Posts Tagged ‘general linear group’

Problem. Let G be a subgroup of \text{GL}(2, \mathbb{C}). Show that if |G| is even and |G| > 2, then G is not simple.

Solution. First notice that we are done if G is abelian because then every subgroup of G would be normal and since |G| > 2 is even, G has a subgroup of order two, which is clearly neither the trivial subgroup nor G.
So suppose, to the contrary, that G is non-abelian and simple and consider the group homomorphism f: G \to \mathbb{C}^{\times} defined by f(g)=\det(g), for all g \in G. Then since \ker f is a normal subgroup of G and G is not simple, either \ker f=\{I_2\}, where I_2 is the identity matrix in \text{GL}(2, \mathbb{C}), or \ker f=G. But we can’t have \ker f=\{I_2\} because then G would be embedded into \mathbb{C}^{\times} making G abelian. Hence \ker f=G, i.e. \det g=1 for all g \in G. Now let h \in G be an element of order two. So h^2=I_2, which implies that h is diagonalizable (because its minimal polynomial divides x^2-1 and so it splits into distinct linear factors). So

h=P \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} P^{-1}

for some P \in \text{GL}(2, \mathbb{C}). Since h^2=I_2, the set of eigenvalues of h, which is \{a,b\}, is a subset of \{-1,1\}. Since \det h=1, we must have a=b=\pm 1 and so h=\pm I_2. Since h has order two, h \ne I_2 and hence h=-I_2. Since h=-I_2 is a central element of G, the subgroup H:=\langle h \rangle is a normal subgroup of order two in G and H \ne G because |G| > 2. So G is not simple and that contradicts our assumption that G is simple. So our assumption is wrong and G is not simple indeed! \Box

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Problem. Let k be a field, n a positive integer. Let G be a finite subgroup of \text{GL}(n,k) such that |G|>1 and suppose also that every g \in G is upper triangular and all the diagonal entries of g are 1.
Show that \text{char}(k) > 0 and |G| is a power of \text{char}(k).

Solution. First, let’s put an order on the set

S:=\{(i,j): \ \ 1 \le i < j \le n\}.

We write (i,j) < (i',j') if i < i' or i=i', \ j < j'. Now let g=[g_{ij}] be any non-identity element of G. Let (r,s) be the smallest element of S such that g_{rs} \ne 0. See that, for any integer m, the (r,s)-entry of g^m is mg_{rs} and the (i,j)-entries of g^m, where (i,j) \in S, \ (i,j) < (r,s), are all zero. But since G is finite, there exists an integer m > 1 such that g^m is the identity matrix and so we must have mg_{rs}=0. Thus m1_k=0 (because g_{rs} \ne 0) and hence p:=\text{char}(k) > 0. So the (i,j)-entries of g^p, where (i,j) \in S and (i,j) \le (r,s), are all zero.
Now if g^p is not the identity matrix, we can replace g with g^p and repeat our argument to find an element (u,v) \in S, \ (u,v) > (r,s), such that all (i,j)-entries of g^{p^2}, where (i,j) \in S, \ (i,j) \le (u,v), are zero. Then, again, if g^{p^2} is not the identity matrix, we repeat the argument for g^{p^2}, etc. Since g has only finitely many entries, there exists some positive integer \ell such that all the (i,j)-entries of g^{p^{\ell}}, where (i,j) \in S, are zero. That means g^{p^{\ell}} is the identity matrix and hence |g| is a power of p. So we have shown that the order of every non-identity element of G is a power of p. Thus |G| is a power of p. \ \Box

Let A be a finite dimensional central simple k-algebra. We proved in this theorem that \text{Nrd}_A: A^{\times} \longrightarrow k^{\times} is a group homomorphism. The image of \text{Nrd}_A is an abelian group because it lies in k^{\times}. So if a,b \in A^{\times}, then \text{Nrd}_A(aba^{-1}b^{-1})=\text{Nrd}_A(aa^{-1}) \text{Nrd}_A(bb^{-1})=1. Therefore (A^{\times})', the commutator subgroup of A^{\times}, is contained in \ker \text{Nrd}_A=\{a \in A: \ \text{Nrd}_A(a) = 1\} and so the following definition makes sense.

Definition. Let A be a finite dimensional central simple algebra. The reduced Whitehead group of A is the factor group \text{SK}_1(A) = (\ker \text{Nrd}_A)/(A^{\times})'.

Remark. Let D be a finite dimensional central division k-algebra of degree n. By the theorem in this post, for every a \in D there exists v \in (D^{\times})' such that \text{Nrd}_D(a)=a^nv. So if \text{Nrd}_D(a)=1, then a^n \in (D^{\times})'. Therefore g^n=1 for all g \in \text{SK}_1(D).

Example 1. \text{SK}_1(M_n(k))=\{1\} if k is a field and either n > 2 or n=2 and |k| > 3.

Proof. Let A=M_n(k). Then A^{\times} = \text{GL}(n,k) and

\ker \text{Nrd}_A = \{a \in M_n(k): \ \det(a) = 1 \}=\text{SL}(n,k).

We proved in here that (\text{GL}(n,k))' = \text{SL}(n,k). Thus \text{SK}_1(A)=\{1\}. \Box

Example 2. \text{SK}_1(\mathbb{H})=\{1\}, where \mathbb{H} is the division algebra of quaternions over \mathbb{R}.

Proof. Let z = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H} and suppose that \text{Nrd}_{\mathbb{H}}(z)=\alpha^2+ \beta^2 + \gamma^2 + \delta^2=1. We need to prove that z is in the commutator subgroup of \mathbb{H}^{\times}. We are going to prove a stronger result, i.e. z = aba^{-1}b^{-1} for some a,b \in \mathbb{H}^{\times}. If z \in \mathbb{R}, then \beta = \gamma = \delta = 0 and z = \alpha = \pm 1. If z = 1, then we can choose a=b=1 and if z=-1, we can choose a=i, \ b=j. So we may assume that z \notin \mathbb{R}. We also have z^2 - 2 \alpha z + 1 = 0 (you may either directly check this or use this fact that every element of a finite dimensional central simple algebra satisfies its reduced characteristic polynomial).  Thus (z-\alpha)^2 = \alpha^2 - 1. Note that since \text{Nrd}_{\mathbb{H}}(z)=1 and z \notin \mathbb{R} we have \alpha^2 < 1. So 1 - \alpha^2 = \alpha_0^2 for some 0 \neq \alpha_0 \in \mathbb{R}. Let

w = \alpha_0^{-1}(z - \alpha). \ \ \ \ \ \ \ \ \ (1)

So, since the real part of z is \alpha, the real part of w is zero and hence

w^2 = -1, \ \ \overline{w} = -w. \ \ \ \ \ \ \ \ \ \ (2).

Since the center of \mathbb{H} is \mathbb{R} and w \notin \mathbb{R}, there exists c \in \mathbb{H} such that u = cw-wc \neq 0. Therefore

uw = -wu = \overline{w}u. \ \ \ \ \ \ \ \ \ \ (3)

Now, by (1), we have z = \alpha + \alpha_0 w and \alpha^2+\alpha_0^2=1. So we can write \alpha = \cos \theta and \alpha_0 = \sin \theta. Let v = \cos(\theta/2) + \sin(\theta/2)w. Then (2) gives us

v^2=z, \ \ \ \overline{v}=\cos(\theta/2) + \sin(\theta/2)\overline{w} = \cos(\theta/2) - \sin(\theta/2)w=v^{-1}. \ \ \ \ \ \ \ \ (4)

We also have u v^{-1} = \overline{v^{-1}}u=vu, by (3) and (4). Thus, by (4) again, uv^{-1}u^{-1}v = v^2=z. \ \Box

To be continued …

We will keep the notation in here and here . For a group G and a,b \in G we let [a,b]=aba^{-1}b^{-1}. Recall that G', the commutator subgroup of G, is the subgroup generated by the set \{[a,b]: \ a,b \in G \}.

Remark 1. Using the second part of Problem 1, it is easy to show that E_{ij}(\alpha \beta)=[E_{ir}(\alpha), E_{rj}(\beta)].

We are now ready to prove that the commutator subgroup of the general linear group GL(n,k) is the special linear group SL(n,k) unless n=2 and k has at most 3 elements.

Problem. Prove that GL(n,k)'=SL(n,k) unless n=2 and |k| \leq 3.

Solution. Clearly for any A,B \in GL(n,k) we have \det (ABA^{-1}B^{-1})=1. So [A,B] \in SL(n,k) and hence GL(n,k)' \subseteq SL(n,k). In order to show that SL(n,k) \subseteq GL(n,k)', we only need to prove that GL(n,k)' contains all elementary matrices because, by Remark 2, SL(n,k) is generated by elementary matrices. So we consider two cases.

Case 1 . n \geq 3 : For any i \neq j choose r \notin \{i,j \}. Then, by the above remark, E_{ij}(\alpha)=[E_{ir}(\alpha), E_{rj}(1)].

Case 2 .  n=2 and |k| > 3 : The equation x(x^2-1)=0 has at most three solutions in the field k and so, since k has more than three elements, we can choose a non-zero element \gamma \in k such that \gamma^2 \neq 1. Thus \gamma^2 - 1 is invertible in k. Now given \alpha \in k, let \beta_1=\alpha (\gamma^2 - 1)^{-1} and \beta_2 = \alpha \gamma^2 (1-\gamma^2)^{-1}. Let

A = \begin{pmatrix} \gamma & 0 \\ 0 & \gamma^{-1} \end{pmatrix}.

A quick calculation shows that

E_{12}(\alpha)=[A, E_{12}(\beta_1)] and E_{21}(\alpha)=[A, E_{21}(\beta_2)]. \Box

Remark 2. In the solution of the above problem, we actually showed that every elementary matrix is in the form [a,b] for some a,b \in SL(n,k). Thus SL(n,k)' = SL(n,k).

We will keep our notations in part (1). We now generalize the result proved in Problem 2.

Problem. Let A \in M_n(k) with \det A = 1. Prove that A is a product of elementary matrices.

Solution. Suppose that A = (a_{ij}), i.e. the (i,j)-entry of A is a_{ij}. The first column of A cannot be all zero because then \det A = 0, which is not true. So if a_{11}=0, then choosing a_{i1} \neq 0, we will have the matrix E_{1i}(1)A whose (1,1)-entry is a_{i1} \neq 0. Then we can replace A by E_{1i}(1)A and so we may assume that a_{11} \neq 0. Now let B_1=\prod_{i=2}^n E_{i1}(-a_{11}^{-1}a_{i1}) and C_1= \prod_{j=2}^n E_{1j}(-a_{11}^{-1}a_{1j}). Then the first row and the first column entries of B_1AC_1 are all zero except for the (1,1)-entry which is a_{11}. We can now do the same with the second row and the second column and continue this process until we get matrices B_1, \cdots , B_{n-1} and C_1, \cdots , C_{n-1} each of which is a product of elementary matrices and

D=B_{n-1}B_{n-2} \cdots B_1 A C_1 C_2 \cdots C_{n-1}

is a diagonal matrix. Note that \det D=1 because \det B_i=\det C_i=1 for all i (see the first part of Problem 1) and \det A=1. Thus by Problem 2, D is a product of elementary matrices. Hence

A=B_1^{-1} \cdots B_{n-1}^{-1}D C_{n-1}^{-1} \cdots C_1^{-1}

is also a product of elementary matrices, by the second part of Problem 1. \Box

Definition 1. The general linear group of order n over a field k is

GL(n,k)=\{A \in M_n(k): \ \det A \neq 0 \}.

Definition 2. The special linear group of order n over a field k is

SL(n,k)=\{A \in M_n(k): \ \det A=1 \}.

Remark 1. It is clear that GL(n,k) is a group with matrix multiplication and that SL(n,k) is a subgroup of GL(n,k). If, as usual, we let k^{\times}:=k \setminus \{0\} be the multiplicative group of the field k, then we can define the group homomorphism f : GL(n,k) \longrightarrow k^{\times} by f(A)=\det A. Obviously f is onto and \ker f = SL(n,k). So SL(n,k) is a normal subgroup of GL(n,k) and GL(n,k)/SL(n,k) \cong k^{\times}.

Remark 2. By the first part of Problem 1, every elementary matrix is in SL(n,k). Thus any product of elementary matrices is also in SL(n,k). On the other hand, by the above problem, every element of SL(n,k) is a product of elementary matrices. So we get the following important result:

SL(n,k), as a group, is generated by all elementary matrices.