1 Introduction
Throughout this paper, X is an infinite set and
$I(X)$
denotes the symmetric inverse semigroup on X, that is, the semigroup (under composition) consisting of all one-to-one partial transformations whose domain,
$\mathop {\mathrm {dom}}\,\alpha $
, and range,
$\mathop {\mathrm {ran}}\alpha $
, are subsets of X (see [Reference Clifford and Preston2, Volume 1, page 29]). In addition, if
$\alpha \in I(X)$
, we write
$$ \begin{align*}g(\alpha)=\vert X\setminus\mathop{\mathrm{dom}}\,\alpha\vert,\quad d(\alpha)=\vert X\setminus\mathop{\mathrm{ran}}\alpha\vert\end{align*} $$
and refer to these cardinal numbers as the gap and defect of
$\alpha $
, respectively.
In [Reference Sanwong and Sullivan11], the authors studied some algebraic properties of the semigroup defined by
$$ \begin{align*}A(X)=\{\alpha\in I(X): g(\alpha)=d(\alpha)\}.\end{align*} $$
In particular, for uncountable X, they described all maximal subsemigroups of
$A(X)$
, some of which involve a maximal subsemigroup of
$G(X)$
, the symmetric group on X. However, little seems to be known about such semigroups of permutations.
In [Reference Hotzel4], Hotzel describes many different types of maximal subsemigroups of
$BL(p,q)$
, the Baer–Levi semigroup defined on X, when
$\vert X\vert =p\geq q\geq \aleph _0$
(for the definition, see [Reference Clifford and Preston2, Section 8.1]). In addition, in [Reference Hotzel4, page 157], he remarks that
${\{\pi \in G(X):\vert A\pi \setminus A\vert <\vert A\vert \}}$
is a maximal subsemigroup of
$G(X)$
if and only if
$\vert A\vert =1$
or
$\aleph _0\leq \vert A\vert \leq \vert X\setminus A\vert $
(see [Reference Clifford and Preston2, Section 3]). However, Hotzel does not prove this assertion in [Reference Hotzel4], albeit a tour-de-force in many other ways, even though he often uses the assumption that
$\aleph _0\leq \vert A\vert \leq \vert X\setminus A\vert $
in [Reference Hotzel4, Section 3] (see Corollary 3.10 and elsewhere). Furthermore, we cannot find any statement like Hotzel’s assertion anywhere else in the literature.
In [Reference Hotzel4, page 154], Hotzel gives a brief summary of what was known about maximal subgroups of
$G(X)$
in 1995. In fact, maximal subgroups of
$G(X)$
have been extensively studied, particularly when X is infinite. For later developments, see [Reference Brazil, Covington, Penttila, Praeger and Woods1, Reference East, Mitchell and Péresse3] and the references therein. In [Reference Hotzel4, page 153], Hotzel remarks that some maximal subgroups of
$G(X)$
are also maximal as subsemigroups of
$G(X)$
. In [Reference East, Mitchell and Péresse3, Section 10], the authors provide several examples of such maximal subsemigroups of
$G(X)$
. However, here we focus on nongroup maximal subsemigroups of
$G(X)$
, since that is what is needed to support the main result in [Reference Sanwong and Sullivan11, Section 4].
In Section 2, we prove Hotzel’s assertion (as quoted above) and, in Section 4, we prove a linear version of it (for interest, and to support a linear version of [Reference Sanwong and Sullivan11, Section 4] which naturally arises from [Reference Mendes Araújo and Mendes-Gonçalves8]). In Section 3, we observe that, in many cases,
$G(X)$
is not isomorphic to
$G(V)$
, the general linear group on an infinite-dimensional vector space, even though their algebraic properties are similar, for example, the description herein of some of their maximal subsemigroups. This is akin to work in [Reference Mendes-Gonçalves and Sullivan10, Reference Sanwong and Sullivan11] (see Section 3 for more details).
2 Infinite symmetric group
In what follows,
$Y=A\,\dot \cup \, B$
means that Y is a disjoint union of A and B, and we let
$\mathop {\mathrm {id}}\nolimits _X$
denote the identity of
$G(X)$
. Also, following standard practice in transformation semigroup theory, we compose mappings from left to right.
We adapt the convention introduced in [Reference Clifford and Preston2, Volume 2, page 241]: namely, if
${\alpha \in G(X)}$
, then we write
$$ \begin{align*}\alpha=\bigg(\begin{array}{@{}c@{}} a_i \cr x_i \end{array}\bigg)\end{align*} $$
and take as understood that the subscript i belongs to some (unmentioned) index set I, that the abbreviation
$\{a_i\}$
denotes
$\{a_i: i\in I\}$
, and that
${X=\{a_i\}=\{x_i\}}$
and
${x_i \alpha ^{-1} = a_i}$
for each i. In addition, if
$X=A\,\dot \cup \, B=C\,\dot \cup \, D$
, where
$\vert A\vert =\vert C\vert $
and
$\vert B\vert =\vert D\vert $
, we often write
$$ \begin{align*}\alpha=\bigg(\begin{array}{@{}cc@{}} A&B\\ C&D \end{array}\bigg),\end{align*} $$
to indicate that
$\alpha \in G(X)$
consists of some (unspecified) bijection from A to C, together with a bijection from B to D.
For all other notation and terminology in semigroup theory, we refer the reader to [Reference Clifford and Preston2, Reference Howie5].
Clearly,
$G(X)$
contains subsemigroups that are not groups. For example, if
$X={\mathbb Z}$
and
$n\alpha =n+1$
for all
$n\in \mathbb {Z}$
, then the cyclic semigroup
$\langle \alpha \rangle $
is a subsemigroup of
$G(X)$
but
$\alpha ^{-1}\notin \langle \alpha \rangle $
, and this idea can be extended to any infinite X. The next result provides infinitely many examples of nongroup maximal subsemigroups of
$G(X)$
.
Proposition 2.1. Let Y be a subset of X such that
$\aleph _0\leq \vert Y\vert \leq \vert X\setminus Y\vert $
. Then the set
$$ \begin{align*}H(Y)=\{\pi\in G(X):\vert Y\pi\setminus Y\vert <\vert Y\vert\}\end{align*} $$
is a maximal subsemigroup of
$G(X)$
that is not a group.
Proof. Let
$\vert Y\vert =m\leq n=\vert X\setminus Y\vert $
and write
$X=Y\,\dot \cup \, Z$
. Clearly,
$$ \begin{align*}H(Y)=\{\pi\in G(X):\vert Y\pi\cap Z\vert <m\}.\end{align*} $$
Suppose that
$\alpha ,\beta \in H(Y)$
. Since
$Y\alpha =(Y\alpha \cap Y)\cup (Y\alpha \cap Z)$
, we have
$$ \begin{align*}Y\alpha\beta\cap Z\subseteq (Y\beta\cap Z)\cup ((Y\alpha\cap Z)\beta\cap Z).\end{align*} $$
So,
$\vert Y\alpha \beta \cap Z\vert \leq \vert Y\beta \cap Z\vert +\vert Y\alpha \cap Z\vert <m+m=m$
and
$\alpha \beta \in H(Y)$
.
Write
$Y=\{a_k\}\,\dot \cup \,\{a_j\}=\{b_j\}\,\dot \cup \,\{c_j\}$
and
$Z=\{x_j\}\,\dot \cup \,\{x_i\}=\{y_k\}\,\dot \cup \,\{y_i\}$
, where
$\vert K\vert <m =\vert J\vert $
and
$\vert I\vert =n$
. Define
$\alpha \in G(X)$
by
$$ \begin{align} \alpha=\bigg(\begin{array}{@{}cccc@{}} a_k&a_j&x_j&x_i\\ y_k&b_j&c_j&y_i \end{array}\bigg). \end{align} $$
Clearly,
$Y\alpha \cap Y=\{b_j\}$
and
$Y\alpha \cap Z=\{y_k\}$
, so
$\alpha \in H(Y)$
. But
$Y\alpha ^{-1}\cap Z=\{x_j\}$
, so
${\alpha ^{-1}\notin H(Y)}$
. That is,
$H(Y)$
is a subsemigroup of
$G(X)$
that is not a group.
To show that
$H(Y)$
is maximal, we let
$h\in G(X)\setminus H(Y)$
and show that every
${g\in G(X)\setminus H(Y)}$
belongs to
$\langle H(Y),h\rangle $
, the semigroup generated by
$H(Y)\cup \{h\}$
: in other words,
$G(X)=\langle H(Y),h\rangle $
. To do this, we consider four cases.
First, note that if
$h,g\notin H(Y)$
, then
$\vert Yh\cap Z\vert \geq m$
and
$\vert Yg\cap Z\vert \geq m$
. But
${\vert Yh\vert =\vert Yg\vert = \vert Y\vert =m}$
, and so
$\vert Yh\cap Z\vert =m=\vert Yg\cap Z\vert $
. In addition, h can be written as
$$ \begin{align*}h=\bigg(\begin{array}{@{}cccc@{}} Y\cap Zh^{-1} & Y\cap Yh^{-1} & Z\cap Zh^{-1} & Z\cap Yh^{-1}\cr Yh\cap Z & Yh\cap Y & Zh\cap Z & Zh\cap Y \end{array}\bigg).\end{align*} $$
By this, we mean that, for each set in the first row, there is a bijection (determined by the permutation h) between it and the set below it. Note that, in general, one or more of the intersections may be empty. However, if
$Zh\cap Y=\emptyset $
, then
$Zh=Zh\cap Z$
, and so
$Z=Z\cap Zh^{-1}$
and
$Y=Y\cap Yh$
, and likewise for other possibilities.
Case 1. Suppose that
$\vert Z\cap Yg^{-1}\vert <m$
and
$\vert Z\cap Yh^{-1}\vert <m$
. Since
$\vert Z\vert =n\geq m$
, we have
${\vert Z\cap Zg^{-1}\vert =n=\vert Z\cap Zh^{-1}\vert }$
. Also,
${\vert Yg^{-1}\vert =\vert Yh^{-1}\vert =\vert Y\vert =m}$
implies that
$\vert Y\cap Yg^{-1}\vert = m= \vert Y\cap Yh^{-1}\vert $
. Write
$$ \begin{align*} Y\cap Yg^{-1}&=B_1\,\dot\cup\, B_2,\quad\vert B_1\vert =m,\ \vert B_2\vert =\vert Z\cap Yh^{-1}\vert,\cr Y\cap Yh^{-1}&=C_1\,\dot\cup\, C_2,\quad\vert C_1\vert =m,\ \vert C_2\vert =\vert Z\cap Yg^{-1}\vert, \end{align*} $$
and consider
$\pi _1,\pi _2\in G(X)$
, defined as follows:
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}ccccc@{}} Y\cap Zg^{-1} & B_1 & Z\cap Yg^{-1} & Z\cap Zg^{-1}&B_2\\ Y\cap Zh^{-1} & C_1 & C_2& Z\cap Zh^{-1}& Z\cap Yh^{-1} \end{array}\bigg),\\ h&=\bigg(\begin{array}{@{}ccccc@{}} Y\cap Zh^{-1} & C_1&C_2 & Z\cap Zh^{-1} & Z\cap Yh^{-1}\cr Yh\cap Z & C_1h&C_2h & Zh\cap Z & Zh\cap Y \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{@{}ccccc@{}} Yh\cap Z & C_1h&C_2h & Zh\cap Z & Zh\cap Y\\ Yg\cap Z & B_1g&Zg\cap Y & Zg\cap Z &B_2g \end{array}\bigg). \end{align*} $$
Clearly,
$\pi _1$
and
$\pi _2$
are well-defined permutations of X (this depends, in part, on our choice of sets and their cardinals). Now
$Y\pi _1\cap Z=Z\cap Yh^{-1}$
, which has cardinal less than m by supposition, so
$\pi _1\in H(Y)$
. Also,
$Y\pi _2\cap Z=\emptyset $
, so
$\pi _2\in H(Y)$
. Moreover,
${g=\pi _1 h\pi _2}$
provided we define
$\pi _2$
as follows. If, for example,
$x\in B_1$
, then
$\pi _2$
maps
$C_1h$
to
$B_1g$
via
$x\pi _1h\mapsto xg$
, and likewise for each set in the first row of
$\pi _2$
.
Case 2. Suppose that
$\vert Z\cap Yg^{-1}\vert <m$
and
$\vert Z\cap Yh^{-1}\vert =m$
. Then
$\vert Y\cap Yg^{-1}\vert =m$
and
${\vert Z\cap Zg^{-1}\vert =n}$
, but now
$\vert Y\cap Yh^{-1}\vert $
and
$\vert Z\cap Zh^{-1}\vert $
may be unequal and less than m and n, respectively. In fact, this case is more complicated than the others. Clearly, if
$m<n$
, then
$\vert Z\cap Zh^{-1}\vert =n$
. On the other hand, if
$m=n$
, then
$\vert Z\cap Zg^{-1}\vert =m\geq \vert Z\cap Zh^{-1}\vert $
. Write
$$ \begin{align*} Y\cap Yg^{-1} &=B_1\,\dot\cup\, B_2,\qquad\quad \vert B_1\vert =m,\ \vert B_2\vert =\vert Y\cap Yh^{-1}\vert,\cr Z\cap Z g^{-1} &=C_1\,\dot\cup\, C_2,\qquad\ \ \,\, \vert C_1\vert =m,\ \vert C_2\vert =\vert Z\cap Zh^{-1}\vert,\cr Z\cap Y h^{-1} &= E_1\,\dot\cup\, E_3,\qquad\quad \vert E_1\vert =m,\ \vert E_3\vert =\vert Z\cap Y g^{-1}\vert,\cr Y\cap Z h^{-1} &=D_1\,\dot\cup\, D_2,\qquad\ \ \, \vert D_1\vert =\vert D_2\vert=m,\cr &=F_1\,\dot\cup\, F_2\,\dot\cup\, F_3,\quad \vert F_1\vert =\vert F_2\vert=m,\ \vert F_3\vert =\vert E_3\vert. \end{align*} $$
Now define
$\pi _1,\pi _2,\pi _3\in G(X)$
as follows:
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}cccccc@{}} Y\cap Zg^{-1} & B_1 &B_2& C_2&C_1&Z\cap Yg^{-1}\cr D_2 & D_1 & Y\cap Yh^{-1}& Z\cap Zh^{-1}& E_1&E_3 \end{array}\bigg),\\ h&=\bigg(\begin{array}{@{}cccccc@{}} D_2&D_1 & Y\cap Y h^{-1} & Z\cap Zh^{-1} & E_1&E_3\\ D_2h & D_1h& Yh\cap Y & Zh\cap Z & E_1h&E_3h \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{@{}cccccc@{}} D_2h & D_1h& Yh\cap Y & Zh\cap Z & E_1h&E_3h\\ F_2 &Z\cap Y h^{-1} & Y\cap Yh^{-1}& Z\cap Zh^{-1}& F_1&F_3 \end{array}\bigg),\\ h&=\bigg(\begin{array}{@{}cccccc@{}} F_2&Z\cap Yh^{-1} & Y\cap Y h^{-1} & Z\cap Zh^{-1} & F_1&F_3\\ F_2h & Zh\cap Y& Yh\cap Y & Zh\cap Z & F_1h&F_3h \end{array}\bigg),\\ \pi_3&=\bigg(\begin{array}{@{}cccccc@{}} F_2h & Zh\cap Y&Yh\cap Y& Zh\cap Z & F_1h&F_3h\\ Yg\cap Z & B_1g&B_2g & C_2g&C_1g &Zg\cap Y \end{array}\bigg). \end{align*} $$
Clearly,
$\pi _1,\pi _2$
and
$\pi _3$
are well-defined permutations of X. Note that
$Y\pi _1\cap Z=\emptyset $
, so
$\pi _1\in H(Y)$
. Also,
$E_1h\subseteq Y$
and
$F_1\subseteq Y$
(and likewise for
$E_3h$
and
$F_3$
), so
$Y\pi _2\cap Z=\emptyset $
and
$\pi _2\in H(Y)$
. In addition,
$F_ih\subseteq Z$
and
$B_jg\subseteq Y$
for
$i=1,2,3$
and
$j=1,2$
. Therefore,
$Y\pi _3\cap Z=\emptyset $
and
$\pi _3\in H(Y)$
. Furthermore, by defining
$\pi _3$
in a suitable manner, we have
$g=\pi _1 h\pi _2 h\pi _3$
.
Case 3. Suppose that
$\vert Z\cap Yg^{-1}\vert =m$
and
$\vert Z\cap Yh^{-1}\vert <m$
. In this event,
${\vert Y\cap Y h^{-1}\vert =m}$
and
$\vert Z\cap Z h^{-1}\vert =n$
, whereas
$\vert Z\cap Zg^{-1}\vert $
and
$\vert Y\cap Yg^{-1}\vert $
are unknown. As before, if
$m<n$
, then
$\vert Z\cap Zg^{-1}\vert =n$
; if
$m=n$
, then
$\vert Z\cap Zh^{-1}\vert =m\geq \vert Z\cap Zg^{-1}\vert $
. Write
$$ \begin{align*} Z\cap Yg^{-1}&=B_1\,\dot\cup\, B_2\,\dot\cup\, B_3,\quad \vert B_1\vert = \vert B_2\vert =m,\ \vert B_3\vert=\vert Z\cap Yh^{-1}\vert,\cr Y\cap Z h^{-1}&=C_1\,\dot\cup\, C_2,\ \ \,\,\quad\quad \vert C_1\vert =m,\ \vert C_2\vert =\vert Y\cap Yg^{-1}\vert,\cr Z\cap Z h^{-1}&=D_2\,\dot\cup\, D_1,\ \ \,\quad\quad \vert D_2\vert =m,\ \vert D_1\vert =\vert Z\cap Zg^{-1}\vert. \end{align*} $$
Now define
$\pi _1,\pi _2\in G(X)$
as follows:
$$ \begin{align*}\pi_1&=\bigg(\begin{array}{@{}cccccc@{}} Y\cap Zg^{-1} & Y\cap Y g^{-1}& B_1& B_2&Z\cap Z g^{-1}&B_3\\ C_1 & C_2 & Y\cap Yh^{-1}& D_2 &D_1 & Z\cap Y h^{-1} \end{array}\bigg),\\ h&=\bigg(\begin{array}{@{}cccccc@{}} C_1 &C_2 & Y\cap Y h^{-1} & D_2&D_1 & Z\cap Y h^{-1}\cr C_1h & C_2h& Yh\cap Y & D_2h&D_1h& Zh\cap Y \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{@{}cccccc@{}} C_1h & C_2h& Yh\cap Y & D_2h&D_1h& Zh\cap Y\\ Yg\cap Z& Yg\cap Y& B_1g& B_2g& Zg\cap Z& B_3g \end{array}\bigg). \end{align*} $$
Clearly,
$\pi _1$
and
$\pi _2$
are permutations of X. Also,
$Y\pi _1\cap Z=\emptyset =Y\pi _2\cap Z$
, so
${\pi _1,\pi _2\in H(Y)}$
. As before,
$g=\pi _1 h\pi _2$
if
$\pi _2$
is defined suitably.
Case 4. Suppose that
$\vert Z\cap Yg^{-1}\vert =m$
and
$\vert Z\cap Yh^{-1}\vert =m$
. In this case, each of the cardinals
$\vert Y\cap Y g^{-1}\vert ,\vert Z\cap Z g^{-1}\vert $
and
$\vert Y\cap Yh^{-1}\vert ,\vert Z\cap Zh^{-1}\vert $
is unknown. If
$m=n$
, write
$$ \begin{align*} Z\cap Yg^{-1}&=B_1\,\dot\cup\, B_2\,\dot\cup\, B_3,\quad \vert B_1\vert =\vert Y\cap Yh^{-1}\vert,\ \vert B_2\vert =\vert Z\cap Zh^{-1}\vert,\ \vert B_3\vert=m,\cr Y\cap Z h^{-1}&=C_1\,\dot\cup\, C_2\,\dot\cup\, C_3,\quad \vert C_1\vert =m,\ \vert C_2\vert =\vert Y\cap Yg^{-1}\vert,\ \vert C_3\vert=\vert Z\cap Zg^{-1}\vert, \end{align*} $$
and define
$\pi _1,\pi _2\in G(X)$
as follows:
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}cccccc@{}} Y\cap Zg^{-1} & Y\cap Y g^{-1}& Z\cap Z g^{-1}&B_1&B_2&B_3\\ C_1 & C_2 & C_3& Y\cap Yh^{-1}& Z\cap Zh^{-1}& Z\cap Y h^{-1} \end{array}\bigg),\\ h&=\bigg(\begin{array}{@{}cccccc@{}} C_1 &C_2 &C_3& Y\cap Y h^{-1} & Z\cap Zh^{-1}& Z\cap Y h^{-1}\cr C_1h & C_2h& C_3h& Yh\cap Y & Zh\cap Z& Zh\cap Y \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{@{}cccccc@{}} C_1h & C_2h& C_3h& Yh\cap Y & Zh\cap Z& Zh\cap Y\\ Yg\cap Z& Yg\cap Y& Zg\cap Z& B_1g&B_2g& B_3g \end{array}\bigg). \end{align*} $$
On the other hand, if
$m<n$
, then
$\vert Z\cap Z g^{-1}\vert =\vert Z\cap Zh^{-1}\vert =n$
. In this case, write
$$ \begin{align*} Z\cap Yg^{-1}&=B_1\,\dot\cup\, B_2,\quad \vert B_1\vert=m,\ \vert B_2\vert =\vert Y\cap Yh^{-1}\vert,\cr Y\cap Z h^{-1}&=C_1\,\dot\cup\, C_2,\quad \vert C_1\vert =m,\ \vert C_2\vert =\vert Y\cap Yg^{-1}\vert, \end{align*} $$
and define
$\pi _1,\pi _2\in G(X)$
as follows:
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}ccccc@{}} Y\cap Zg^{-1} & Y\cap Y g^{-1}& Z\cap Z g^{-1}&B_1&B_2\\ C_1 & C_2 & Z\cap Zh^{-1}& Z\cap Y h^{-1}& Y\cap Yh^{-1} \end{array}\bigg),\\h&=\bigg(\begin{array}{@{}ccccc@{}} C_1 &C_2 & Z\cap Zh^{-1}& Z\cap Y h^{-1}& Y\cap Y h^{-1}\cr C_1h & C_2h & Zh\cap Z& Zh\cap Y& Yh\cap Y \end{array}\bigg),\\\pi_2&=\bigg(\begin{array}{@{}ccccc@{}} C_1h & C_2h & Zh\cap Z& Zh\cap Y& Yh\cap Y\\ Yg\cap Z& Yg\cap Y& Zg\cap Z& B_1g&B_2g \end{array}\bigg). \end{align*} $$
In both cases,
$\pi _1,\pi _2\in G(X)$
and
$Y\pi _1\cap Z=\emptyset =Y\pi _2\cap Z$
. Hence,
$\pi _1,\pi _2\in H(Y)$
and, as before,
$g=\pi _1 h\pi _2$
for a suitably defined
$\pi _2$
.
We note that the
$\alpha $
defined in (2.1) belongs to
$H(Y)$
but
$$ \begin{align*}\alpha\notin H(Z)=\{\beta\in G(X):\vert Z\beta\cap Y\vert<n\},\end{align*} $$
even when
$n=m$
. Also, there are
$2^n$
subsets of X with cardinal n, and hence there are
$2^n$
partitions of X into two subsets, each with cardinal n. Consequently, there are at least
$2^n$
distinct nongroup subsemigroups of
$G(X)$
that are maximal.
Hotzel’s claim follows easily from Proposition 2.1.
Theorem 2.2. Let A be a subset of X. Then the set
$$ \begin{align*}H(A)=\{\pi\in G(X):\vert A\pi\setminus A\vert <\vert A\vert\}\end{align*} $$
is a maximal subsemigroup of
$G(X)$
if and only if
$\vert A\vert =1$
or
$\aleph _0\leq \vert A\vert \leq \vert X\setminus A\vert $
.
Proof. First, suppose that
$\vert A\vert =m$
for some
$m{\kern-1pt}\in{\kern-1pt} \mathbb {N}\setminus \{1\}$
, and write
$A{\kern-1pt}={\kern-1pt}\{a_1,a_2,\dots ,a_m\}$
. Let
$X\setminus A=\{x_1,x_2,\dots ,x_m\}\,\dot \cup \,\{y_i\}$
, with
$\vert I\vert =n\geq \aleph _0>m$
, and define
$\alpha ,\beta \in G(X)$
by
$$ \begin{align*}\alpha=\bigg(\begin{array}{@{}ccccccccc@{}} a_1&a_2&\dots&a_m&x_1&x_2&\dots&x_m&y_i\cr a_1&x_2&\dots&x_m&x_1&a_2&\dots&a_m&y_i \end{array}\bigg),\end{align*} $$
$$ \begin{align*}\beta=\bigg(\begin{array}{@{}ccccccccc@{}}a_1&a_2&\dots&a_m&x_1&x_2&\dots&x_m&y_i\cr x_1&a_2&\dots&a_m&a_1&x_2&\dots&x_m&y_i \end{array}\bigg).\end{align*} $$
It is easy to verify that
$\vert A\alpha \setminus A\vert =m-1$
and
$\vert A\beta \setminus A\vert =1$
, and hence
$\alpha ,\beta \in H(A)$
. But
$\vert A(\alpha \beta )\setminus A\vert =m$
, and so
$\alpha \beta \notin H(A)$
. Therefore,
$H(A)$
is not a semigroup when
${2\leq \vert A\vert <\aleph _0}$
.
On the other hand, if A is infinite but
$\vert A\vert>\vert X\setminus A\vert $
, then
$\vert A\vert =\vert X\vert $
. Given that
$\pi \in G(X)$
,
$A\pi \setminus A\subseteq X\setminus A$
. Therefore,
$\vert A\pi \setminus A\vert \leq \vert X\setminus A\vert <\vert A\vert $
, and so
$H(A)=G(X)$
. Thus, we have just proved that
$\vert A\vert =1$
or
$\aleph _0\leq \vert A\vert \leq \vert X\setminus A\vert $
when
$H(A)$
is a maximal subsemigroup of
$G(X)$
.
Conversely, assume that
$\vert A\vert =1$
and write
$A=\{a\}$
. Clearly,
$$ \begin{align*}H(A)=\{\pi\in G(X): A\pi\setminus A=\emptyset\}=\{\pi\in G(X): a\pi=a\}. \end{align*} $$
It is not difficult to see that
$H(A)$
is a subsemigroup of
$G(X)$
. In fact, it is a subgroup of
$G(X)$
. Next, we prove that
$H(A)$
is a maximal subsemigroup. To do this, we let
${h\in G(X)\setminus H(A)}$
and we show that
$g\in \langle H(A),h\rangle $
for every
$g\in G(X)\setminus H(A)$
. Since
$h,g\notin H(A)$
, it follows that
$ah=b$
and
$ag=c$
for some
$b,c\in X\setminus \{a\}$
. Also, there exist
$d,e\in X\setminus \{a\}$
such that
$dh=a$
and
$eg=a$
. Write
$X=\{a,e\}\,\dot \cup \, Y=\{a,d\}\,\dot \cup \, Z$
and define
$\pi _1,\pi _2$
in
$G(X)$
by
$$ \begin{align*}\pi_1=\bigg(\begin{array}{@{}ccc@{}} a&e&Y\\ a&d&Z \end{array}\bigg),\quad \pi_2=\bigg(\begin{array}{@{}ccc@{}} a&b&Zh\\ a&c&Yg \end{array}\bigg).\end{align*} $$
Clearly,
$\pi _1,\pi _2\in H(A)$
and
$g=\pi _1h\pi _2$
if
$\pi _2$
is suitably defined. Thus,
$H(A)$
is a maximal subsemigroup of
$G(X)$
if
$\vert A\vert =1$
. By Proposition 2.1, this is also true if
$\aleph _0\leq \vert A\vert \leq \vert X\setminus A\vert $
.
3 An isomorphism problem
In [Reference Mendes-Gonçalves and Sullivan9], the authors proved that the Baer–Levi semigroup
$BL(p,q)$
of type
$(p,q)$
defined on an infinite set is never isomorphic to its linear counterpart
$GS(m,n)$
defined on an infinite-dimensional vector space (for the definitions, see [Reference Mendes-Gonçalves and Sullivan9]). This is surprising since
$BL(p,q)$
and
$GS(m,n)$
have many algebraic properties in common. Likewise, in [Reference Mendes-Gonçalves and Sullivan10], the same authors showed that the symmetric inverse semigroup
$I(X)$
defined on an arbitrary set is almost never isomorphic to the analogous semigroup
$I(V)$
defined on an arbitrary vector space V over a field F. In Section 2, we provided a family of nongroup maximal subsemigroups of
$G(X)$
. Before we do the same for
$G(V)$
in Section 4, we observe that, although
$G(X)$
and
$G(V)$
are similar in character, these groups are almost never isomorphic. In fact, it is well known that the centre of
$G(X)$
is
$\{\mathop {\mathrm {id}}\nolimits _X\}$
and the centre of
$G(V)$
is
$\{k\mathop {\mathrm {id}}\nolimits _V:k\in F\setminus \{0\}\}$
. Thus,
$G(X)$
and
$G(V)$
are never isomorphic if
$\vert F\vert>2$
.
Theorem 3.1. Any semigroup S can be embedded in
$T(V)$
, the semigroup of all linear transformations of some vector space V with dimension
$\vert S\vert $
, if S contains an identity (or
$\vert S\vert +1 $
if S does not contain an identity).
Proof. Write
$S^1=\{ a_i \}$
. Let F be any field and let
$F_i$
be a copy of F for each
$i\in I$
. As in [Reference Hungerford6, page 182, Remark (c)], we let V be the vector space
$\sum F_i$
over F whose basis can be identified in a natural way with
$\{ a_i \}$
: that is,
$\sum F_i$
is the set of all
$(r_i)_{i\in I}$
, where
$r_i\in F_i$
and at most finitely many
$r_i$
are nonzero.
For each
$x\in S$
, let
$\rho _{x}:S^1 \to S^1,a_i\mapsto a_ix$
, be a mapping of the basis
$\{a_i\}$
into itself (note that
$\rho _x$
may not be injective). Hence,
$\rho _{x}$
can be extended by linearity to an element of
$T(V)$
that we also denote by
$\rho _x$
. Clearly, for each
$x,y\in S$
,
$\rho _{xy}$
and
$\rho _x\circ \rho _y$
agree on the basis
$\{a_i\}$
, and hence they agree on all of V. That is, the mapping
$\rho :S\to T(V), x\to \rho _x$
, is a homomorphism. Moreover, since
$1\in \{a_i\}$
,
$\rho _x=\rho _y$
implies that
$x=y$
, so
$\rho $
is injective and the result follows.
It is well known that, if
$\vert X\vert =m\geq \aleph _0$
, then
$\vert G(X)\vert =2^m$
(compare [Reference Mendes-Gonçalves and Sullivan9, page 479]). To determine the cardinal of
$G(V)$
when
$\mathop {\mathrm {dim}}\, V=n\geq \aleph _0$
, we first recall [Reference Jacobson7, Volume II, page 245]: if V is a vector space over a field F and
$\mathop {\mathrm {dim}}\, V = n \geq \aleph _0$
, then
$\vert V\vert =n\times \vert F\vert $
. From this, we deduce (as in [Reference Mendes-Gonçalves and Sullivan9, page 480]) that
$\vert T(V)\vert =\vert V\vert ^n$
and thus
$$ \begin{align*}\vert T(V)\vert = \begin{cases} 2^n & \text{ if } \vert F \vert \leq n,\cr \vert F \vert^n & \text{ if } \vert F \vert> n.\cr \end{cases}\end{align*} $$
Lemma 3.2. If
$\mathop {\mathrm {dim}}\, V=n\geq \aleph _0$
, then the cardinal of
$G(V)$
is
$\vert V\vert ^n$
.
Proof. Suppose that
$\{a_i\}$
is a basis for V and
$\vert F\vert \geq 3$
. In this case, if
$k_i\in F\setminus \{0,1\}$
for each i, then
$\{k_ia_i\}$
is a basis for V that differs from
$\{a_i\}$
. Now each bijection
$k_ia_i\mapsto a_i$
extends by linearity to some
$\pi _k\in G(V)$
, and
$\pi _k\neq \pi _{k'}$
for
$k\neq k'$
. Note that there are
$\vert F \vert ^n$
bases for V of the form
$\{k_ia_i\}$
. Since
$\vert G(A)\vert =2^n$
for each basis A of V,
$$ \begin{align*}\vert G(V)\vert\geq 2^n.\vert F\vert^n=(n.\vert F\vert)^n=\vert V\vert^n.\end{align*} $$
But
$\vert V \vert ^n=\vert T(V)\vert \geq \vert G(V)\vert $
, and equality follows. Now suppose that
$\vert F\vert =2$
. For each fixed
$i_0\in I$
, write
$I'=I\setminus \{i_0\}$
and choose
$j_0\in I'$
. Then
$\{a_i:i\in I'\}\,\dot \cup \,\{a_{i_0}+a_{j_0}\}$
is a basis for V and so the number of bases for V is at least n. Hence,
$$ \begin{align*}\vert G(V)\vert\geq n.2^n=(n.2)^n=\vert V\vert^n,\end{align*} $$
and thus we also have equality in the case
$\vert F \vert =2$
.
In passing, we observe that, if
$\vert F\vert =2$
and
$2^m\not = 2^n$
, where
$m,n\geq \aleph _0$
as above, then
$G(X)$
is not isomorphic to
$G(V)$
.
4 General linear group
In Section 2, we provided a family of nongroup maximal subsemigroups of the symmetric group
$G(X)$
on an infinite set X. Here, we do the same for
$G(V)$
, the general linear group on an infinite-dimensional vector space V. When
$\alpha \in G(V)$
, we take the notation displayed at the start of Section 2 to mean that
$\alpha $
is the extension by linearity to the whole of V of a bijection between bases
$\{a_i\}$
and
$\{x_i\}$
for V. The subspace U of V generated by a linearly independent subset
$\{u_i\}$
of V is denoted by
$\langle u_i\rangle $
, and we write
$\mathop {\mathrm {dim}}\, U=\vert I\vert $
. Observe that, given that
$\alpha \in G(V)$
and
$U\leq V$
,
$$ \begin{align*}U=\langle u_i\rangle\oplus\langle u_j\rangle\quad \text{if and only if } U\alpha=\langle u_i\alpha\rangle\oplus\langle u_j\alpha\rangle.\end{align*} $$
Our next result is the linear analogue of Proposition 2.1. In the set case, the complement of Y in X is unique and this makes the definition of appropriate mappings in
$G(X)$
and the proof of maximality simpler. But the problem here is that, given a subspace W of V, we may not fix a complementary subspace U of W in V and define all linear mappings in
$G(V)$
necessary for the proof of our result by considering the images of W and of the fixed U. The concept of quotient space plays an important role, since it simplifies the task, given all possible choices of complementary subspaces. In fact, we are mainly concerned with codimensions of subspaces in a vector space, and we know that
$\mathop {\mathrm {codim}} W=\mathop {\mathrm {dim}}\, V/W$
.
Proposition 4.1. Let W be a subspace of V with
$\aleph _0\leq \mathop {\mathrm {dim}}\, W=m\leq \mathop {\mathrm {codim}} W$
. If
$$ \begin{align*}H(W)=\{\alpha\in G(V):\mathop{\mathrm{dim}}\, W\alpha/(W\alpha\cap W) <m\},\end{align*} $$
then
$H(W)$
is a maximal subsemigroup of
$G(V)$
that is not a group.
Proof. Suppose that
$\alpha ,\beta \in H(W)$
and let
$W\alpha \cap W=\langle a_i\rangle $
. Write
$W\alpha =\langle a_i\rangle \oplus \langle b_j\rangle $
, where
$\vert J\vert =\mathop {\mathrm {dim}}\, W\alpha /(W\alpha \cap W) <m$
. Since
$\alpha \in G(V)$
, there exist unique
$w_i$
,
$w_j$
in W such that
$a_i=w_i\alpha $
and
$b_j=w_j\alpha $
for every i and every j. It is not difficult to see that
$W=\langle w_i\rangle \oplus \langle w_j\rangle $
. On the other hand,
$\langle a_i\rangle \subseteq W$
and we may write
$W=\langle a_i\rangle \oplus \langle a_\ell \rangle $
. Moreover,
$\{a_i\}\,\dot \cup \,\{a_\ell \}\,\dot \cup \,\{b_j\}$
is a linearly independent subset of V and
$$ \begin{align*}V=\langle a_i\rangle\oplus\langle a_\ell\rangle\oplus\langle b_j\rangle\oplus \langle b_t\rangle,\end{align*} $$
where T may be empty. Also, if
$v_\ell ,v_t\in V$
are such that
$v_\ell \alpha =a_\ell $
and
$v_t\alpha =b_t$
, for each
$\ell $
and each t, then
$V=\langle w_i\rangle \oplus \langle w_j\rangle \oplus \langle v_\ell \rangle \oplus \langle v_t\rangle $
. Note that
$$ \begin{align*}\alpha=\bigg(\begin{array}{@{}cccc@{}} w_i&w_j&v_\ell&v_t\cr a_i&b_j&a_\ell&b_t\end{array}\bigg).\end{align*} $$
Write
$\langle a_i\beta \rangle \cap W=\langle c_r\rangle $
and
$\langle a_i\beta \rangle =\langle c_r\rangle \oplus \langle c_s\rangle $
. Similarly, let
$\langle b_j\beta \rangle \cap W=\langle d_x\rangle $
and
$\langle b_j\beta \rangle =\langle d_x\rangle \oplus \langle d_y\rangle $
. Then
$$ \begin{align*}W\beta=\langle a_i\beta\rangle\oplus\langle a_\ell\beta\rangle=\langle c_r\rangle\oplus\langle c_s\rangle\oplus\langle a_\ell\beta\rangle\end{align*} $$
and
$$ \begin{align*}W\alpha\beta=\langle a_i\beta\rangle\oplus\langle b_j\beta\rangle=\langle c_r\rangle\oplus\langle c_s\rangle\oplus\langle d_x\rangle\oplus\langle d_y\rangle.\end{align*} $$
Since
$\langle c_r\rangle \oplus \langle d_x\rangle \subseteq W$
, we have
$\mathop {\mathrm {dim}}\, W\alpha \beta /(W\alpha \beta \cap W)\leq \vert S\vert +\vert Y\vert $
. But we also have
$\vert J\vert =\mathop {\mathrm {dim}} W\alpha / (W\alpha \cap W)<m$
and
$\vert Y\vert \leq \vert J\vert $
. Also,
$$ \begin{align*}\vert S\vert= \mathop{\mathrm{dim}}\,\langle a_i\beta\rangle/(\langle a_i\beta\rangle\cap W)\leq \mathop{\mathrm{dim}}\, W\beta/(W\beta\cap W)<m.\end{align*} $$
Thus,
$\vert S\vert +\vert Y\vert <m+m=m$
, and
$\alpha \beta \in H(W)$
.
Now write
$W\kern1.5pt{=}\kern1.5pt\langle w_k\rangle \oplus \langle w_j\rangle \kern1.5pt{=}\kern1.5pt\langle v_j\rangle \oplus \langle u_j\rangle $
and
${V\kern1.5pt{=}\kern1.5pt W\oplus \langle a_j\rangle \oplus \langle a_i\rangle \kern1.5pt{=}\kern1.5pt W\oplus \langle b_k\rangle \oplus \langle b_i\rangle }$
, where
$\vert K\vert <\vert J\vert =m\leq n=\vert I\vert =\mathop {\mathrm {codim}} W$
. Define
$\alpha \in G(V)$
by
$$ \begin{align*}\alpha=\bigg(\begin{array}{@{}cccc@{}} w_k&w_j&a_j&a_i\\ b_k&v_j&u_j&b_i \end{array}\bigg).\end{align*} $$
Clearly,
$W\alpha =\langle b_k\rangle \oplus \langle v_j\rangle $
and so
$\mathop {\mathrm {dim}}\, W\alpha /(W\alpha \cap W)=\vert K\vert <m$
. Thus,
$\alpha \in H(W)$
. But
$W\alpha ^{-1}=\langle w_j\rangle \oplus \langle a_j\rangle $
, and hence
$\mathop {\mathrm {dim}}\, W\alpha ^{-1}/(W\alpha ^{-1}\cap W)=\vert J\vert =m$
and
$\alpha ^{-1}\notin H(W)$
. In other words, we have just shown that
$H(W)$
is a subsemigroup of
$G(V)$
that is not a group.
To show that
$H(W)$
is maximal, we let
$h\in G(V)\setminus H(W)$
and we show that, for every
$g\in G(V)\setminus H(W)$
, we have
$g\in \langle H(W),h\rangle $
. Given that
$h,g\in G(V)\setminus H(W)$
, we have
$\mathop {\mathrm {dim}}\, Wh/(Wh\cap W)\geq m$
and
$\mathop {\mathrm {dim}}\, Wg/(Wg\cap W)\geq m$
. But
$\mathop {\mathrm {dim}}\, Wh=\mathop {\mathrm {dim}}\, Wg=\mathop {\mathrm {dim}}\, W=m$
, since
$h,g\in G(V)$
. Therefore,
$$ \begin{align*}\mathop{\mathrm{dim}}\, Wh/(Wh\cap W)=\mathop{\mathrm{dim}}\, Wg/(Wg\cap W)=m.\end{align*} $$
Write
$Wg^{-1}\cap W=\langle a_ig^{-1}\rangle $
and
$Wg^{-1}=\langle a_ig^{-1}\rangle \oplus \langle b_\ell g^{-1}\rangle $
. Then
$W=\langle a_i\rangle \oplus \langle b_\ell \rangle $
. Also, we may write
$W=\langle a_ig^{-1}\rangle \oplus \langle u_jg^{-1}\rangle $
and
$V=\langle a_ig^{-1}\rangle \oplus \langle u_jg^{-1}\rangle \oplus \langle b_\ell g^{-1}\rangle \oplus \langle u_k g^{-1}\rangle $
. Clearly,
$\vert L\vert +\vert K\vert =\mathop {\mathrm {codim}} W$
and
$\vert I\vert +\vert J\vert =\vert I\vert +\vert L\vert =m$
. Since
$W=\langle a_i\rangle \oplus \langle b_\ell \rangle $
and
$Wg=\langle (a_ig^{-1})g\rangle \oplus \langle (u_jg^{-1})g\rangle =\langle a_i\rangle \oplus \langle u_j\rangle $
,
$$ \begin{align*}m=\mathop{\mathrm{dim}}\, Wg/(Wg\cap W)\leq \vert J\vert\leq \mathop{\mathrm{dim}}\, Wg=m,\end{align*} $$
and hence
$\vert J\vert =m$
.
Proceeding similarly, write
$Wh^{-1}{\kern-1pt}\cap{\kern-1pt} W=\langle c_ph^{-1}\rangle $
,
$Wh^{-1}{\kern-1pt}={\kern-1pt}\langle c_ph^{-1}\rangle \oplus \langle d_q h^{-1}\rangle $
and
$W=\langle c_ph^{-1}\rangle \oplus \langle v_rh^{-1}\rangle $
. Then
$W=\langle c_p\rangle \oplus \langle d_q\rangle $
and we may write
$$ \begin{align*}V=\langle c_ph^{-1}\rangle\oplus\langle v_rh^{-1}\rangle\oplus\langle d_q h^{-1}\rangle\oplus\langle v_s h^{-1}\rangle.\end{align*} $$
As before, we may conclude that
$\vert P\vert +\vert Q\vert =\vert P\vert +\vert R\vert =m$
,
$\vert Q\vert +\vert S\vert =\mathop {\mathrm {codim}} W$
and
$\vert R\vert =m$
.
Case 1. Suppose that
$\mathop {\mathrm {dim}}\, Wh^{-1}/(Wh^{-1}\cap W)<m$
and
$\mathop {\mathrm {dim}}\, Wg^{-1}/(Wg^{-1}\cap W)<m$
. Then
$\vert L\vert =\mathop {\mathrm {dim}}\, Wg^{-1}/(Wg^{-1}\cap W)<m\leq \mathop {\mathrm {codim}} W$
, so
$\vert K\vert =\mathop {\mathrm {codim}} W$
and
$\vert I\vert =m$
. Since
$\vert J\vert =\vert I\vert $
, we may write
$\{u_jg^{-1}\}$
as
$\{u_ig^{-1}\}$
.
Analogously,
$\vert Q\vert <m=\vert P\vert =\vert R\vert $
and
$\vert S\vert =\mathop {\mathrm {codim}} W$
. Therefore, we may write
$\{c_ph^{-1}\}$
,
$\{v_rh^{-1}\}$
and
$\{v_sh^{-1}\}$
as
$\{c_ih^{-1}\}$
,
$\{v_ih^{-1}\}$
and
$\{v_kh^{-1}\}$
, respectively. Since
$\vert Q\vert <\vert I\vert $
,
$\vert L\vert <\vert I\vert $
and
$\vert I\vert =m\geq \aleph _0$
, we may write
$$ \begin{align*}\langle a_ig^{-1}\rangle=\langle w_ig^{-1}\rangle\oplus\langle w_qg^{-1}\rangle,\quad \langle c_ih^{-1}\rangle=\langle y_ih^{-1}\rangle\oplus\langle y_\ell h^{-1}\rangle.\end{align*} $$
Now define
$\pi _1,\pi _2$
in
$G(V)$
by
$$ \begin{align*}\pi_1=\bigg(\begin{array}{@{}ccccc@{}}u_ig^{-1}&w_ig^{-1}&b_\ell g^{-1}&u_k g^{-1}&w_q g^{-1}\cr v_ih^{-1}&y_ih^{-1}&y_\ell h^{-1}&v_k h^{-1}&d_q h^{-1} \end{array}\bigg),\quad \pi_2=\bigg(\begin{array}{ccccc}v_i&y_i&y_\ell&v_k&d_q\cr u_i&w_i&b_\ell&u_k&w_q \end{array}\bigg).\end{align*} $$
Clearly,
$g=\pi _1 h\pi _2$
. It is not difficult to see that
${W\pi _1=\langle v_ih^{-1}\rangle \oplus \langle y_ih^{-1}\rangle \oplus \langle d_qh^{-1}\rangle }$
. But
$\langle v_ih^{-1}\rangle \oplus \langle y_ih^{-1}\rangle \subseteq W$
, and so
$\mathop {\mathrm {dim}}\, W\pi _1/(W\pi _1\cap W)\leq \vert Q\vert <m$
. On the other hand,
$W\pi _2=\langle w_i\rangle \oplus \langle b_\ell \rangle \oplus \langle w_q\rangle \subseteq W$
, and so
$\mathop {\mathrm {dim}}\, W\pi _2/(W\pi _2\cap W)=0<m$
. Thus,
${\pi _1,\pi _2\in H(W)}$
.
Case 2. Suppose that
$\mathop {\mathrm {dim}}\, Wh^{-1}/(Wh^{-1}\cap W)=m$
and
$\mathop {\mathrm {dim}}\, Wg^{-1}/(Wg^{-1}\cap W)<m$
. Then
$\vert Q\vert =m$
,
$\vert L\vert <m$
, and this inequality implies that
$\vert I\vert =m\leq \mathop {\mathrm {codim}} W=\vert K\vert $
, but
$\vert P\vert $
is unknown (at most m). Also,
$\vert Q\vert +\vert S\vert =\vert I\vert +\vert S\vert =\mathop {\mathrm {codim}} W$
and we may write
$\{u_j g^{-1}\}$
,
$\{v_rh^{-1}\}$
and
$\{d_qh^{-1}\}$
as
$\{u_i g^{-1}\}$
,
$\{v_ih^{-1}\}$
and
$\{d_ih^{-1}\}$
, respectively. Also, let
$$ \begin{align*} \langle a_ig^{-1}\rangle&=\langle w_ig^{-1}\rangle\oplus\langle w_pg^{-1}\rangle,\cr \langle u_kg^{-1}\rangle&=\langle y_ig^{-1}\rangle\oplus\langle y_sg^{-1}\rangle,\cr \langle d_ih^{-1}\rangle&=\langle x_ih^{-1}\rangle\oplus\langle x_\ell h^{-1}\rangle,\cr \langle v_ih^{-1}\rangle&=\langle z_ih^{-1}\rangle\oplus\langle z^*_ih^{-1}\rangle=\langle z'_ih^{-1}\rangle\oplus\langle z"_ih^{-1}\rangle\oplus\langle z"'_\ell h^{-1}\rangle, \end{align*} $$
and define
$\pi _1,\pi _2,\pi _3$
in
$G(V)$
by
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}cccccc@{}} u_ig^{-1}&w_ig^{-1}&w_p g^{-1}&y_s g^{-1}&y_i g^{-1}&b_\ell g^{-1}\cr z^*_ih^{-1}&z_ih^{-1}&c_p h^{-1}&v_s h^{-1}&x_i h^{-1}&x_\ell h^{-1} \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{@{}cccccc@{}} z^*_i&z_i&c_p&v_s&x_i&x_\ell\\ z"_ih^{-1}&d_ih^{-1}&c_ph^{-1}&v_s h^{-1}&z'_ih^{-1}&z"'_\ell h^{-1} \end{array}\bigg),\\ \pi_3&=\bigg(\begin{array}{@{}cccccc@{}} z"_i&d_i&c_p&v_s&z'_i&z"'_\ell\\ u_i&w_i&w_p&y_s&y_i&b_\ell \end{array}\bigg). \end{align*} $$
Clearly,
$g=\pi _1 h\pi _2 h\pi _3$
. Also,
$$ \begin{align*} W\pi_1&=\langle z^*_ih^{-1}\rangle\oplus\langle z_ih^{-1}\rangle\oplus\langle c_ph^{-1}\rangle,\cr W\pi_2&=\langle c_ph^{-1}\rangle\oplus\langle z'_ih^{-1}\rangle\oplus\langle z"'_\ell h^{-1}\rangle,\cr W\pi_3&=\langle w_i\rangle\oplus\langle w_p\rangle, \end{align*} $$
and so, for
$i=1,2,3$
,
$W\pi _i\subseteq W$
and
$\mathop {\mathrm {dim}}\, W\pi _i/(W\pi _i\cap W)=0<m$
, that is,
$\pi _i\in H(W)$
.
Case 3. Suppose that
$\mathop {\mathrm {dim}}\, Wh^{-1}/(Wh^{-1}\cap W)<m$
and
$\mathop {\mathrm {dim}}\, Wg^{-1}/(Wg^{-1}\cap W)=m$
. In this event,
$\vert I\vert \leq m=\vert J\vert =\vert L\vert \leq \mathop {\mathrm {codim}} W=\vert L\vert +\vert K\vert $
. On the other hand,
$\vert Q\vert <m=\vert P\vert =\vert R\vert \leq \mathop {\mathrm {codim}} W=\vert S\vert $
. Thus, we may write
$\{b_\ell g^{-1}\}$
,
$\{c_ph^{-1}\}$
and
$\{v_rh^{-1}\}$
as
$\{b_j g^{-1}\}$
,
$\{c_jh^{-1}\}$
and
$\{v_jh^{-1}\}$
, respectively. Also, write
$$ \begin{align*} \langle b_jg^{-1}\rangle&=\langle w_jg^{-1}\rangle\oplus\langle z_jg^{-1}\rangle\oplus\langle w_qg^{-1}\rangle,\cr \langle v_jh^{-1}\rangle&=\langle x_jh^{-1}\rangle\oplus\langle x_i h^{-1}\rangle,\cr \langle v_sh^{-1}\rangle&=\langle y_jh^{-1}\rangle\oplus\langle y_kh^{-1}\rangle, \end{align*} $$
and define
$\pi _1,\pi _2$
in
$G(V)$
by
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}cccccc@{}} u_jg^{-1}&a_ig^{-1}&w_j g^{-1}&z_j g^{-1}&u_k g^{-1}&w_q g^{-1}\cr x_jh^{-1}&x_ih^{-1}&c_j h^{-1}&y_j h^{-1}&y_k h^{-1}&d_q h^{-1} \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{@{}cccccc@{}} x_j&x_i&c_j&y_j&y_k&d_q\\ u_j&a_i&w_j&z_j&u_k&w_q \end{array}\bigg). \end{align*} $$
Clearly,
$g=\pi _1 h\pi _2$
. Also,
$$ \begin{align*} W\pi_1&=\langle x_jh^{-1}\rangle\oplus\langle x_ih^{-1}\rangle,\cr W\pi_2&=\langle w_j\rangle\oplus\langle w_q\rangle, \end{align*} $$
and hence
$\mathop {\mathrm {dim}}\, W\pi _1/(W\pi _1\cap W)=\mathop {\mathrm {dim}}\, W\pi _2/(W\pi _2\cap W)=0<m$
. In other words,
$\pi _1, \pi _2\in H(W)$
.
Case 4. Suppose that
$\mathop {\mathrm {dim}}\, Wh^{-1}/(Wh^{-1}\cap W)=m$
and
$\mathop {\mathrm {dim}}\, Wg^{-1}/(Wg^{-1}\cap W)=m$
. Then
$\vert I\vert \leq m=\vert L\vert =\vert J\vert \leq \vert L\vert +\vert K\vert =\mathop {\mathrm {codim}} W$
and
$\vert P\vert \leq m=\vert Q\vert =\vert R\vert \leq \vert Q\vert +\vert S\vert =\mathop {\mathrm {codim}} W$
. Thus, we may write
$\{b_\ell g^{-1}\}$
,
$\{d_qh^{-1}\}$
and
$\{v_rh^{-1}\}$
as
$\{b_j g^{-1}\}$
,
$\{d_jh^{-1}\}$
and
$\{v_jh^{-1}\}$
, respectively.
If
$m<\mathop {\mathrm {codim}} W$
, then
$\vert K\vert =\mathop {\mathrm {codim}} W=\vert S\vert $
. Write
$\{v_sh^{-1}\}$
as
$\{v_kh^{-1}\}$
and
$$ \begin{align*}\langle b_jg^{-1}\rangle=\langle w_jg^{-1}\rangle\oplus\langle w_pg^{-1}\rangle,\quad \langle v_jh^{-1}\rangle=\langle x_jh^{-1}\rangle\oplus\langle x_i h^{-1}\rangle.\end{align*} $$
Now define
$\pi _1,\pi _2$
in
$G(V)$
by
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}ccccc@{}} u_jg^{-1}&a_ig^{-1}&u_k g^{-1}&w_p g^{-1}&w_j g^{-1}\cr x_jh^{-1}&x_ih^{-1}&v_k h^{-1}&c_p h^{-1}&d_j h^{-1} \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{ccccc} x_j&x_i&v_k&c_p&d_j\\ u_j&a_i&u_k&w_p&w_j \end{array}\bigg). \end{align*} $$
If
$m=\mathop {\mathrm {codim}} W$
, then
$\vert K\vert \leq m$
and
$\vert S\vert \leq m$
. Write
$$ \begin{align*} \langle b_jg^{-1}\rangle&=\langle w_pg^{-1}\rangle\oplus\langle w_sg^{-1}\rangle\oplus\langle w_jg^{-1}\rangle,\cr \langle v_jh^{-1}\rangle&=\langle x_jh^{-1}\rangle\oplus\langle x_i h^{-1}\rangle\oplus\langle x_k h^{-1}\rangle, \end{align*} $$
and define
$\pi _1,\pi _2$
in
$G(V)$
by
$$ \begin{align*} \pi_1&=\bigg(\begin{array}{@{}cccccc@{}} u_jg^{-1}&a_ig^{-1}&u_k g^{-1}&w_p g^{-1}&w_s g^{-1}&w_j g^{-1}\cr x_jh^{-1}&x_ih^{-1}&x_k h^{-1}&c_p h^{-1}&v_s h^{-1}&d_j h^{-1} \end{array}\bigg),\\ \pi_2&=\bigg(\begin{array}{@{}cccccc@{}} x_j&x_i&x_k&c_p&v_s&d_j\\ u_j&a_i&u_k&w_p&w_s&w_j \end{array}\bigg). \end{align*} $$
It is easy to see that, in both cases,
$W\pi _1,W\pi _2\subseteq W$
, and so
$\pi _1,\pi _2\in H(W)$
. Moreover,
$g=\pi _1 h\pi _2$
.
As for the set case, it is not difficult to see that
$H(W)$
is a nongroup maximal subsemigroup of
$G(V)$
if and only if
$\aleph _0\leq \mathop {\mathrm {dim}}\, W\leq \mathop {\mathrm {codim}} W$
. In fact, as we prove in our next result, the linear version of Hotzel’s claim holds.
Theorem 4.2. Let W be a subspace of V. Then the set
$$ \begin{align*}H(W)=\{\pi\in G(V):\mathop{\mathrm{dim}}\, W\pi/(W\pi\cap W)<\mathop{\mathrm{dim}}\, W\}\end{align*} $$
is a maximal subsemigroup of
$G(V)$
if and only if
$\mathop {\mathrm {dim}}\, W=1$
or
$\aleph _0\leq \mathop {\mathrm {dim}}\, W\leq \mathop {\mathrm {codim}} W$
.
Proof. By Proposition 4.1, if
$\aleph _0\leq \mathop {\mathrm {dim}}\, W\leq \mathop {\mathrm {codim}} W$
, then
$H(W)$
is a maximal subsemigroup of
$G(V)$
. Now assume that
$\mathop {\mathrm {dim}}\, W=1$
and let
$W=\langle w\rangle $
, with
$w\not =0$
. Since
$\mathop {\mathrm {dim}}\, W\pi =1$
for each
$\pi \in G(V)$
, it follows that
$\mathop {\mathrm {dim}}\, W\pi /(W\pi \cap W)<1$
if and only if
$W\pi =W$
, and hence
$$ \begin{align*}H(W)=\{\pi\in G(V): w\pi=k\pi\text{\ for some } k\in F\setminus\{0\}\}.\end{align*} $$
It is easy to see that
$H(W)$
is a subsemigroup of
$G(V)$
that is a group. Given
$h,g\in G(V)\setminus H(W)$
,
$$ \begin{align*}Wh\cap W=\{0\}=Wg\cap W.\end{align*} $$
Therefore, there exist nonzero
$u,v,a,b\notin \langle w\rangle $
such that
$wh=u$
,
$wg=v$
,
$ah=w$
and
$bg=w$
. Write
$V=\langle w\rangle \oplus \langle a\rangle \oplus \langle a_i\rangle =\langle w\rangle \oplus \langle b\rangle \oplus \langle b_i\rangle $
, and define
$\pi _1,\pi _2\in G(V)$
by
$$ \begin{align*}\pi_1=\bigg(\begin{array}{@{}ccc@{}} w&b&b_i\\ w&a&a_i \end{array}\bigg),\quad \pi_2=\bigg(\begin{array}{@{}ccc@{}} w&u&a_ih\\ w&v&b_ig \end{array}\bigg).\end{align*} $$
Clearly,
$\pi _1,\pi _2\in H(W)$
, since
$w\pi _i=w$
for
$i=1,2$
. Also,
$$ \begin{align*}\pi_1 h\pi_2=\bigg(\begin{array}{@{}ccc@{}} w&b&b_i\\ v&w&b_ig \end{array}\bigg)=g.\end{align*} $$
Thus, if
$\mathop {\mathrm {dim}}\, W=1$
, then
$H(W)$
is a maximal subsemigroup of
$G(V)$
.
Conversely, suppose that
$\mathop {\mathrm {dim}}\, W=m$
with
$m\in \mathbb {N}\setminus \{1\}$
, and write
$W=\langle w_1,\dots , w_m\rangle $
. Since V is infinite-dimensional, we may write
$V=\langle w_1,\dots , w_m\rangle \oplus \langle u_1,\dots , u_m\rangle \oplus \langle v_i\rangle $
, where
$\vert I\vert =\mathop {\mathrm {dim}}\, V$
. Now define
$\alpha ,\beta \in G(V)$
by
$$ \begin{align*}\alpha=\bigg(\begin{array}{@{}ccccccccc@{}}w_1&w_2&\dots&w_m&u_1&u_2&\dots&u_m&v_i\cr w_1&u_2&\dots&u_m&u_1&w_2&\dots&w_m&v_i \end{array}\bigg),\end{align*} $$
$$ \begin{align*}\beta=\bigg(\begin{array}{@{}ccccccccc@{}}w_1&w_2&\dots&w_m&u_1&u_2&\dots&u_m&v_i\cr u_1&w_2&\dots&w_m&w_1&u_2&\dots&u_m&v_i \end{array}\bigg).\end{align*} $$
Clearly,
$\mathop {\mathrm {dim}}\, W\alpha /(W\alpha \cap W)=\mathop {\mathrm {dim}}\,\langle u_2,\dots ,u_m\rangle =m-1$
and
$\mathop {\mathrm {dim}}\, W\beta /(W\beta \cap W)=\mathop {\mathrm {dim}}\,\langle u_1\rangle =1$
, so
$\alpha ,\beta \in H(W)$
. But
$$ \begin{align*}\alpha\beta=\bigg(\begin{array}{@{}ccccccccc@{}}w_1&w_2&\dots&w_m&u_1&u_2&\dots&u_m&v_i\cr u_1&u_2&\dots&u_m&w_1&w_2&\dots&w_m&v_i \end{array}\bigg),\end{align*} $$
and hence
$\mathop {\mathrm {dim}}\, W\alpha \beta /(W\alpha \beta \cap W)=\mathop {\mathrm {dim}}\,\langle u_1,u_2,\dots ,u_m\rangle =m$
. Therefore,
$\alpha \beta \notin H(W)$
and
$H(W)$
is not a semigroup. Next, assume that W is infinite-dimensional and
$\mathop {\mathrm {dim}}\, W>\mathop {\mathrm {codim}} W$
. Given that
$\pi \in G(V)$
,
$\mathop {\mathrm {dim}}\, W\pi /(W\pi \cap W)\leq \mathop {\mathrm {codim}} W<\mathop {\mathrm {dim}}\, W$
. Thus,
${H(W)=G(V)}$
. In other words, we have just proved that
$H(W)$
is not a maximal subsemigroup of
$G(V)$
when
$\mathop {\mathrm {dim}}\, W\in \mathbb {N}\setminus \{1\}$
or W is infinite-dimensional but
$\mathop {\mathrm {dim}}\, W>\mathop {\mathrm {codim}} W$
.
