1 Introduction
Kiefer and Ryzhikov [Reference Kiefer and Ryzhikov3] have recently proved that a finite irreducible semigroup of
$n\times n$
matrices over the field of rational numbers has cardinality at most
$3^{n^2}$
. This bound is a major improvement over previous bounds [Reference Bumpus, Haase, Kiefer, Stoienescu and Tanner1], and their proof has a number of new ideas. However, the proof can be greatly simplified by exploiting the known representation theory of finite semigroups. I present here a short proof using ideas very close to Minkowski’s proof of an analogous result for finite groups. Kiefer and Ryzhikov instead reduce to Minkowski’s theorem in a combinatorial tour de force. A key point in my proof is Rhodes’s result that irreducible finite semigroups of matrices (with zero) act faithfully on the left and right of their (
$0$
-)minimal ideal [Reference Rhodes6].
2 The main result
A semigroup of matrices is irreducible if it has no nontrivial invariant subspace. Let S be an irreducible subsemigroup of
$n\times n$
-matrices over
$\mathbb Q$
. Without loss of generality, I assume that
$0\in S$
, since adjoining it doesn’t change irreducibility. I tacitly assume
$S\neq 0$
, to avoid trivialities.
An ideal of a semigroup with zero is
$0$
-minimal if it contains no proper nonzero ideal. A semigroup with zero is called generalized group mapping if it has a
$0$
-minimal ideal I such that S acts faithfully on both the left and right of I, in which case I must be
$0$
-simple and the unique
$0$
-minimal ideal of S. A semigroup S with zero is
$0$
-simple if
$S^2\neq 0$
and S has no proper nonzero ideals. See [Reference Krohn, Rhodes, Tilson and Arbib4] or [Reference Rhodes and Steinberg7, Chapter 4]. Each
$0$
-simple semigroup S contains a maximal subgroup G in
$S\setminus \{0\}$
(i.e., a subsemigroup maximal for the property of being a group), and any two maximal subgroups are isomorphic [Reference Rhodes and Steinberg7, Appendix A].
Recall that two elements of a semigroup are
$\mathscr J$
-equivalent (respectively,
$\mathscr R$
-equivalent and
$\mathscr L$
-equivalent) if they generate the same principal two-sided (respectively, right and left) ideal. One puts
$\mathscr H=\mathscr R\cap \mathscr L$
. It is well known that each maximal subgroup is an
$\mathscr H$
-class [Reference Rhodes and Steinberg7, Appendix A]. Finite semigroups are stable, meaning that
$$\begin{align*}sx\mathrel{\mathscr{J}} s\iff sx\mathrel{\mathscr{R}} s\ \text{and}\ xs\mathrel{\mathscr{J}} s\iff xs\mathrel{\mathscr{L}} s.\end{align*}$$
See [Reference Rhodes and Steinberg7, Theorem A.2].
Lemma 2.1 (Rhodes [Reference Rhodes6, Lemma 1.5])
Let S be a finite irreducible semigroup containing
$0$
over a field K. Then S is generalized group mapping.
Proof Let I be any
$0$
-minimal ideal of S. The K-span A of S is a simple algebra [Reference Clifford and Preston2, Theorem 5.7]. The K-span of I is an ideal of A, and hence is A by simplicity. But A contains the identity matrix
$1$
[Reference Clifford and Preston2, Lemma 5.32]. Thus,
$1$
is in the K-span of I, say
$1=\sum _{i=1}^k c_is_i$
with
$c_i\in K$
,
$s_i\in I$
. If s and t act the same on the left of I, then
$$\begin{align*}s=s1=\sum_{i=1}^k c_iss_i= \sum_{i=1}^k c_its_i=t1=t\end{align*}$$
and dually for the right. Thus, S is generalized group mapping.
The following is a special case of Proposition 3.28 of Chapter 8 of [Reference Krohn, Rhodes, Tilson and Arbib4].
Proposition 2.2 Let S be a finite generalized group mapping semigroup with
$0$
-minimal ideal I. Let G be a maximal subgroup of
$I\setminus \{0\}$
. Then a homomorphism
${\varphi \colon S\to T}$
is injective if and only if
$\varphi (I\setminus \{0\})\cap \varphi (0)=\emptyset $
and
$\varphi |_G$
is injective.
Proof For the nontrivial direction, assume
$\varphi (s)=\varphi (s')$
. Note that
$s=s'$
in S if and only if
$xsy=xs'y$
for all
$x,y\in I$
(see Lemma 4.6.23 [Reference Rhodes and Steinberg7]), since if
$s\neq s'$
, there exists
$x\in I$
with
$xs\neq xs'$
by faithfulness of the right action, in which case by faithfulness of the left action there is
$y\in I$
with
$xsy\neq xs'y$
. So, let
${x,y\in I}$
. Since
$\varphi (xsy)=\varphi (xs'y)$
, the hypotheses on
$\varphi $
guarantee that either
$xsy=0=xs'y$
or
$xsy,xs'y\in I\setminus \{0\}$
. In the latter case,
$xsy\mathrel {\mathscr J} x\mathrel {\mathscr {J}} y\mathrel {\mathscr J} xs'y$
, and hence by stability,
$xsy\mathrel {\mathscr H} xs'y$
. But then by Green’s lemma [Reference Rhodes and Steinberg7, Lemma A.3.1], there exist
$a,b$
so that
${r\mapsto arb}$
is a bijection from the
$\mathscr H$
-class of
$xsy,xs'y$
to G. Then
$\varphi (axsyb)=\varphi (axs'yb)$
implies that
$axsyb=axs'yb$
since
$\varphi |_G$
is injective. But then
$xsy=xs'y$
, as was required. Thus,
$\varphi $
is injective.
Finally, we record here a well-known lemma, at least for groups.
Lemma 2.3 Let S be a finite subsemigroup of
$M_n(\mathbb Q)$
. Then S is conjugate to a subsemigroup of
$M_n(\mathbb Z)$
.
Proof Note that
$\mathbb Q^n$
does not contain any free abelian subgroup of rank greater than n, because it does not contain a set of more than n linearly independent vectors. Also, recall that the rank of a subgroup of a free abelian group is never larger than the rank of the ambient group. Let
$b_1,\ldots , b_n$
be a basis for
$\mathbb Q^n$
. By finiteness of S, the subgroup
$L=\langle \{b_1,\ldots , b_n\}\cup sb:1\cup \cdots \cup sb:n\rangle $
is an S-invariant finitely generated subgroup of
$\mathbb Q^n$
, and hence is free abelian. Moreover, it has rank n and spans
$\mathbb Q^n$
since it contains
$b_1,\ldots , b_n$
. Then S is represented by integer matrices with respect to any integral basis of L.
Now I prove the upper bound of [Reference Kiefer and Ryzhikov3]. I will use an old result of Minkowski that the kernel of the map
$GL_n(\mathbb Z)\to GL_n(\mathbb Z/p\mathbb Z)$
is torsion-free for p an odd prime. For completeness, I’ll sketch the proof.
Theorem 2.4 (Minkowski [Reference Minkowski5])
Let p be an odd prime. Then
$\ker (\mathrm {GL}_n(\mathbb Z)\to \mathrm {GL}_n(\mathbb Z/p\mathbb Z))$
is torsion-free.
Proof If it is not torsion-free, it has an element A of prime order q. So
$A^q=1$
and
$A=1+p^kC$
where
$C\not \equiv 0\bmod p$
and
$k\geq 1$
. Therefore,
$$\begin{align*}1=(1+p^kC)^q=1+qp^kC+\sum_{i=2}^q\binom{q}{i}p^{ik}C^i,\end{align*}$$
and so
$qC=-\sum _{i=2}^q\binom {q}{i}p^{k(i-1)}C^i\equiv 0\bmod p$
. Thus,
$q=p$
as
$C\not \equiv 0\bmod p$
. Dividing both sides by p yields
$C=-\sum _{i=2}^{p-1}\binom {p}{i}p^{k(i-1)-1}C^i-p^{k(p-1)-1}C^p$
. Since
$p\mid \binom {p}{i}$
for
${1<i<p}$
, this contradicts
$C\not \equiv 0\bmod p$
as
$p> 2$
implies
$k(p-1)-1>1$
.
To improve the bound a little bit, I shall also use that the kernel has only
$2$
-torsion when
$p=2$
(see [Reference Serre8]). Recall that a semigroup is aperiodic if all its maximal subgroups are trivial.
Theorem 2.5 Let
$S\subseteq M_n(\mathbb Q)$
be irreducible. Then
$|S|\leq 3^{n^2}$
. If the maximal subgroup G of the unique (
$0$
-)minimal ideal of S has odd order (e.g., if S is aperiodic), then
${|S|\leq 2^{n^2}}$
.
Proof Without loss of generality, assume that S contains
$0$
and, by Lemma 2.3, that
$S\subseteq M_n(\mathbb Z)$
. By Lemma 2.1, S is generalized group mapping with unique
$0$
-minimal ideal I. Let G be the maximal subgroup of I and
$e\neq 0$
its identity. Let
$p=2$
if G has odd order, and
$p=3$
if G has even order. Suppose that e has rank r. Then
$e\mathbb Z^n$
is a free abelian group of rank r. Indeed, tensoring with
$\mathbb Q$
is exact, so
$\mathbb Q\otimes _{\mathbb Z}e\mathbb Z^n\cong e\mathbb Q^n\cong \mathbb Q^r$
. Since
$e\mathbb Z^n$
, as a subgroup of
$\mathbb Z^n$
, is free abelian, it must then have rank r. Moreover,
$\mathbb Z^n=e\mathbb Z^n\oplus (1-e)\mathbb Z^n$
. Thus, by choosing an integral basis for
$\mathbb Z^n$
adapted to this direct sum, we may assume without loss of generality that
$$\begin{align*}e=\begin{bmatrix} 1_r & 0_{r,n-r}\\ 0_{n-r,r} & 0_{n-r}\end{bmatrix}.\end{align*}$$
Then the canonical projection
$\pi \colon M_n(\mathbb Z)\to M_n(\mathbb Z/p\mathbb Z)$
satisfies
$\pi (e)\neq 0$
, and thus
$\pi $
separates
$0$
from I (as I is
$0$
-minimal and
$\pi ^{-1}(0)\cap I$
is an ideal). Observe that G then sits inside of
$$\begin{align*}\begin{bmatrix} GL_r(\mathbb Z) & 0_{r,n-r}\\ 0_{n-r,r} & 0_{n-r}\end{bmatrix}\end{align*}$$
and so
$\pi |_G$
can be identified with the restriction of the projection
$GL_r(\mathbb Z)\to GL_r(\mathbb Z/p\mathbb Z)$
. Minkowski’s Theorem (or its
$p=2$
variant) then implies
$\pi |_G$
is injective. Therefore,
$\pi |_S$
is injective by Proposition 2.2, and so
$|S|\leq p^{n^2}$
, as required.
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