Groups and semigroups generated by a single unitary orbit

Abstract

We investigate the structure of the multiplicative semigroup generated by the set of matrices that are unitarily equivalent to a given invertible matrix A. In particular, we give necessary and sufficient conditions for such a semigroup to be the special linear group or the general linear group or other classical matrix groups. Furthermore, we use the above to determine the subgroups of the general linear group that are normalized by the unitary group.

Appendix

In this section, we give elementary accessible proofs of Propositions 1.1 and 1.2. We shall simply bridge the gap between what is easy to find in the literature regarding normal subgroups of linear groups and the statements in Propositions 1.1 and 1.2. Our starting points are the following four facts which may be found in [4]. As usual, the centre of a group G is denoted by Z(G).

(1) If $G= {\mathbf{SU}}_n({{\mathbb {C}}})$ then G/Z(G) is simple.

(2) If $G={\mathbf{SL}}_n({{\mathbb {F}}})$ then G/Z(G) is simple unless $n=2$ and $|{\mathbb {F}}|=2$ or 3.

(3) The commutator subgroup of ${\mathbf{U}}_n({{\mathbb {C}}})$ or ${\mathbf{SU}}_n({{\mathbb {C}}})$ is ${\mathbf{SU}}_n({{\mathbb {C}}})$.

(4) The commutator subgroup of ${\mathbf{GL}}_n({{\mathbb {F}}})$ or ${\mathbf{SL}}_n({{\mathbb {F}}})$ is ${\mathbf{SL}}_n({{\mathbb {F}}})$ unless $n=2$ and $|{\mathbb {F}}|=2$ or 3.

The commutator group of a group G will be denoted, as usual, by [G, G] or $G'$. For a subgroup H of G, by [G, H], we denote the subgroup generated by commutators of the form $g^{-1}h^{-1}gh$ where $g \in G$ and $h \in H$. A group G is said to be simple modulo its centre if G/Z(G) is simple.

Proposition 4.1

Let G be a group with centre C and assume that G/C is simple and that the commutator subgroup [G, G] includes the centre. Then every proper normal subgroup of G is included in the centre.

Proof

Assume that H is a normal subgroup that is not included in the centre C. Then HC is a normal subgroup that properly includes the centre, and so HC/C is normal subgroup of G/C. By simplicity of G/C, we conclude that $HC/C = G/C$, i.e. $HC=G$. This implies that $[H,H]=[G,G]$. Therefore $C \subseteq [G,G] = [H,H] \subseteq H$ and so $H=HC=G$. This proves the assertion. $\square$

Proposition 4.2

(a) Every proper normal subgroup in ${\mathbf{SU}}_n({{\mathbb {C}}})$ is included in its centre.

(b) If $n >2$ or $|{\mathbb {F}}| >3$ then every proper normal subgroup of ${\mathbf{SL}}_n({{\mathbb {F}}})$ is included in its centre.

Proof

This follows immediately from Proposition 4.1 and the facts (1)–(4) above. $\square$

Next we identify the normal subgroups of the general linear group and the general unitary group. For normal subgroups of ${\mathbf{GL}}_n({{\mathbb {F}}})$, another proof of the assertion below is in [5].

Proposition 4.3

(a) The normal subgroups of ${\mathbf{U}}_n({{\mathbb {C}}})$ are the subgroups in the centre, i.e., subgroups of scalar matrices or subgroups of the form ${\mathrm{det}}^{-1} (\Gamma )$, where $\Gamma$ is a subgroup of the circle group ${\mathbf{S}}^1.$

(b) If $n >2$ or $|{\mathbb {F}}| >3$, then the normal subgroups of ${\mathbf{GL}}_n({{\mathbb {F}}})$ are the subgroups in the centre or subgroups of the form ${\mathrm{det}}^{-1} (\Gamma )$, where $\Gamma$ is a subgroup of ${{\mathbb {F}}}^{\dagger }$, the multiplicative group of nonzero elements of ${\mathbb {F}}$.

Proof

Let G be any of the groups above, let C be the centre of G, and assume that H is a normal subgroup of G. Then HC is a normal subgroup of G. Furthermore $[H,H] =[HC,HC]$, $[G,H]=[G,HC]$ and $H \subseteq [G,H]$. In addition, [G, H] is a normal subgroup of G and hence also a normal subgroup of [G, G]. By the facts (2)-(3) above [G, G] is ${\mathbf{SU}}_n({{\mathbb {C}}})$ (respectively ${\mathbf{SL}}_n({{\mathbb {F}}})$). By Proposition 4.2, either [G, H] in included in the centre of [G, G] or $[G,H]= [G,G]$. We shall later see that the first alternative implies that H is included in the set of scalar matrices. We consider the second alternative first. From it, we get that $H \supseteq [G,G]$. However ${\mathrm{det}}: G \longrightarrow {{\mathbb {C}}}^{\dagger }$ is an epimorphism with kernel [G, G] and image ${\mathbf{S}}^1$, the circle group, (respectively ${{\mathbb {C}}}^{\dagger }$). It then follows that $H={\mathrm{det}}^{-1} (\Gamma )$, where $\Gamma$ is a subgroup of ${\mathbf{S}}^1 ($respectively a subgroup of ${{\mathbb {C}}}^{\dagger }$).

Next assume the first alternative, i.e, [G, H] in included in the group of scalar matrices of determinant 1. In other words, $SAS^{-1}A^{-1} = \lambda I$ for every $A \in H$ and $S \in G$. This may be written as

$$\begin{aligned} SAS^{-1} = \lambda A \end{aligned}$$

(1)

for every $A \in H$ and $S \in G$ (where $\lambda$ depends on A and S.) We shall show that this implies A is a scalar.

In the unitary case, we may assume that $A = {\mathrm{diag}} (a_{11}, \ldots a_{nn})$. It is easy to see that there is a permutation matrix S such that $SAS^{-1}$ is not a scalar multiple of A unless A is a scalar or $n=2$ and $a_{22}=-a_{11}$. In the latter case, it is clear that there exists a unitary matrix S such that $SAS^{-1}=\begin{pmatrix} 0&{}a_{11} \\ a_{11}&{}0 \end{pmatrix}$ which is not a scalar multiple of A.

In the general linear group, if $A=[a_{ij}]$ is not a scalar, then there is a matrix $B=[b_{ij}]$ similar to A with $b_{11}=0$. To see this, take a nonzero vector ${\mathbf{x}}$ which is not an eigenvalue of A and write the matrix of the linear transformation ${\mathbf{x}} \mapsto A{\mathbf{x}}$ with respect to a basis whose first two terms are ${\mathbf{x}}$ and $A {\mathbf{x}}.$ Therefore we may assume that A itself satisfies the condition that $a_{11}=0$. Applying the above to $A-I$, we conclude that A is similar to a matrix $C=[c_{ij}]$ with $c_{11}=1$. This shows that C is not a scalar multiple of A as desired. $\square$