Lie Algebras of Generalized Quaternion Groups 1 Introduction 2

Lie Algebras of Generalized Quaternion Groups Samantha Clapp Advisor: Dr. Brandon Samples

Abstract Every ﬁnite group has an associated Lie algebra. Its Lie algebra can be viewed as a subspace of the group algebra with certain bracket conditions imposed on the elements. If one calculates the character table for a ﬁnite group, the structure of its associated Lie algebra can be described. In this work, we consider the family of generalized quaternion groups and describe its associated Lie algebra structure completely.

1 Introduction

The Lie algebra of a group is a useful tool because it is a vector space where linear algebra is available. It is interesting to consider the Lie algebra structure associated to a specific group or family of groups. A Lie algebra is simple if its dimension is at least two and it only has {0} and itself as ideals. Some examples of simple algebras are the classical Lie algebras: sl(n), sp(n) and o(n) as well as the five exceptional finite dimensional simple Lie algebras. A direct sum of simple lie algebras is called a semi-simple Lie algebra. Therefore, it is also interesting to consider if the Lie algebra structure associated with a particular group is simple or semi-simple. In fact, the Lie algebra structure of a finite group is well known and given by a theorem of Cohen and Taylor [1]. In this theorem, they specifically describe the Lie algebra structure using character theory. That is, the associated Lie algebra structure of a finite group can be described if one calculates the character table for the finite group. In particular, the quaternions are a way to generalize complex numbers to higher dimensions. The quaternion group has many scientific applications; for example, aerospace and robotics engineers use them to model the positions and orientations of planes and robots. In this paper, we will begin with basic definitions and examples of Lie algebras of quaternions and then describe the Lie algebra structure of the generalized quaternion group completely.

2 Preliminaries

In order to describe the Lie algebra structure associated with the generalized quaternion group, we must ﬁrst consider the deﬁnition of a Lie algebra.

Definition 2.1. Let L be a vector space over a field F . Then L is a Lie algebra if it possesses a bracket operation [ , ]: L × L → L that satisfies the following axioms:

i. The bracket operation is bilinear.

ii. [x, x] = 0 for all x ∈ L.

iii. [x, [y, z]] + [y, [x, z]] + [z, [x, y]] = 0 (x, y, z ∈ L). 1 Note that (i.) and (ii.) imply anticommutativity, [x, y] = −[y, x], and (iii.) is called the Jacobi identity.

With the definition of a Lie algebra, we can consider a few examples. The first is the set of all x1 3 3 vectors in R under the bracket operation of the cross product. Let x, y, z ∈ R with x = x2 , x3 y1 z1 x2y3 − x3y2 y = y2 , and z = z2 . The cross product is defined as the following x × y = x3y1 − x1y3 . y3 z3 x1y2 − x2y1 x2y3 − x3y2 3 Note that x and y are general vectors and x × y = x3y1 − x1y3 ∈ R ; therefore, the set of x1y2 − x2y1 vectors in R3 is closed under the cross product. Next we will consider the three axioms.

i. First, let a, b ∈ R. Then we have the calculation for bilinearity as follows,     (ax2 + by2)z3 − (ax3 + by3)z2 ax2z3 + by2z3 − ax3z2 − by3z2 [ax + by, z] = (ax3 + by3)z1 − (ax1 + by1)z3 = ax3z1 + by3z1 − ax1z3 − by1z3 (ax1 + by1)z2 − (ax2 + by2)z1 ax1z2 + by1z2 − ax2z1 − by2z1   a(x2z3 − x3z2) + b(y2z3 − y3z2) = a(x3z1 − x1z3) + b(y3z1 − y1z3) = a[x, z] + b[y, z], a(x1z2 − x2z1) + b(y1z2 − y2z1)     x2(ay3 + bz3) − x3(ay2 + bz2) x2ay3 + x2bz3 − x3ay2 − x3bz2 and [x, ay + bz] = x3(ay1 + bz1) − x1(ay3 + bz3) = x3ay1 + x3bz1 − x1ay3 − x1bz3 x1(ay2 + bz2) − x2(ay1 + bz1) x1ay2 + x1bz2 − x2ay1 − x2bz1   (ax2y3 − x3y2) + b(x2z3 − x3z2) = a(x3y1 − x1y3) + b(x3z1 − x1z3) = a[x, y] + b[x, z]. a(x1y2 − x2y1) + b(x1z2 − x2z1)

Therefore, the cross product satisﬁes bilinearity. ii. Next, consider an element bracketed with itself:

    x2x3 − x3x2 0 [x, x] = x3x1 − x1x3 = 0 . x1x2 − x2x1 0

So we have that the second axiom is satisﬁed. iii. Consider the Jacobi identity:

         y2z3 − y3z2 z2x3 − z3x2 x2y3 − x3y2 [x, [y, z]] + [y, [z, x]] + [z[x, y]] = x, y3z1 − y1z3 + y, z3x1 − z1x3 + z, x3y1 − x1y3 y1z2 − y2z1 z1x2 − z2x1 x1y2 − x2y1     x2(y2z3 − y3z2) − x3(y2z3 − y3z2) y2(z2x3 − z3x2) − y3(z2x3 − z3x2) = x3(y3z1 − y1z3) − x1(y3z1 − y1z3) + y3(z3x1 − z1x3) − y1(z3x1 − z1x3) x1(y1z2 − y2z1) − x2(y1z2 − y2z1) y1(z1x2 − z2x1) − y2(z1x2 − z2x1)     z2(x2y3 − x3y2) − z3(x2y3 − x3y2) 0 + z3(x3y1 − x1y3) − z1(x3y1 − x1y3) =  0  z1(x1y2 − x2y1) − z2(x1y2 − x2y1) 0.

Consequently, the cross product satisﬁes the Jacobi identity.

2 Since we have shown that all of the properties for a Lie algebra are satisﬁed by the cross product, then we have that the vectors of R3 with the bracket operation of the cross product is indeed an example of a Lie algebra.

Next, consider the set of all 2 × 2 matrices of trace zero with entries in C with the bracket operation being ab − ba is a special linear algebra known as sl(2). Consider the matrices a a b b c c a = 1 2 , b = 1 2 , and c = 1 2 , such that the entries are in . Note a, b, c ∈ sl(2) a3 −a1 b3 −b1 c3 −c1 C and

a a b b b b a a ab − ba = 1 2 1 2 − 1 2 1 2 a3 −a1 b3 −b1 b3 −b1 a3 −a1

a b + a b a b − a b b a + b a b a − b a = 1 1 2 3 1 2 2 3 − 1 1 2 3 1 2 2 1 a3b1 − a1b3 a3b2 + a1b1 b3a1 − b1a3 b3a2 + b1a1

a b + a b − b a − b a a b − a b − b a + b a = 1 1 2 3 1 1 2 3 1 2 2 3 1 2 2 1 ∈ sl(2). a3b1 − a1b3 − b3a1 + b1a3 a3b2 + a1b1 − b3a2 − b1a1

Thus, we see that sl(2) is closed under the bracket operation. Now consider the three axioms for a Lie algebra.

i. First, let m, n ∈ C. The bilinearity calculation is as follows,

ma + nb ma + nb c c [ma + nb, c] = 1 1 2 2 , 1 2 ma3 + nb3 −ma1 − nb1 c3 −c1 ma c + nb c + ma c + nb c ma c + nb c − ma c + nb c = 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 ma3c1 + nb3c1 + ma1c3 + nb1c3 ma3c2 + nb3c2 + ma1c1 + nb1c1 c ma + c nb + c ma + nc b c ma + c nb − c ma − c nb − 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 c3ma1 + c3nb1 − c1ma3 + c1nb3 c3ma2 + c3nb2 + c1ma1 + c1nb1 m(a c + a c ) + n(b c + b c ) m(a c − a c ) + n(b c + b c ) = 1 1 2 3 1 1 2 3 1 2 2 1 1 2 2 1 m(a3c1 + a1c3) + n(b3c1 + b1c3) m(a3c2 + a1c1) + n(b3c2 + b1c1) m(c a + c a ) + n(c b + c b ) m(c a − c a + n(c b 1 − c b ) − 1 1 2 3 1 1 2 3 1 2 2 ) 1 2 2 1 m(c3a1 − c1ma3) + n(c3b1 + c1b3) m(c3a2 + c1a1) + n(c3b2 + c1b1) = m[a, c] + n[b, c],

a a mb + nc mb + nc and [a, mb + nc] = 1 2 , 1 1 2 2 a3 −a1 mb3 + nc3 −mb1 − nc1 a mb + a nc + a mb + na c a mb + a nc − a mb − a nc = 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 a3ma1 + a3nc1 − a1mb3 + a1nc3 a3mb2 + a3nc2 + a1mb1 + a1nc1 mb a + nc a + mb a + nc a mb a + nc a − mb a + nc a − 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 mb3a1 + nc3a1 + mb1a3 + nc1a3 mb3a2 + nc3a2 + mb1a1 + nc1a1 m(a b + a b ) + n(a c + a c ) m(a b − a b ) + n(a c − a c ) = 1 1 2 3 1 1 2 3 1 2 2 1 1 2 2 1 m(a3a1 − a1b3) + n(a3c1 + a1c3) m(a3b2 + a1b1) + n(a3c2 + a1c1) m(b a + b a ) + n(c a + c a ) m(b a − b a ) + n(c a + c a ) − 1 1 2 3 1 1 2 3 1 2 2 1 1 2 2 1 = m[a, b] + n[a, c]. m(b3a1 + b1a3) + n(c3a1 + c1a3) m(b3a2 + b1a1) + n(c3a2 + c1a1)

ii. Next, consider an element bracketed with itself:

a a a a a a a a 0 0 [a, a] = 1 2 1 2 − 1 2 1 2 = . a3 −a1 a3 −a1 a3 −a1 a3 −a1 0 0

3 iii. Consider the Jacobi identity calculation:

b b c c c c b b [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = a, 1 2 1 2 − 1 2 1 2 b3 −b1 c3 −c1 c3 −c1 b3 −b1 c c a a a a c c + b, 1 2 1 2 − 1 2 1 2 c3 −c1 a3 −a1 a3 −a1 c3 −c1 a a b b b b a a + c, 1 2 1 2 − 1 2 1 2 a3 −a1 b3 −b1 b3 −b1 a3 −a1 b c − c b 2b c − 2b c c a − a c 2c a − 2c a = a, 2 3 2 3 1 2 2 1 + b, 2 3 2 3 1 2 2 1 2b3c1 − 2b1c3 b3c2 − c3b2 2c3a1 − 2c1a3 c3a2 − a3c2 a b − b a 2a b − 2a b + c, 2 3 2 3 1 2 2 1 2a3b1 − 2a1b3 a3b2 − b3a2 a a b c − c b 2b c − 2b c b c − c b 2b c − 2b c a a = 1 2 2 3 2 3 1 2 2 1 − 2 3 2 3 1 2 2 1 1 2 a3 −a1 2b3c1 − 2b1c3 b3c2 − c3b2 2b3c1 − 2b1c3 b3c2 − c3b2 a3 −a1 b b c a − a c 2c a − 2c a c a − a c 2c a − 2c a b b + 1 2 2 3 2 3 1 2 2 1 − 2 3 2 3 1 2 2 1 1 2 b3 −b1 2c3a1 − 2c1a3 c3a2 − a3c2 2c3a1 − 2c1a3 c3a2 − a3c2 b3 −b1 c c a b − b a 2a b − 2a b c c a b − b a 2a b − 2a b + 1 2 2 3 2 3 1 2 2 1 − 1 2 2 3 2 3 1 2 2 1 c3 −c1 2a3b1 − 2a1b3 a3b2 − b3a2 c3 −c1 2a3b1 − 2a1b3 a3b2 − b3a2 2a b c − 2a b c − 2a b c + 2a b c 4a b c − 4a b c + 2a b c − 2a c b = 2 3 1 2 1 3 3 1 2 3 2 1 1 1 2 1 2 1 2 3 2 2 3 2 2a3b2c3 − 2a3c2b3 − 4a1b3c1 + 4a1b1c3 2a3b1c2 − 2a3b2c1 − 2a2b3c1 + 2a2b1c3 2b c a − 2b c a − 2b c a + 2b c a 4b c a − 4b c a + 2b c a − 2b a c + 2 3 1 2 1 3 3 1 2 3 2 1 1 1 2 1 2 1 2 3 2 2 3 2 2b3c2a3 − 2b3a2c3 − 4b1c3a1 + 4b1c1a3 2b3c1a2 − 2b3c2a1 − 2b2c3a1 + 2b2c1a3 2c a b − 2c a b − 2c a b + 2c a b 4c a b − 4c a b + 2c a b − 2c b a + 2 3 1 2 1 3 3 1 2 3 2 1 1 1 2 1 2 1 2 3 2 2 3 2 2c3a2b3 − 2c3b2a3 − 4c1a3b1 + 4c1a1b3 2c3a1b2 − 2c3a2b1 − 2c2a3b1 + 2c2a1b3 0 0 = . 0 0

Thus, we have shown that sl(2) satisﬁes all of the conditions for a Lie algebra.

Since we want to look specifically at the structure of the generalized quaternions, we must consider the Lie algebra of a finite group. Let G be a finite group. Then the Lie algebra of G suggested by Plesken, denoted L(G), is a linear span of elementsg ˆ = g − g−1 ∈ G(C) for every g ∈ G. The Lie algebra bracket is given by defining [ˆg, hˆ] =g ˆhˆ − hˆgˆ which extends linearly to all of L(G). Also note that gd−1 = −gˆ and

[ˆg, hˆ] = ghc − gh[−1 − g[−1h + g\−1h−1. It follows that L(G) is closed under the Lie bracket operation. In fact one can show that L(G) is a Lie algebra.

Since we are describing the Lie algebra structure of the generalized quaternions, it follows that we need to define the generalized quaternions. Definition 2.2. The generalized quaternion group of order 4n is defined as 2n 4 n 2 −1 −1 Q4n = a, b | a = b = 1, a = b , b ab = a , where n ≥ 2. 4 With this definition, we can consider the specific example of the quaternion group of order eight. The restrictions on generators for Q8 are as follows: 2 4 −1 −1 Q8 = a, b | a = b , b = 1, a ba = b . We can write the eight elements in the following form: 1, a, a2, a3, b, ab, a2b, a3b. Using the Lie algebra defined by Plesken, we have that 1 and a2 both are their own inverses and a is 3 3 2 the inverse of a , b is the inverse of a b, and ab is the inverse of a b, and therefore L(Q8) is generated by three elements,a, ˆ ˆb, and abb where

aˆ = a − a3, ˆb = b − a3b, and abb = ab − a2b.

Hence, dim(L(Q8)) = 3. Now, let c = ab. We have that [ˆa, ˆb] = 4ˆc, [ˆc, aˆ] = 4ˆb, [ˆb, cˆ] = 4ˆa. Note that

0 1 0 0 1 0 , , 0 0 1 0 and 0 −1 is a basis for sl(2) label them A, B, and C, respectively. We want to show that L(Q8) is isomorphic to sl(2), so consider the following association:

0 −2 φ(ˆa) → −2A + 2B = 2 0

0 2i φ(ˆb) → 2iA + 2iB = 2i 0

−2i 0 φ(ˆc) → −2C = . 0 2i It follows,

0 −2 0 2i 0 2i 0 −2 [φ(ˆa), φ(ˆb)] = − 2 0 2i 0 2i 0 2 0

−4i 0 4i 0 −8i 0 = − = 0 4i 0 −4i 0 8i

= 4φ(ˆc) = φ([ˆa, ˆb]),

−2i 0 0 −2 0 −2 −2i 0 [φ(ˆc), φ(ˆa)] = − 0 2i 2 0 2 0 0 2i

0 4i 0 −4i 0 8i = − = 4i 0 −4i 0 8i 0

= 4φ(ˆb) = φ([ˆc, aˆ]),

0 2i −2i 0 −2i 0 0 2i [φ(ˆb), φ(ˆc)] = − 2i 0 0 2i 0 2i 2i 0

0 −4 0 4 0 −8 = − = 4 0 −4 0 8 0

= 4φ(ˆa) = φ([ˆb, cˆ]).

5 ∼ Hence L(Q8) = sl(2).

In order to describe the associated Lie algebra of a group, it is helpful to deﬁne a representation.

Definition 2.3. A representation of a finite group G on a finite-dimensional complex vector space V is a group homomorphism φ : G → GL(V ).

It is important to note that we can encode essential information about representations in a condensed form through character values. In fact, complex representations of finite groups are determined, up to isomorphism, by their characters. Specifically, we define a character as follows.

Deﬁnition 2.4. A character of V is a complex valued-function deﬁned on G that assigns to each element, g ∈ G, the trace of the linear transformation that g induces on V , that is, χ(g) = T r(φ(g)).

With the deﬁnition of a representation, we can consider describing the Lie algebra structure of a particular group. In fact, if we can calculate certain character and indicator values, we are able to describe the structure of the Lie algebra for the entire group. That is, we can know what types of Lie algebras, the number of each type of Lie algebra, as well as dimensions of each piece that form the irreducible components of the Lie algebra. This information can be determined through the following theorem.

Theorem 2.5 (Cohen & Taylor, 2007). The Lie algebra L(G) admits the decomposition M M M L(G) = o(χ(1)) ⊕ sp(χ(1)) ⊕ 0gl(χ(1)) χ∈R χ∈Gp χ∈C

where R, Gp, and C are the sets of irreducible character of real, sympletic, and complex types, respectively, and where the prime signiﬁes that there is just one summand gl(χ(1)) for each pair {χ, χ} from C.

With this theorem, it is important to know more about the character values and indicator values for a group.

Deﬁnition 2.6. The character table of G is a m × m table whose rows correspond to irreducible group representations and columns correspond to the conjugacy classes of each element.

Table 1 is the general from of a character table. Note that each entry of the character table would be of the form χi(gm) = T r(φ(gm)) where each χi is a diﬀerent representation of the group.

In order to use the Theorem 2.5, we also need to know how to calculate the indicator value. 6 G C1 C2 C3 ... χ1 χ2 χ3 . .

Table 1: Character Table

Definition 2.7. Let G be a group with a finite-dimensional continuous complex representation with an associated character χ. Then the Frobenius-Schur indicator is defined to be 1 X χ(g2). |G| g∈G

Now we can consider the example of Q8. First, we will calculate one of the indicator values 3 2 3 2 for Q8. The conjugacy classes for Q8 are the following: {e}, {a, a }, {a }, {ab, a b}, {b, a b}. Note that the conjugacy classes are important because every element in a conjugacy class has the same character value, and consequently, the trace of the square of the representation will also be the same for each element. Therefore, we only need to calculate the trace of the square of the representation for one element in each conjugacy class and then multiply this trace by the number of elements in each conjugacy class. Therefore, consider the following matrix representation correspondence:

2 h−1 0 i 4 h1 0i 2 h−1 0 i 2 h−1 0 i φ(a) = 0 −1 , φ(a) = 0 1 , φ(b) = 0 −1 , and φ(ab) = 0 −1 , and our trace values are T r(φ(e)2) = 2, T r(φ(a)2) = −2, T r(φ(a)4) = 2, T r(φ(ab)2) = −2, and T r(φ(b)2) = −2.

Thus, our indicator value is the following, 1 X 1 χ(g2) = (2 + 2(−2) + 2 + 2(−2) + 2(−2)) = −1. |G| 8 g∈G

The remaining four indicator values for Q8 are 1; therefore, we have the indicator values for Q8 arranged as [1, 1, 1, 1, -1]. Note we have the character values for Q8 in Table 2. Since by Theorem 2.5, we want χ(1), we use the values of the ﬁrst column of the character table. It follows that we have four ones for the indicator values that pair with four ones from the character table, which by Theorem 2.5, give us four cases of o(2), that are zero dimensional orthogonal representations. Next we have a −1 indicator value that pairs with a 2 character value. This gives us one case of sp(2) which is a three dimensional simplectic representation that is equal to sl(2). Thus, we have that L(Q8) = sp(2), so L(Q8) = sl(2).

Now consider the example of Q16. The indicator values are [1, 1, 1, 1, 1, -1, -1] for Q16. Note the character values for Q16 are in Table 3. 7 Q8 C1 C2 C3 C4 C5 χ1 1 1 1 1 1 χ2 1 1 1 -1 -1 χ3 1 1 -1 -1 1 χ4 1 1 -1 1 -1 χ5 2 -2 0 0 0

Table 2: Character Table for Q8

Q8 C1 C2 C3 C4 C5 C6 C7 χ1 1 1 1 1 1 1 1 χ2 1 -1 -1 1 1 1 -1 χ3 1 -1 1 1 1 -1 1 χ4 1 1 -1 1 1 -1 -1 χ5 2 0√ 0 -2 2 0√ 0 χ6 2 0 √2 0 -2 0 -√2 χ7 2 0 - 2 0 -2 0 2

Table 3: Character Table for Q16

Note that we have four ones for the indicator values that pair with four ones from the character table which by Theorem 2.5 give us four cases of o(1), which are zero dimensional orthogonal representation. Next we have a 1 indicator value that pairs with a 2 character value. This gives one case of o(2), which is a one dimensional orthogonal representation. Since we are working over the complex numbers, this piece is simply a one dimensional complex representation. Next we have two, −1 indicator values that pair with two, 2 character values. This gives us one case of sp(2) which is a three dimensional simplectic representation. That is L(Q16) = sp(2) ⊕ sp(2) ⊕ o(2) ⇒ L(Q16) = sl(2) ⊕ sl(2) ⊕ C.

3 Main Results

In order to describe the generalized quaternions completely, we need the character and indicator values. It follows that we must consider matrix representations for the group. Consider the following proposition: ωn 0 0 −1 iπ Proposition 3.1. The matrices a = and b = , where ωn = e n , are 0 ωn 1 0 generators for the generalized quaternion group. −1 0 0 1 1 0 Proof. Note that b2 = , b3 = , and b4 = . Consider 0 −1 −1 0 0 1

2n 2n i2π 2n ωn 0 ωn 0 e 0 1 0 a = = 2n = i2π = . 0 ωn 0 ωn 0 e 0 1 8 Thus we have that the ﬁrst condition b4 = a2n = I is satisﬁed. Note that

n n iπ n ωn 0 ωn 0 e 0 −1 0 a = = n = iπ = . 0 ωn 0 ωn 0 e 0 −1

Therefore, we have that the second condition an = b2 is satisﬁed. Now consider

0 1 ω 0 0 −1 0 ω 0 −1 ω 0 b−1ab = n = n = n . −1 0 0 ωn 1 0 −ωn 0 1 0 0 ωn

Thus the third condition b−1ab = a−1 is satisﬁed. Therefore, we have that the matrices a and b are generators for the generalized quaternion group.

We get other irreducible representations for the quaternions from the powers of this representation as proven in the following proposition. r ωn 0 0 −1 Proposition 3.2. Let φr : Q4n → GL(2, C) be given by a → and b → . 0 ωn 1 0 Then φr is a group homomorphism for each {r ∈ N | 1 ≤ r ≤ n − 1}.

j k j k Proof. Consider a b where {j ∈ N | 0 ≤ j ≤ n − 1} and k ∈ {0, 1}, thus a b ∈ Q4n and l m l m a b where {l ∈ N | 0 ≤ l ≤ n − 1} and m ∈ {0, 1}, thus a b ∈ Q4n. We have the following:

jr k lr m j k l m ωn 0 0 −1 ωn 0 0 −1 j k l m φr(a b )φr(a b ) = = φr(a b a b ). 0 ωn 1 0 0 ωn 1 0

Note that this representation does not give us all of the irreducible representations. Thus we shall also consider another representation given in the following proposition. ωn 0 0 1 i2π Proposition 3.3. The matrices a = , and b = , where ωn = e n , are 0 ωn 1 0 generators for the generalized quaternion group. 1 0 0 1 1 0 Proof. Note that b2 = , b3 = , and b4 = . Consider 0 1 1 0 0 1

2n 2n i4π 2n ωn 0 ωn 0 e 0 1 0 a = = 2n = i4π = . 0 ωn 0 ωn 0 e 0 1

Thus we have that the ﬁrst condition b4 = a2n = I is satisﬁed. Note that

n n i2π n ωn 0 ωn 0 e 0 1 0 a = = n = i2π = . 0 ωn 0 ωn 0 e 0 1

9 Therefore, we have that the second condition an = b2 is satisﬁed. Now consider

0 1 ω 0 0 1 0 ω 0 1 ω 0 b−1ab = n = n = n . 1 0 0 ωn 1 0 ωn 0 1 0 0 ωn

Thus the third condition b−1ab = a−1 is satisﬁed. Therefore, we have that the matrices a and b are generators for the generalized quaternion group.

Note that if we use the odd powers of the representation from Proposition 3.1 up to 2n − 1 n−2 and use all of the powers of the representation from Proposition 3.3 up to 2 we get all of the irreducible representations. Moreover, it will be useful to us to describe the form for the group elements as seen in the following proposition.

Proposition 3.4. The elements {1, a, a2, ..., a2n−1, b, ab, a2b, ..., a2n−1b} represent the 4n dis- tinct group elements of Q4n.

Proof. Let j, k ∈ Z such that 0 ≤ j ≤ 2n − 1 and 0 ≤ k ≤ 2n − 1. It follows from the definition of the generalized quaternions that j 6= k, aj 6= ak, since a is defined to be of order a2n. Suppose aj = akb, thus aj−k = b. Since a is a diagonal matrix the powers of a will be a diagonal matrix. However, b is an off diagonal matrix, consequently aj 6= akb. Now suppose ajb = akb, it follows that aj = ak, which brings us to the first case. Therefore ajb 6= akb. Further we can describe the conjugacy classes for the group.

Proposition 3.5. Each group has n + 3 conjugacy classes, with representatives [ {1} , {an} , ak, a−k , akb : k even , and akb : k odd . 1≤k

Proof. Let {j ∈ Z | 0 ≤ J ≤ n − 1} and k ∈ {0, 1}.

First consider the identity element. It follows that ajbk(1)(ajbk)−1 = ajbk(1)(ajbk)−1 = 1. Therefore, 1 forms a conjugacy class.

Next consider the element an. There are two cases: k = 0 and k = 1.

Case 1: Let k = 0, then we have aj(an)a−j = aj+n−j = an.

Case 2: Let k = 1 and consider

(ajb)−1anajb = b−1a−janajb = b−1abb−1abb−1a...b−1ab = a−n = an

10 Therefore, there is a conjugacy class consisting of only an.

Next we have the element aj. Let {m ∈ Z | 0 ≤ m ≤ n − 1}. Again there are two cases: k = 0 and k = 1.

Case 1: Let k = 0. It follows that amaja−m = am+j−m = aj.

Case 2: Let k = 1, then we have (amb)−1ajamb = b−1a−majamb = b−1amb = a−j.

Therefore, we have that aj and a−j form a conjugacy class.

Now consider the elements of the form ajb such that j is even. Let {m ∈ Z | 0 ≤ m ≤ n − 1}. Again there are two cases: k = 0 and k = 1.

Case 1: Let k = 0. We have

amajba−m = amajbb−1amb = amajamb = aj+2mb.

Note that j + 2m is even.

Case 2: Let k = 1. It follows that

b−1a−majbamb = b−1ajbamb = a−j+mamb = a−j+2mb.

Note that −j + 2m is even.

Thus we have that all elements of the form ajb, where j is even form a conjugacy class.

Consider elements of the form ajb such that j is odd. Let {m ∈ Z | 0 ≤ m ≤ n − 1}. Again there are two cases: k = 0 and k = 1.

Case 1: Let k = 0. We have

amajba−m = amajbb−1amb = amajamb = aj+2mb.

Note that j + 2m is odd.

11 Case 2: Let k = 1. Consider the following b−1a−majbamb = b−1ajbamb = a−j+mamb = a−j+2mb. Note that −j + 2m is odd. Thus we have that all elements of the form ajb, where j is odd form a conjugacy class.

Based on the Proposition 3.4 we have accounted for all of the elements of Q4n. Now we have that the conjugacy classes are [ {1} , {an} , ak, a−k , akb : k even , and akb : k odd . 1≤k

Proof. Let n be a number of the form 2j where j ∈ N and j ≥ 1. Since every element in a conjugacy class shares the same character value, it follows that χ(g2) would be the same for each element in the same conjugacy classes and by Proposition 3.5 we know what the conjugacy classes for Q4n.

First consider the case akb : k even .

Since the trace is same for every element in the conjugacy class consider the case where 0 −1 0 −1 −1 0 k = 0. We have a0b = b. It follows that φ(b)2 = = . There- 1 0 1 0 0 −1 fore, T r(φ(b)2) = −2. Also note that there are n elements in this conjugacy class.

Next consider the class akb : k odd . Note that

kπi ! kπi ! k e n 0 0 −1 0 −e n φ(a b) = −kπi = −kπi and 0 e n 1 0 e n 0

kπi ! kπi ! 0 k 2 0 −e n 0 −e n −e 0 −1 0 φ(a b) = −kπi −kπi = 0 = . e n 0 e n 0 0 −e 0 −1

12 Thus T r(φ(akb)2) = −2. Also note that there are n elements in this conjugacy class.

Next consider the conjugacy class {a, a−1}. It follows that

πi ! e n 0 φ(a) = −πi and 0 e n

πi ! πi ! 2πi ! 2 e n 0 e n 0 e n 0 φ(a) = −πi −πi = −2πi . 0 e n 0 e n 0 e n

2 2πi −2πi 2 −2 Therefore, T r(φ(a) ) = e n + e n Note for the next conjugacy class {a , a }, we get

2πi ! e n 0 φ(a) = −2πi and 0 e n

2πi ! 2πi ! 4πi ! 2 e n 0 e n 0 e n 0 φ(a) = −2πi −2πi = −4πi . 0 e n 0 e n 0 e n

2 4πi −4πi Therefore, T r(φ(a) ) = e n +e n It follows that our conjugacy classes and the correspond- ing traces are:

−1 2πi −2πi a, a e n + e n 2 −2 4πi −4πi a , a e n + e n 3 −3 6πi −6πi a , a e n + e n . . . . 2n−2 −(2n−2) (2n−1)πi −(2n−1)πi a , a e n + e n . We can write the identity as

0(πi) ! e n 0 φ(e) = −0(πi) and 0 e n 0(πi) ! 0(πi) ! 0(2πi) ! 2 e n 0 e n 0 e n 0 φ(e) = −0(πi) −0(πi) = −0(2πi) . 0 e n 0 e n 0 e n

2 0(2πi) 0(−2πi) Thus T r(φ(e) ) = e n + e n .

We also have the conjugacy class {an}. That is

nπi ! n e n 0 φ(a ) = −nπi and 0 e n

nπi ! nπi ! 2nπi ! n 2 e n 0 e n 0 e n 0 φ(a ) = −nπi −nπi = −2nπi . 0 e n 0 e n 0 e n

13 n 2 2nπi −2nπi Thus T r(φ(a ) ) = e n + e n .

Now consider the following sum:

0(2πi) 0(−2πi) 2πi −2πi 4πi −4πi 6πi −6πi (2n−2)πi −(2n−2)πi 2nπi −2nπi e n + e n + e n + e n + e n + e n + e n + e n + ... + e n + e n + e n + e n n n X 2πir −2πir X r −r = e n + e n = ω + ω = 0 + 0 = 0. r=0 r=0 Thus when calculating the indicator value we have

1 X 1 −4n χ(g2) = (0 − 2n − 2n) = = −1. |g| 4n n g∈G

Proposition 3.7. The indicator value for Q of the representations generated by a = 4n ωn 0 0 1 2πi and b = , where ωn = e n , is always -1. 0 ωn 1 0

Proof. Let n be a number of the form 2j, where j ∈ N and j ≥ 1. Since every element in a conjugacy class shares the same character value, it follows that χ(g2) would be the same for each element in the same conjugacy classes and by Proposition 3.5 we know what the conjugacy classes for Q4n.

First consider the class akb : k even .

Since the trace is same for every element in the conjugacy class consider the case where 0 1 0 1 1 0 k = 0. We have a0b = b. It follows that φ(b)2 = = . Therefore, 1 0 1 0 0 1 T r(φ(b)2) = 2. Also note that there are n elements in this conjugacy class.

Next consider the class akb : k odd . Note that

2kπi ! 2kπi ! k e n 0 0 1 0 e n φ(a b) = −2kπi = −2kπi and 0 e n 1 0 e n 0

2kπi ! 2kπi ! 0 k 2 0 e n 0 −e n e 0 1 0 φ(a b) = −2kπi −2kπi = 0 = . e n 0 e n 0 0 e 0 1

Thus T r(φ(akb)2) = 2. Also note that there are n elements in this conjugacy class.

14 Next consider the conjugacy class {a, a−1}. It follows that

2πi ! e n 0 φ(a) = −2πi and 0 e n

2πi ! 2πi ! 4πi ! 2 e n 0 e n 0 e n 0 φ(a) = −2πi −2πi = −4πi . 0 e n 0 e n 0 e n

2 4πi −4πi 2 −2 Therefore, T r(φ(a) ) = e n + e n Note for the next conjugacy class {a , a }, we get

4πi ! e n 0 φ(a) = −4πi and 0 e n

4πi ! 4πi ! 16πi ! 2 e n 0 e n 0 e n 0 φ(a) = −4πi −4πi = −16πi . 0 e n 0 e n 0 e n

2 16πi −16πi Therefore, T r(φ(a) ) = e n + e n It follows that our conjugacy classes and the corre- sponding traces are:

−1 4πi −4πi a, a e n + e n 2 −2 16πi −16πi a , a e n + e n 3 −3 36πi −36πi a , a e n + e n . . . . 2n−2 −(2n−2) (2n−1)2πi −(2n−1)2πi a , a e n + e n .

We can write the identity as

0(2πi) ! e n 0 φ(e) = −0(2πi) and 0 e n 0(2πi) ! 0(2πi) ! 0(4πi) ! 2 e n 0 e n 0 e n 0 φ(e) = −0(2πi) −0(2πi) = −0(4πi) . 0 e n 0 e n 0 e n

2 0(4πi) 0(−4πi) Thus T r(φ(e) ) = e n + e n .

We also have the conjugacy class {an}. That is

2nπi ! n e n 0 φ(a ) = −2nπi and 0 e n

2nπi ! 2nπi ! 4nπi ! n 2 e n 0 e n 0 e n 0 φ(a ) = −2nπi −2nπi = −4nπi . 0 e n 0 e n 0 e n

15 n 2 4nπi −4nπi Thus T r(φ(a ) ) = e n + e n .

Now consider the following sum:

0(4πi) 0(−4πi) 4πi −4πi 16πi −16πi 36πi −36πi (2n−2)2πi −(2n−2)2πi e n + e n + e n + e n + e n + e n + e n + e n + ... + e n + e n n n 4nπi −4nπi X 4πir −4πir X 2r −2r + e n + e n = e n + e n = ω + ω = 0 + 0 = 0. r=0 r=0 Thus when calculating the indicator value we have

1 X 1 4n χ(g2) = (0 + 2n + 2n) = = 1. |g| 4n n g∈G

Recall from our earlier examples that L(Q8) = sp(2) ⇒ L(Q8) = sl(2) and L(Q16) = sp(2) ⊕ sp(2) ⊕ o(2) ⇒ L(Q16) = sl(2) ⊕ sl(2) ⊕ C. In looking at these examples and examples of greater n values we see a pattern, which we describe with the following theorem.

j Theorem 3.8. Let n = 2 , where j ∈ N. The Lie algebra of Q4n is

2j−1 2j−1−1 M M L(Q4n) = sp(2) ⊕ o(2) k=1 k=1

2j−1 2j−1−1 M M = sl(2) ⊕ C. k=1 k=1

Proof. We know that the odd powers of representations from Proposition 3.1 from 1 to 2n−1 n−2 and that all powers of the representations from 1 to 2 from Proposition 3.3 form all the irreducible representations for Qn. By Proposition 3.6 we have that all of the representations from Proposition 3.1 have an indicator of -1. Note that the taking the odd powers from 1 n j j−1 to 2n − 1 gives us 2 representations. However, since n is of the form 2 , j ∈ N we have 2 indicator values of -1.

Similarly by Proposition 3.7 we have that all of the representations from Proposition 3.3 n−2 2j −2 j−1 have an indicator of 1. Note that since we have 2 = 2 = 2 − 1. Hence there are 2j−1 − 1 indicator values of 1. Note that we always have 4 one dimensional representations. The identity for these representations is simply the number 1. In considering the indicator values for these representations, it is know that for all four the indicator value is 1.

Note that for all the matrix representations χ(1) is χ of the identity element which will be the trace of the identity matrix. All of our matrix representations are two dimensional, we

16 have that our character value is T r(χ(1)) = 2. Since we have 2j−1 representations from the representations described in Proposition 3.1 and 2j−1 − 1 representations from the representations described in 3.3, we have a total of 2j−1 + 2j−1 − 1 = 2j − 1 = n − 1 character values that are 2. For all four of the one dimensional representations our character value will be 1 since T r(1) = 1.

It follows that using Theorem 2.5 that the four indicator values of 1 from the one dimensional representations pair with the four Character values or 1, to give four cases of o(1) which are zero dimensional representations. From the two dimensional matrix representations we 2j−1 indicator values of -1, that pair with 2j−1 character values of 2, which from Theorem 2.5 gives us 2j−1 cases of sp(2) = sl(2). From the remaining two dimensional matrix representations we have 2j−1 −1 indicator values of 1 which pair with the remaining 2j−1 −1 character values which are all 2s, thus we have 2j−1 − 1 cases of o(2) = C.

References

[1] A. M. Cohen and D. E. Taylor. On a certain lie algebra deﬁned by a ﬁnite group. Mathematical Association of America, 1.3:633–639, 2007.