ 0 = 1 xρ 1 = ( 1) j. xρ 2 = ( 1) i xρ 3 = ( 1) i+j since G is ableian each element is its own conjugacy class. Let χ i be the character of ρ i then putting this together we get the following character table; (1,1) (a,1) (1,b) (a,b) χ 0 1 1 1 1 χ 1 1 1-1 -1 χ 2 1-1 1-1 χ 3 1-1 -1 1 Example 12. Consider the group D 8 as a subgroup of S 4 that permutes the four corners of a square such that, a = (1234),b = (12)(34) and D 8 =< a,b : a 4 = b 2 = 1,b 1 ab = a 1 >. 1. Find the character table of D 8 by determining its irreducible representations over C and conjugacy classes. 2. Let π be the corresponding permutation character of D 8. Find the value of π on the elements of D 8 and express π as a sum of irreducible characters. Solution 9. 1. From Example 8. we have the non-isomorphic irreducible representations of D 8 and from Remark 1. we can find the conjugacy classes. Taking representatives from each class, 1, a, a 2,b, ab, and putting this together we get the following character table. Page 17 of 22

e a a 2 b ab χ 0 1 1 1 1 1 χ 1 1 1 1-1 -1 χ 2 1-1 1 1-1 χ 3 1-1 1-1 1 χ 4 2 0-2 0 0 2. We calculated the in Example 8. the values of the permutation character over the representatives of the conjugacy classes of D 8. e a a 2 b ab π 4 0 0 0 2 By Theorem 9. we need to calculate the inner product of π and the irreducible models to find out which of the irreducible characters make up the permutation character. We have, π,χ 0 = 1 π,χ 1 = 0 π,χ 2 = 0 π,χ 3 = 1 π,χ 4 = 1. Therefore by Theorem! we have π = χ 0 + χ 3 + χ 4. 0.5.3 Normal Subgroups and Lifted characters We now need to recall the notion of a normal subgroup of G i.e. a subset of G, call it N such that for all g G, g N = N g. Also recall that we can factor a group G by N to get the set G/N = {g N : g G}. We have the following theorem. Theorem 13. Let N G (a normal subgroup of G), let χ be a character of G/N. Then let χ : G C be such that χ(g ) = χ(n g ) then χ is a character of G and χ and χ have the same degree. Definition 20. If N G and χ is a character of G/N then χ of G such that χ(g ) = χ(n g ) is called the lift of χ to G. Page 18 of 22

Definition 21. Let G be a subgroup of G generated as follows G = {x G ( g,h G)(x = g 1 h 1 g h)}. Then G is called the derived subgroup of G. Theorem 14. The linear characters of G are the lifts to G of the irreducible characters of G/G. Therefore the number of distinct linear characters of G is G/G and so divides G. Example 13. Let G = Q 8 1. Find the five conjugacy classes of G. 2. Find G, and construct all the linear characters of G. 3. Complete the character table to G using orthogonality. Solution 10. 1. It is easy to check that Remark 1 applies to the quaternion group of even order. Therefore the conjugacy classes of Q 8 are {1},{a, a 3 },{a 2 },{b, a 2 b},{ab, a 3 b}. 2. To find G first we observe that if g,h < a > then we get x = 1. Suppose therefore we have g = a,h = a k b. It is easy to check that (a i b) 1 = a j +2 b therefore we have g 1 h 1 g h = a 1 (a k b) 1 = a 3 a k+2 ba k+1 b = a k+1 bba (k+1) = a k+2 a 2 a (k+1) = a 2 We can see from the equations above that putting g = a 2 or a 3 would lead to the same result (as we would just be adding to the LHS of the second last line and subtracting from the RHS). Alternatively we could have g = a k b,h = a in which case we get, g 1 h 1 g h = (a k b) 1 a 1 a k ba = a k+2 ba k+3 ba = a k+2 ba k+2 b = 1. Page 19 of 22

Lastly we could have g = a i b,h = a j b in which case we get g 1 h 1 g h = (a i b) 1 (a j b) 1 a i ba j b = a i+2 ba j +2 ba i ba j b = a i+2 bba (j +2) a i bba j = a 2i 2j. It is easy to check that 2i 2j mod 4 is 2 or 0. This covers all possibilities for x therefore G = {1, a 2 }. Therefore Q 8 /Q 8 = {Q 8,Q 8 a,q 8 b,q 8 ab} which is isomorphic to C 2 C 2 which we found the irreducible characters for in Example 9. Therefore by Theorem 13. these are the 4 linear characters of Q 8. 3. Since there are 5 conjugacy classes by Theorem 11. we know there must be 5 irreducible representations. Since all the linear characters correspond to 1 dimensional CQ 8 -modules using Theorem 8. we can see that the dimension of the CQ 8 -module corresponding to our missing character, χ 5 is 2. Therefore it is clear to see that χ 5 (1) = 2 (since it is the trace of a 2 dimensional identity matrix). Combining the row and column orthogonality form Theorem 13. and Example 8. which gives us the values of C Q8 (g ) over the representatives of the conjugacy classes we can complete the character table, we get; e a a 2 b ab χ 0 1 1 1 1 1 χ 1 1 1 1-1 -1 χ 2 1-1 1 1-1 χ 3 1-1 1-1 1 χ 4 2 0-2 0 0 If we compare this to the character table of D 8 we found in Example 12. we can see they are the same. This is not an obvious result as they are different groups. Page 20 of 22

0.6 ACKNOWLEDGEMENTS I would like to thank Sue Sierra for supervising my project. I would also like to thank EPSRC for funding it. Page 21 of 22

REFERENCES [1] G. James and M.W. Liebeck. Representations and Characters of Groups. Cambridge mathematical textbooks. Cambridge University Press, 2001. 22