## Finite subgroups of even order > 2 of GL(n,C) are not simple

Posted: May 1, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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! $\Box$

## A finite p-subgroup of GL(n,k)

Posted: February 27, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields, Linear Algebra
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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 …

## Commutator subgroup of GL(n,k)

Posted: January 29, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $SL(n,k)' = SL(n,k).$

## Elementary matrices (2)

Posted: January 29, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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.