Symmetric Properties of (b,c)-Inverses

Abstract

Let b and c be two elements in a semigroup S. The $(b,c)$-inverse is an important outer inverse because it unifies many common generalized inverses. This paper is devoted to presenting some symmetric properties of $(b,c)$-inverses and $(c,b)$-inverses. We first find that S contains a $(b,c)$-invertible element if and only if it contains a $(c,b)$-invertible element. Then, for four given elements $a,b,c,d$ in S, we prove that a is $(b,c)$-invertible and d is $(c,b)$-invertible if and only if $abd$ is invertible along c and $dca$ is invertible along b. Inspired by this result, the $(b,c)$-invertibility is characterized by one-sided invertible elements. Furthermore, we show that a is inner $(b,c)$-invertible and d is inner $(c,b)$-invertible if and only if c is inner $(a,d)$-invertible and b is inner $(d,a)$-invertible.

1. Introduction

An element a in a semigroup S is said to be regular if there exists $x\in S$ such that $axa=a$, in which case x is called an inner inverse (or a $\left\{1\right\}$-inverse) of a. Recall that an involution ∗ of S is a self-map such that ${\left(a^{\ast }\right)}^{\ast }=a$ and ${\left(ab\right)}^{\ast }=b^{\ast }{a}^{\ast }$ for all $a,b\in S$. If there exists x satisfying $axa=a$, $xax=x$, ${\left(ax\right)}^{\ast }=ax$ and ${\left(xa\right)}^{\ast }=xa$, then it is the unique solution of the previous four equations and is called the Moore–Penrose inverse [1] of a (denoted by $a^{†}$).

An element a in a semigroup S is Drazin invertible [2] if there exists $x\in S$ such that

$$x{a}^{m+1}=a^m\mathrm{for}\mathrm{some}m\in {\mathbb{N}}^{+},a{x}^2=x,ax=xa.$$

If such x exists, then it is unique and called the Drazin inverse of a (denoted by $a^D$). The smallest integer m that makes the above equations hold is called the Drazin index of a and denoted by $\mathrm{ind}\left(a\right)$. If $\mathrm{ind}\left(a\right)=1$, x is called the group inverse of a and denoted by $a^{\#}$.

Let S be any semigroup and $a,b\in S$. Mary [3] defined that the inverse of a along b as the unique element y satisfying the following relations:

$$y\in bS\cap bS,yab=b,bay=b.$$

In this case, a is said to be invertible along b, and y is denoted by $a^{\left|\right|b}$. If, moreover, $a{a}^{\left|\right|b}a=a$, then $a^{\left|\right|b}$ is called the inner inverse of a along b. He also proved that the Moore–Penrose inverse of an element a is equal to $a^{\left|\right|a^{\ast }}$, and the group inverse of a is equal to $a^{\left|\right|a}$. The set of all elements which are invertible along b is denoted by $S^{\left|\right|b}$.

Let S be any semigroup and $a,b,c\in S$. Drazin [4] defined the $(b,c)$-inverse of a to be the unique element y satisfying

$$y\in bS\cap Sc,yab=b,cay=c.$$

In this case, a is said to be $(b,c)$-invertible, and y is denoted by $a^{\left|\right|(b,c)}$. When $b=c$, we can see that $a^{\left|\right|(b,b)}=a^{\left|\right|b}$. To see the difference between inverses along an element and $(b,c)$-inverses, we consider the semigroup ${\mathbb{C}}^{2\times 2}$. Let $a=b=\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$ and $c=\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$. Then $a^{\left|\right|(b,c)}=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\ne \left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]=a^{\left|\right|b}$.

Later, Drazin [5] also defined the one-sided version of the $(b,c)$-inverse in a semigroup S. If $b\in Scab$, or equivalently if there exists y such that $y\in Sc$ and $yab=b$, then a is said to be left $(b,c)$-invertible. Such y is called a left $(b,c)$-inverse of a. Dually, a is said to be right $(b,c)$-invertible if $c\in cabS$, or equivalently if there exists z such that $z\in bS$ and $caz=c$. Such z is called a right $(b,c)$-inverse of a. Drazin proved that a is $(b,c)$-invertible if and only if a is left and right $(b,c)$-invertible. Given any semigroup S and $b,c\in S$, we denote the sets of all left $(b,c)$-invertible elements, right $(b,c)$-invertible elements and $(b,c)$-invertible elements in S by $S_l^{\left|\right|(b,c)}$, $S_r^{\left|\right|(b,c)}$ and $S^{\left|\right|(b,c)}$, respectively.

The motivation of this paper comes from the following facts.

Lemma 1

(Theorem 7 in [3]). Let S be any semigroup and $a,b\in S$. Then a is invertible along b if and only if $ab$ is group invertible with $b\in Sab$ if and only if $ba$ is group invertible with $b\in baS$, in which case,

$$a^{\left|\right|b}=b{\left(ab\right)}^{\#}={\left(ba\right)}^{\#}b.$$

Lemma 2

(Corollary 2.7 in [6]). Let S be any semigroup and $a,b\in S$. Then a is inner invertible along b if and only if a is invertible along b and b is invertible along a.

These are two interesting results with nice symmetry. However, in general cases, the $(b,c)$-invertibility of a does not imply that $ab$, $ac$, $ba$ and $ca$ are group invertible (see Example 2.1 in [7]), and a being $(b,c)$-invertible with $a{a}^{\left|\right|(b,c)}a=a$ does not imply that $b\in S^{\left|\right|(a,a)}$ and $c\in S^{\left|\right|(a,a)}$ (see the case of $b=a$ and $c=a^{\ast }$).

Wu and Chen [7] had done some interesting work on the case of $a\in S^{\left|\right|(b,c)}\cap S^{\left|\right|(c,b)}$. They characterize $a\in S^{\left|\right|(b,c)}\cap S^{\left|\right|(c,b)}$ by using group invertible elements and invertible elements, respectively. We find that $S^{\left|\right|(b,c)}\ne \varnothing$ implies $S^{\left|\right|(c,b)}\ne \varnothing$, so it seems more natural to consider the situation $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$, which of course includes the case of $a\in S^{\left|\right|(b,c)}\cap S^{\left|\right|(c,b)}$. This paper focuses on the case of $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$.

In Section 2, we prove that

$$a\in S^{\left|\right|(b,c)}\mathrm{and}d\in S^{\left|\right|(c,b)}\iff abd\in S^{\left|\right|c}\mathrm{and}dca\in S^{\left|\right|b},$$

which allows us to transform many questions on the $(b,c)$-invertibility and $(c,b)$-invertibility into those on the invertibility along b and c. As an application of this observation, the (one-sided) $(b,c)$-invertibility is characterized by one-sided invertible elements.

If $a\in S^{\left|\right|(b,c)}$ such that $a{a}^{\left|\right|(b,c)}a=a$, then $a^{\left|\right|(b,c)}$ is called the inner $(b,c)$-inverse of a. In Section 3, we consider some symmetric properties of inner $(b,c)$-inverses. We prove that a is inner $(b,c)$-invertible and d is inner $(c,b)$-invertible if and only if c is inner $(a,d)$-invertible and b is inner $(d,a)$-invertible. Especially, a is both inner $(b,c)$-invertible and inner $(c,b)$-invertible if and only if both b and c are inner invertible along a if and only if a is inner invertible along b and c, in which case

$$a^{\left|\right|(b,c)}=a^{\left|\right|b}a{a}^{\left|\right|c}\mathrm{and}{a}^{\left|\right|(c,b)}=a^{\left|\right|c}a{a}^{\left|\right|b}.$$

At last, under the assumption that $a{a}^{\left|\right|c}a=a$, we characterize the product $a^{\left|\right|b}a{a}^{\left|\right|c}$ by equations and prove that $a^{\left|\right|b}a{a}^{\left|\right|c}$ is equal to the $(b,ba{a}^{\left|\right|c})$-inverse of a, which generalizes some results on the DMP inverse.

2. Characterizations of $\mathit{a}\in {\mathbf{S}}^{\left|\right|(\mathit{b},\mathit{c})}$ and $\mathit{d}\in {\mathbf{S}}^{\left|\right|(\mathit{c},\mathit{b})}$

We first recall two basic characterizations of $(b,c)$-invertibility, which will be frequently used in the sequel discussion.

Lemma 3

(Theorem 2.2 in [4]). Let S be any semigroup and $a,b,c\in S$. Then a is $(b,c)$-invertible if and only if $c\in cabS$ and $b\in Scab$. In this case, $a^{\left|\right|(b,c)}=sc=bt$, where $c=cabt$ and $b=scab$.

Lemma 4

(Proposition 6.1 in [4]). Let S be any semigroup and $a,b,c\in S$. Then a is $(b,c)$-invertible if and only if there exists $y\in S$ satisfying that

$$yay=y,S{}^{1}y=S{}^{1}c,yS{}^1=bS{}^1,$$

where $S^1$ stands for the monoid generated by S.

From previous two Lemmas, we can immediately obtain a connection between $S^{\left|\right|(b,c)}$ and $S^{\left|\right|(c,b)}$.

Proposition 1.

Let S be any semigroup and $b,c\in S$. Then $S^{\left|\right|(b,c)}\ne \varnothing$ if and only if $S^{\left|\right|(c,b)}\ne \varnothing$.

Proof. 

If $a\in S^{\left|\right|(b,c)}$, then $cabS{}^1=cS{}^1$ and $S{}^{1}cab=S{}^{1}b$ by Lemma 3. From Proposition 3.3 in [8], we know that $cab$ is regular. Thus $cab$ is the $(c,b)$-inverse of ${\left(cab\right)}^{-}$ by Lemma 4, for any inner inverse ${\left(cab\right)}^{-}$ of $cab$.

By symmetry, the converse statement is also true. □

Proposition 2.

Let S be any semigroup and $b,c\in S$. If $S^{\left|\right|(b,c)}\ne \varnothing$, then the mapping $\varphi :x\mapsto c{x}^{-}b$ is a bijection from $\left\{a^{\left|\right|(b,c)}\right|a\in S^{\left|\right|(b,c)}\}$ to $\left\{d^{\left|\right|(c,b)}\right|d\in S^{\left|\right|(c,b)}\}$, for any inner inverse $x^{-}$ of x.

Proof. 

At first, we prove that $\varphi$ is well defined. Suppose that $x=a^{\left|\right|(b,c)}$ for some $a\in S^{\left|\right|(b,c)}$. For any inner inverse $x^{-}$ of x, we know that $a^{\left|\right|(b,c)}$ is the $(b,c)$-inverse of $x^{-}$ by Lemma 4. Then

$$c{x}^{-}b=\left(ca{a}^{\left|\right|(b,c)}\right)x^{-}b=ca\left[{\left(x^{-}\right)}^{\left|\right|(b,c)}{x}^{-}b\right]=cab.$$

Next we prove that $\varphi$ is a bijection. Define another mapping $\psi :y\mapsto b{y}^{-}c$ from $\left\{d^{\left|\right|(c,b)}\right|d\in S^{\left|\right|(c,b)}\}$ to $\left\{a^{\left|\right|(b,c)}\right|a\in S^{\left|\right|(b,c)}\}$. Similarly, $\psi$ is well-defined. Then we obtain

$$\psi \varphi \left(x\right)=b{\left(cab\right)}^{-}c=a^{\left|\right|(b,c)}=x,$$

where the last second equality holds because of Theorem 2.7 in [9]. Similarly, $\varphi \psi \left(y\right)=y$. Thus, $\varphi$ is a bijection. □

Let $a,b,c$ be elements in a semigroup S such that a is $(b,c)$-invertible. We wonder what conditions are needed to ensure that d is $(c,b)$-invertible. To handle this question, we consider the following Lemma.

Lemma 5.

Let S be any semigroup and $b,c,d,u,v\in S$. If $uS{}^1=bS{}^1$ and $S{}^{1}v=S{}^{1}c$, then

(1)

d is left $(b,c)$-invertible if and only if d is left $(u,v)$-invertible;

(2)

d is right $(b,c)$-invertible if and only if d is right $(u,v)$-invertible;

(3)

(Remark 2.2(i) in [10]) d is $(b,c)$-invertible if and only if d is $(u,v)$-invertible, in which case, $a^{\left|\right|(b,c)}=a^{\left|\right|(u,v)}$.

Proof. (1) Suppose that $b=ug$, $c=hv$, $u=bt$ and $v=sc$ for some $g,h,s,t\in S{}^1$.

If d is left $(b,c)$-invertible, then there exists $x\in S$ such that $b=xcdb$. It follows that

$$u=bt=xcdbt=xcdu=xhvdu\in Svdu.$$

Conversely, suppose that $u=yvdu$ for some $y\in S$. We have that

$$b=ug=yvdug=yvdb=yscdb\in Scdb.$$

(2) It can be proved similarly.

(3) It can be proved by combining (1) and (2). □

Proposition 3.

Let S be any semigroup and $a,b,c,d\in S$. If a is $(b,c)$-invertible, then we have the following:

(1)

d is left $(c,b)$-invertible if and only if d is left invertible along $cab$;

(2)

d is right $(c,b)$-invertible if and only if d is right invertible along $cab$;

(3)

d is $(c,b)$-invertible if and only if d is invertible along $cab$, in which case, $d^{\left|\right|(c,b)}=d^{\left|\right|cab}$.

Proof. 

If $a\in S^{\left|\right|(b,c)}$, then $cabS{}^1=cS{}^1$ and $S{}^{1}cab=S{}^{1}b$ by Lemma 3. Taking $u=v=cab$ and exchanging the position of b and c in Lemma 5, then the proposition follows. □

Now we give the main result of this section, which presents a necessary and sufficient condition for any semigroup S and $a,b,c,d\in S$ such that $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$.

Theorem 1.

Let S be any semigroup and $a,b,c,d\in S$. Then $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$ if and only if $abd\in S^{\left|\right|c}$ and $dca\in S^{\left|\right|b}$. In this case,

$$a^{\left|\right|(b,c)}=bd{\left(abd\right)}^{\left|\right|c}={\left(dca\right)}^{\left|\right|b}dc,$$

$$d^{\left|\right|(c,b)}={\left(abd\right)}^{\left|\right|c}ab=ca{\left(dca\right)}^{\left|\right|b}.$$

Proof. 

If $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$, then we know that

$$c\in cabS,b\in Scab,b\in bdcS,\mathrm{and}c\in Sbdc$$

by Lemma 3. It follows that

$$c\in cabS\subseteq cabdcS\mathrm{and}c\in Sbdc\subseteq Scabdc,$$

which means that $abd\in S^{\left|\right|c}$. Similarly, $dca\in S^{\left|\right|b}$.

Conversely, if $abd\in S^{\left|\right|c}$ and $dca\in S^{\left|\right|b}$, then we have

$$c=cabd{\left(abd\right)}^{\left|\right|c}\in cabS\mathrm{and}b={\left(dca\right)}^{\left|\right|b}dcab\in Scab.$$

So $a\in S^{\left|\right|(b,c)}$ by Lemma 3. Similarly, $d\in S^{\left|\right|(c,b)}$. The formulae of $a^{\left|\right|(b,c)}$ and $d^{\left|\right|(c,d)}$ follow from Lemma 3. □

From above proof, we can see that the one-sided version of Theorem 1 is also true. We list it below and omit its proof.

Proposition 4.

Let S be any semigroup and $a,b,c,d\in S$. Then

(1)

$a\in S_l^{\left|\right|(b,c)}$ and $d\in S_l^{\left|\right|(c,b)}$ if and only if $abd\in S_l^{\left|\right|c}$ and $dca\in S_l^{\left|\right|b}$;

(2)

$a\in S_r^{\left|\right|(b,c)}$ and $d\in S_r^{\left|\right|(c,b)}$ if and only if $abd\in S_r^{\left|\right|c}$ and $dca\in S_r^{\left|\right|b}$.

Let S be any semigroup and $a,b\in S$. Lemma 1 shows that $a\in S^{\left|\right|b}$ if and only if $b\in Sab$ and $ab\in S^{\#}$ if and only if $b\in baS$ and $ba\in S^{\#}$, in which case $a^{\left|\right|(b,b)}=b{\left(ab\right)}^{\#}={\left(ba\right)}^{\#}b$. By Theorem 1, we can also characterize the $(b,c)$-inverse and $(c,b)$-inverse by the group inverses.

Proposition 5.

Let S be any semigroup and $a,b,c,d\in S$. If $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$, then $abdc,bdca,dcab$ and $cabd$ are group invertible. In this case,

$$a^{\left|\right|(b,c)}=bdc{\left(abdc\right)}^{\#}=bd{\left(cabd\right)}^{\#}c=b{\left(dcab\right)}^{\#}dc={\left(bdca\right)}^{\#}bdc,$$

$$d^{\left|\right|(c,b)}=cab{\left(dcab\right)}^{\#}=ca{\left(bdca\right)}^{\#}b=c{\left(abdc\right)}^{\#}ab={\left(cabd\right)}^{\#}cab.$$

Proof. 

If $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$, then $abd\in S^{\left|\right|c}$ and $dca\in S^{\left|\right|b}$ by Theorem 1. According to Lemma1, $abdc,bdca,dcab$ and $cabd$ are group invertible with

$${\left(abd\right)}^{\left|\right|c}=c{\left(abdc\right)}^{\#}={\left(cabd\right)}^{\#}c\mathrm{and}{\left(dca\right)}^{\left|\right|b}=b{\left(dcab\right)}^{\#}={\left(bdca\right)}^{\#}b.$$

Substituting them into the formulae for $a^{\left|\right|(b,c)}$ and $d^{\left|\right|(c,b)}$ in Theorem 1, the formulae in terms of the group inverses follow. □

Proposition 6.

Let S be any semigroup and $a,b,c,d\in S$. If u is any one of $abdc$, $bdca$, $dcab$, $cabd$, then the following conditions are equivalent:

(1)

$a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$;

(2)

u is group invertible, $a\in S_l^{\left|\right|(b,c)}$ and $d\in S_l^{\left|\right|(c,b)}$;

(3)

u is group invertible, $a\in S_r^{\left|\right|(b,c)}$ and $d\in S_r^{\left|\right|(c,b)}$;

(4)

u is Drazin invertible, $a\in S_l^{\left|\right|(b,c)}$ and $d\in S_l^{\left|\right|(c,b)}$;

(5)

u is Drazin invertible, $a\in S_r^{\left|\right|(b,c)}$ and $d\in S_r^{\left|\right|(c,b)}$.

Proof. (1) ⇒ (2). By Proposition 5.

(2) ⇒ (4). It is obvious.

(4) ⇒ (1). If u is Drazin invertible, then $abdc$ and $dcab$ are Drazin invertible by Cline’s formula [11]. Meanwhile, from $b\in Scab$ and $c\in Sbdc$, we know that

$$Sdcab\subseteq Scab\subseteq Sbdcab\subseteq Scabdcab\subseteq Sbdcabdcab\subseteq S{\left(dcab\right)}^2\subseteq Sdcab.$$

It follows that $\mathrm{ind}\left(dcab\right)=1$, which means that $dcab$ is group invertible. Similarly, $abdc$ is group invertible.

Noting that $b\in Scab\subseteq Sbdcab\subseteq Sdcab$ and $c\in Sbdc\subseteq Scabdc\subseteq Sabdc$, we have

$$b=bdcab{\left(dcab\right)}^{\#}\in bdcS\mathrm{and}c=cabdc{\left(abdc\right)}^{\#}\in cabS.$$

(1) ⇒ (3) ⇒ (5) ⇒ (1) can be proved dually. □

Let R be any associative ring with 1 and $a,b\in R$ such that b is regular with an inner inverse $b^{-}$. Theorem 3.2 in [12] proved that a is invertible along b if and only if $ab+1-b^{-}b$ is invertible if and only if $ba+1-b{b}^{-}$ is invertible. Denoting the set of all invertible (resp., left and right invertible) elements in R by $R^{-1}$ (resp., $R_l^{-1}$ and $R_r^{-1}$), we characterize the (one-sided) $(b,c)$-inverse and (one-sided) $(c,b)$-inverse by using (one-sided) invertible elements as follows.

Proposition 7.

Let R be any associative ring with 1 and $a,b,c,d\in R$ such that b and c are regular. If $b^{-}$ is an inner inverse of b and $c^{-}$ is an inner inverse of c, denote

$$u=cabd+1-c{c}^{-},v=bdca+1-b{b}^{-},$$

$$s=abdc+1-c^{-}c,t=dcab+1-b^{-}b.$$

Then

(1)

$a\in R_l^{\left|\right|(b,c)}$ and $d\in R_l^{\left|\right|(c,b)}$ if and only if $u\in R_l^{-1}$ and $v\in R_l^{-1}$ if and only if $s\in R_l^{-1}$ and $t\in R_l^{-1}$, in which case $u_l^{-1}cab$ is a left $(c,b)$-inverse of d and $v_l^{-1}bdc$ is a left $(b,c)$-inverse of a, where $u_l^{-1}$ and $v_l^{-1}$ are left inverses of u and v, respectively;

(2)

$a\in R_r^{\left|\right|(b,c)}$ and $d\in R_r^{\left|\right|(c,b)}$ if and only if $u\in R_r^{-1}$ and $v\in R_r^{-1}$ if and only if $s\in R_r^{-1}$ and $t\in R_r^{-1}$, in which case $bdc{s}_r^{-1}$ is a right $(b,c)$-inverse of a and $cab{t}_r^{-1}$ is a right $(c,b)$-inverse of d, where $s_r^{-1}$ and $t_r^{-1}$ are right inverses of s and t, respectively;

(3)

$a\in R^{\left|\right|(b,c)}$ and $d\in R^{\left|\right|(c,b)}$ if and only if $u\in R^{-1}$ and $v\in R^{-1}$ if and only if $s\in R^{-1}$ and $t\in R^{-1}$, in which case,

$$a^{\left|\right|(b,c)}=v^{-1}bdc=bdc{s}^{-1}\mathrm{and}{d}^{\left|\right|(c,b)}=u^{-1}cab=cab{t}^{-1}.$$

Proof. (1) By Proposition 4, $a\in R_l^{\left|\right|(b,c)}$ and $d\in R_l^{\left|\right|(c,b)}$ if and only if $abd\in R_l^{\left|\right|c}$ and $dca\in R_l^{\left|\right|b}$. Additionally, $abd\in R_l^{\left|\right|c}$ and $dca\in R_l^{\left|\right|b}$ if and only if $u\in R_l^{-1}$ and $v\in R_l^{-1}$ by Theorem 3.2 in [13], which is equivalent to $s\in R_l^{-1}$ and $t\in R_l^{-1}$ by Jacobson’s lemma.

If $u\in R_l^{-1}$, multiplying by c on the right of $u=cabd+1-c{c}^{-}$ yields that $uc=cabdc$. It follows that $c=u_l^{-1}uc=u_l^{-1}cabdc\in Rbdc$, which means that $u_l^{-1}cab$ is a left $(c,b)$-inverse of d. Similarly, one can prove that $v_l^{-1}bdc$ is a left $(b,c)$-inverse of a.

(2) Similarly by using Theorem 3.4 in [13].

(3) Combining (1) and (2), it follows. □

If $a\in S^{\left|\right|(b,c)}$, we showed in the proof of Proposition 1 that ${\left(cab\right)}^{-}\in S^{\left|\right|(c,b)}$ for any inner inverse ${\left(cab\right)}^{-}$ of $cab$. Suppose that b, c and $cab$ are regular. then $a\in R^{\left|\right|(b,c)}$ if and only if $u=cab{\left(cab\right)}^{-}+1-c{c}^{-}\in R^{-1}$ and $v={\left(cab\right)}^{-}cab+1-b^{-}b\in R^{-1}$ by replacing d by ${\left(cab\right)}^{-}$ in Proposition 7. However, characterizing the left $(b,c)$-invertibility of a only requires that b, $cab$ are regular and v is left invertible.

Proposition 8.

Let R be any associative ring with 1 and $a,b,c\in R$ such that b and $cab$ are regular. If $b^{-}$ is an inner inverse of b and ${\left(cab\right)}^{-}$ is an inner inverse of $cab$, then the following conditions are equivalent:

(1)

a is left $(b,c)$-invertible;

(2)

$v={\left(cab\right)}^{-}cab+1-b^{-}b\in R_l^{-1}$;

(3)

$t=b{\left(cab\right)}^{-}ca+1-b{b}^{-}\in R_l^{-1}$.

In this case, $t_l^{-1}b{\left(cab\right)}^{-}c$ is a left $(b,c)$-inverse of a, where $t_l^{-1}$ is a left inverse of t.

Proof. (1) ⇒ (2). If a is left $(b,c)$-invertible, then $Rcab=Rb$. It follows that $b{\left(cab\right)}^{-}cab=b$. Then we have

$$\begin{array}{ccc}& & [b^{-}b+1-{\left(cab\right)}^{-}cab][{\left(cab\right)}^{-}cab+1-b^{-}b]\hfill \\ & =& b^{-}b{\left(cab\right)}^{-}cab+b^{-}b(1-b^{-}b)+(1-{\left(cab\right)}^{-}cab){\left(cab\right)}^{-}cab\hfill \\ & & +(1-{\left(cab\right)}^{-}cab)(1-b^{-}b)\hfill \\ & =& b^{-}b+0+0+1-b^{-}b-{\left(cab\right)}^{-}cab+{\left(cab\right)}^{-}cab{b}^{-}b\hfill \\ & =& 1.\hfill \end{array}$$

So $v={\left(cab\right)}^{-}cab+1-b^{-}b$ is left invertible.

(2) ⇒ (3). By Jacobson’s lemma.

(3) ⇒ (1). Multiplying by b on the right of $t=b{\left(cab\right)}^{-}ca+1-b{b}^{-}$ yields that $tb=b{\left(cab\right)}^{-}cab$. It follows that

$$b=t_l^{-1}tb=t_l^{-1}b{\left(cab\right)}^{-}cab\in Rcab.$$

Then $t_l^{-1}b{\left(cab\right)}^{-}c$ is a left $(b,c)$-inverse of a. □

Dually, we have a characterization for right $(b,c)$-invertibility as follows.

Proposition 9.

Let R be any associative ring with 1 and $a,b,c\in R$ such that c and $cab$ are regular. If $c^{-}$ is an inner inverse of c and ${\left(cab\right)}^{-}$ is an inner inverse of $cab$, then the following conditions are equivalent:

(1)

a is right $(b,c)$-invertible;

(2)

$u=cab{\left(cab\right)}^{-}+1-c{c}^{-}\in R_r^{-1}$;

(3)

$s=ab{\left(cab\right)}^{-}c+1-c^{-}c\in R_r^{-1}$.

In this case, $b{\left(cab\right)}^{-}c{s}_r^{-1}$ is a right $(b,c)$-inverse of a, where $s_r^{-1}$ is a right inverse of s.

Combining Propositions 8 and 9, we have the following characterization for $(b,c)$-invertibility.

Theorem 2.

Let R be any associative ring with 1 and $a,b,c\in R$ such that b, c and $cab$ are regular. If $b^{-},c^{-},{\left(cab\right)}^{-}$ are inner inverses of $b,c,cab$, respectively, then the following conditions are equivalent:

(1)

a is $(b,c)$-invertible;

(2)

$u=cab{\left(cab\right)}^{-}+1-c{c}^{-}\in R_r^{-1}$ and $v={\left(cab\right)}^{-}cab+1-b^{-}b\in R_l^{-1}$;

(3)

$s=ab{\left(cab\right)}^{-}c+1-c^{-}c\in R_r^{-1}$ and $t=b{\left(cab\right)}^{-}ca+1-b{b}^{-}\in R_l^{-1}$.

In this case,

$$a^{\left|\right|(b,c)}=t_l^{-1}b{\left(cab\right)}^{-}c=b{\left(cab\right)}^{-}c{s}_r^{-1},$$

where $t_l^{-1}$ is a left inverse of t and $s_r^{-1}$ is a right inverse of s.

3. Symmetric Properties of Inner $(b,c)$-Invertible Elements

Let S be any semigroup and $a,b,c\in S$. If $a\in S^{\left|\right|(b,c)}$ such that $a{a}^{\left|\right|(b,c)}a=a$, then $a^{\left|\right|(b,c)}$ is called the inner $(b,c)$-inverse of a. For arbitrary $a\in S^{\left|\right|(b,c)}$, it is easy to verify that $a^{\left|\right|(b,c)}$ is the inner $(b,c)$-inverse of $a{a}^{\left|\right|(b,c)}a$. Theorem 2.13 in [14] proved that a is inner $(b,c)$-invertible if and only if $b\in Sab$, $c\in caS$ and $a\in abS\cap Sca$.

Let R be any associative ring with 1 and $a,b,c\in R$. Theorem 3.16 in [15] proved that a is inner $(b,c)$-invertible if and only if a is regular, $R=a^{\circ }\oplus bR$ and $R={}^{\circ }a\oplus Rc$. We give a characterization for inner $(b,c)$-invertible elements as follows.

Proposition 10.

Let S be any semigroup and $a,b,c\in S$. Then the following conditions are equivalent:

(1)

a is inner $(b,c)$-invertible;

(2)

a is $(b,c)$-invertible and $a\in abS$;

(3)

a is $(b,c)$-invertible and $S\in Sca$.

Proof. (1) ⇒ (2). Suppose that $a^{\left|\right|(b,c)}=bt$ for some $t\in S$. Then

$$a=a{a}^{\left|\right|(b,c)}a=abta\in abS.$$

(2) ⇒ (1). Assume that $a=aby$ for some $y\in S$. Then

$$a{a}^{\left|\right|(b,c)}a=a{a}^{\left|\right|(b,c)}aby=aby=a.$$

(1) ⇔ (3) can be proved similarly. □

Let S be any semigroup and $a,d\in S$. Lemma 2 shows that $a\in S^{\left|\right|b}$ and $b\in S^{\left|\right|a}$ if and only if a is inner invertible along b. It follows immediately that a is inner invertible along b if and only if b is inner invertible along a. We consider to generalize this fact to the case of $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$.

Proposition 11.

Let S be any semigroup and $a,b,c\in S$. If $a\in S^{\left|\right|(b,c)}$ and $d\in S^{\left|\right|(c,b)}$, then c is inner $(a^{\prime },d^{\prime })$-invertible and b is inner $(d^{\prime },a^{\prime })$-invertible, where $a^{\prime }=a{a}^{\left|\right|(b,c)}a$ and $d^{\prime }=d{d}^{\left|\right|(c,b)}d$.

Proof. 

Let $a^{\prime }=a{a}^{\left|\right|(b,c)}a$ and $d^{\prime }=d{d}^{\left|\right|(c,b)}d$. We first prove that c is $(a^{\prime },d^{\prime })$-invertible. In fact, supposing that $a^{\left|\right|(b,c)}=sc$ for some $s\in S$,

$$\begin{array}{cc}\hfill a{a}^{\left|\right|(b,c)}a& =a{a}^{\left|\right|(b,c)}a{a}^{\left|\right|(b,c)}a\hfill \\ & =asca{a}^{\left|\right|(b,c)}a\hfill \\ & =as{d}^{\left|\right|(c,b)}dca{a}^{\left|\right|(b,c)}a\hfill \\ & =as{d}^{\left|\right|(c,b)}d{d}^{\left|\right|(c,b)}dca{a}^{\left|\right|(b,c)}a\in Sd{d}^{\left|\right|(c,b)}dca{a}^{\left|\right|(b,c)}a.\hfill \end{array}$$

Similarly, $d{d}^{\left|\right|(c,b)}d\in d{d}^{\left|\right|(c,b)}dca{a}^{\left|\right|(b,c)}aS$.

Meanwhile, we have

$$\begin{array}{cc}\hfill c{c}^{\left|\right|(a^{\prime },d^{\prime })}c& =d^{\left|\right|(c,b)}dc{c}^{\left|\right|(a^{\prime },d^{\prime })}c\hfill \\ & =d^{\left|\right|(c,b)}d{d}^{\left|\right|(c,b)}dc{c}^{\left|\right|(a^{\prime },d^{\prime })}c\hfill \\ & =d^{\left|\right|(c,b)}dc\hfill \\ & =c.\hfill \end{array}$$

By symmetry, we have that b is inner $(d^{\prime },a^{\prime })$-invertible. □

Lemma 6.

Let S be any semigroup and $a,b,c,d\in S$. If $a\in S^{\left|\right|(b,c)}$, $d\in S^{\left|\right|(c,b)}$, $c\in S^{\left|\right|(a,d)}$ and $b\in S^{\left|\right|(d,a)}$, then

$$a^{\left|\right|(b,c)}a=b{b}^{\left|\right|(d,a)},a{a}^{\left|\right|(b,c)}=c^{\left|\right|(a,d)}c,$$

$$d{d}^{\left|\right|(c,b)}=b^{\left|\right|(d,a)}b,d^{\left|\right|(c,b)}d=c{c}^{\left|\right|(a,d)}.$$

Proof. 

If $a\in S^{\left|\right|(b,c)}$, $d\in S^{\left|\right|(c,b)}$, $c\in S^{\left|\right|(a,d)}$ and $b\in S^{\left|\right|(d,a)}$, then we have

$$a^{\left|\right|(b,c)}a=a^{\left|\right|(b,c)}ab{b}^{\left|\right|(d,a)}=b{b}^{\left|\right|(d,a)}.$$

Similarly, $a{a}^{\left|\right|(b,c)}=c^{\left|\right|(a,d)}c$, $d{d}^{\left|\right|(c,b)}=b^{\left|\right|(d,a)}b$ and $d^{\left|\right|(c,b)}d=c{c}^{\left|\right|(a,d)}$. □

Now we have the main result of this section.

Theorem 3.

Let S be any semigroup and $a,b,c,d\in S$. Then the following conditions are equivalent:

(1)

a is inner $(b,c)$-invertible and d is inner $(c,b)$-invertible;

(2)

c is inner $(a,d)$-invertible and b is inner $(d,a)$-invertible;

(3)

$a\in S^{\left|\right|(b,c)}$, $d\in S^{\left|\right|(c,b)}$ and $b\in S^{\left|\right|(d,a)}$;

(4)

$a\in S^{\left|\right|(b,c)}$, $d\in S^{\left|\right|(c,b)}$ and $c\in S^{\left|\right|(a,d)}$.

Proof. (1) ⇒ (2). If a is inner $(b,c)$-invertible and d is inner $(c,b)$-invertible, then c is inner $(a,d)$-invertible and b is inner $(d,a)$-invertible by Proposition 11.

(2) ⇒ (1). It is similar to the proof of (1) ⇒ (2).

(1) ⇒ (3). If a is inner $(b,c)$-invertible and d is inner $(c,b)$-invertible, then $a{a}^{\left|\right|(b,c)}a=a$ and $d{d}^{\left|\right|(c,b)}d=d$. It follows that b is $(d,a)$-invertible by Proposition 11.

(3) ⇒ (1). If $a\in S^{\left|\right|(b,c)}$, $d\in S^{\left|\right|(c,b)}$ and $b\in S^{\left|\right|(d,a)}$, then $a{a}^{\left|\right|(b,c)}a=ab{b}^{\left|\right|(d,a)}=a$ and $d{d}^{\left|\right|(c,b)}d=b^{\left|\right|(d,a)}bd=d$ by Lemma 6.

The equivalence of (1) and (4) can be proved similarly. □

Corollary 1.

Let S be any semigroup and $a,b\in S$. Then a is inner invertible along b if and only if b is inner invertible along a.

If $a\in S^{\left|\right|b}$ and $b\in S^{\left|\right|a}$, then $a^{\left|\right|b}={\left(ba\right)}^{\#}b$ and $b^{\left|\right|a}=a{\left(ba\right)}^{\#}$ by Lemma 1. It follows that

$$a^{\left|\right|b}{b}^{\left|\right|a}={\left(ba\right)}^{\#}ba{\left(ba\right)}^{\#}={\left(ba\right)}^{\#}.$$

By symmetry, $b^{\left|\right|a}{a}^{\left|\right|b}={\left(ab\right)}^{\#}$. We generalized this result to the case of $(b,c)$-inverses.

Proposition 12.

Let S be any semigroup and $a,b,c,d\in S$. If $a\in S^{\left|\right|(b,c)}$, $d\in S^{\left|\right|(c,b)}$, $b\in S^{\left|\right|(d,a)}$ and $c\in S^{\left|\right|(a,d)}$, then $abdc,bdca,dcab,cabd\in S^{\#}$ with

$${\left(dcab\right)}^{\#}=b^{\left|\right|(d,a)}{a}^{\left|\right|(b,c)}{c}^{\left|\right|(a,d)}{d}^{\left|\right|(c,b)},$$

$${\left(cabd\right)}^{\#}=d^{\left|\right|(c,b)}{b}^{\left|\right|(d,a)}{a}^{\left|\right|(b,c)}{c}^{\left|\right|(a,d)},$$

$${\left(abdc\right)}^{\#}=c^{\left|\right|(a,d)}{d}^{\left|\right|(c,b)}{b}^{\left|\right|(d,a)}{a}^{\left|\right|(b,c)},$$

$${\left(bdca\right)}^{\#}=a^{\left|\right|(b,c)}{c}^{\left|\right|(a,d)}{d}^{\left|\right|(c,b)}{b}^{\left|\right|(d,a)}.$$

Proof. 

If $a\in S^{\left|\right|(b,c)}$, $d\in S^{\left|\right|(c,b)}$, $b\in S^{\left|\right|(d,a)}$ and $c\in S^{\left|\right|(a,d)}$, then $abdc,dcab\in S^{\#}$ with $a^{\left|\right|(b,c)}=bdc{\left(abdc\right)}^{\#}$ and $c^{\left|\right|(a,d)}={\left(abdc\right)}^{\#}abd$ by Proposition 5, then we have

$$\begin{array}{ccc}& & b^{\left|\right|(d,a)}{a}^{\left|\right|(b,c)}{c}^{\left|\right|(a,d)}{d}^{\left|\right|(c,b)}\hfill \\ & =& b^{\left|\right|(d,a)}bdc{\left(abdc\right)}^{\#}{\left(abdc\right)}^{\#}abd{d}^{\left|\right|(c,b)}\hfill \\ & =& dc{\left(abdc\right)}^{\#}{\left(abdc\right)}^{\#}ab\hfill \\ & =& {\left(dcab\right)}^{\#},\hfill \end{array}$$

where the last equality follows by Cline’s formula [11]. The remaining three equalities can be verified similarly. □

Proposition 13.

Let S be any semigroup and $a,b,c\in S$. Then the following conditions are equivalent:

(1)

a is both inner $(b,c)$-invertible and inner $(c,b)$-invertible;

(2)

both b and c are inner invertible along a;

(3)

a is inner invertible along b and c.

In this case,

$$a^{\left|\right|(b,c)}=b{b}^{\left|\right|a}{a}^{\left|\right|c}=a^{\left|\right|b}{c}^{\left|\right|a}c=a^{\left|\right|b}a{a}^{\left|\right|c}$$

and

$$a^{\left|\right|(c,b)}=c{c}^{\left|\right|a}{a}^{\left|\right|b}=a^{\left|\right|c}{b}^{\left|\right|a}b=a^{\left|\right|c}a{a}^{\left|\right|b}.$$

Proof. (1) ⇔ (2). Taking $a=d$ in Theorem 3, then the equivalence between (1) and (2) follows.

(2) ⇔ (3). By Corollary 13.

In this case, noting that $a^{\left|\right|(b,c)}a=b{b}^{\left|\right|a}=a^{\left|\right|b}a$ and $a{a}^{\left|\right|(b,c)}=c^{\left|\right|a}c=a{a}^{\left|\right|c}$ by Lemma 6, we have

$$a^{\left|\right|(b,c)}=a^{\left|\right|(b,c)}a{a}^{\left|\right|(b,c)}=a^{\left|\right|(b,c)}{c}^{\left|\right|a}c=a^{\left|\right|(b,c)}a{a}^{\left|\right|c}=b{b}^{\left|\right|a}{a}^{\left|\right|c}=a^{\left|\right|b}a{a}^{\left|\right|c}=a^{\left|\right|b}{c}^{\left|\right|a}c.$$

Similarly, we can obtain the formula of $a^{\left|\right|(c,b)}$. □

Let S be any semigroup and $a\in S$. Theorem 4.4 in [16] proved that a is core invertible if and only if a is $(a,a^{\ast })$-invertible, and a is dual core invertible if and only if a is $(a^{\ast },a)$-invertible. Taking $b=a$ and $c=a^{\ast }$ in Proposition 13, we have the following result.

Corollary 2

(Theorem 5.6 in [17]). Let S be any semigroup and $a\in S$. Then a is both core invertible and dual core invertible if and only if a is both groups are invertible and Moore–Penrose invertible. In this case, $a^{\#}a{a}^{†}$ is the core inverse of a and $a^{†}a{a}^{\#}$ is the dual core inverse of a.

The reason why the $(b,c)$-inverse of a is equal to $a^{\left|\right|(b,b)}a{a}^{\left|\right|(c,c)}$ in Theorem 13 is based on the following fact.

Proposition 14.

Let S be any semigroup and $a,b,c\in S$. If $a\in S^{\left|\right|c}\cap S^{\left|\right|(c,b)}\cap S^{\left|\right|b}$, then $a{a}^{\left|\right|(c,b)}a\in S^{\left|\right|(b,c)}$ with

$${\left(a{a}^{\left|\right|(c,b)}a\right)}^{\left|\right|(b,c)}=a^{\left|\right|b}a{a}^{\left|\right|c}.$$

Proof. 

It is clear that $a^{\left|\right|b}a{a}^{\left|\right|c}\in bS\cap Sc$. We have

$$a^{\left|\right|b}a{a}^{\left|\right|c}a{a}^{\left|\right|(c,b)}ab=a^{\left|\right|b}a{a}^{\left|\right|(c,b)}ab=a^{\left|\right|b}ab=b$$

and

$$ca{a}^{\left|\right|(c,b)}a{a}^{\left|\right|b}a{a}^{\left|\right|c}=ca{a}^{\left|\right|(c,b)}a{a}^{\left|\right|c}=ca{a}^{\left|\right|c}=c.$$

So $a{a}^{\left|\right|(c,b)}a\in S^{\left|\right|(b,c)}$ with ${\left(a{a}^{\left|\right|(c,b)}a\right)}^{\left|\right|(b,c)}=a^{\left|\right|b}a{a}^{\left|\right|c}$. □

If a is invertible along b and c, then the $(c,b)$-invertibility can be characterized by $a^{\left|\right|b}a{a}^{\left|\right|c}$.

Proposition 15.

Let S be any semigroup and $a,b,c\in S$. If a is invertible along b and c, then

(1)

$a\in S_l^{\left|\right|(c,b)}$ if and only if $S{}^1{a}^{\left|\right|b}a{a}^{\left|\right|c}=S{}^{1}c$;

(2)

$a\in S_r^{\left|\right|(c,b)}$ if and only if $a^{\left|\right|b}a{a}^{\left|\right|c}S{}^1=bS{}^1$;

(3)

$a\in S^{\left|\right|(c,b)}$ if and only if $a^{\left|\right|b}a{a}^{\left|\right|c}S{}^1=bS{}^1$ and $S{}^1{a}^{\left|\right|b}a{a}^{\left|\right|c}=S{}^{1}c$.

Proof. (1) Noting that $S{}^1{a}^{\left|\right|b}=S{}^{1}b$ and $a^{\left|\right|c}S{}^1=cS{}^1$, we have $a\in S_l^{\left|\right|(c,b)}$ if and only if $a\in S_l^{\left|\right|(a^{\left|\right|c},a^{\left|\right|b})}$ by Lemma 5. Additionally, $a\in S_l^{\left|\right|(a^{\left|\right|c},a^{\left|\right|b})}$ if and only if $S{}^1{a}^{\left|\right|b}a{a}^{\left|\right|c}=S{}^1{a}^{\left|\right|c}=S{}^{1}c$ by definition.

(2) Can be proved similarly.

(3) Combining (1) and (2). □

Let $A\in {\mathbb{C}}^{n\times n}$. Malik and Thome [18] defined the matrix $A^{D,†}=A^{D}A{A}^{†}$ to be the DMP inverse of A and $A^{†,D}=A^{†}A{A}^D$ to be the dual DMP inverse of A. Later, Mehdipour and Salemi [19] defined the matrix $A^{c†}=A^{†}A{A}^{D}A{A}^{†}$ to be the CMP inverse of A. We know that $A^{†}=A^{\left|\right|A^{\ast }}$ and $A^D=a^{\left|\right|A^m}$, where $m=\mathrm{ind}\left(A\right)$, it is natural to consider the properties of $a^{\left|\right|b}a{a}^{\left|\right|c}$, $a^{\left|\right|c}a{a}^{\left|\right|b}$ and $a^{\left|\right|c}a{a}^{\left|\right|b}a{a}^{\left|\right|c}$, under the assumption that $a{a}^{\left|\right|c}a=a$.

Proposition 16.

Let S be any semigroup and $a,b,c\in S$. If a is invertible along b and c such that $a{a}^{\left|\right|c}a=a$, then

(1)

$a^{\left|\right|b}a{a}^{\left|\right|c}$ is the unique solution of the following equations

$$xax=x,bax=ba{a}^{\left|\right|c},xa=a^{\left|\right|b}a;$$

(2)

$a^{\left|\right|c}a{a}^{\left|\right|b}$ is the unique solution of the following equations

$$xax=x,ax=a{a}^{\left|\right|b},xab=a^{\left|\right|c}ab;$$

(3)

$a^{\left|\right|c}a{a}^{\left|\right|b}a{a}^{\left|\right|c}$ is the unique solution of the following equations

$$xax=x,axa=a{a}^{\left|\right|b}abax=ba{a}^{\left|\right|c},xab=a^{\left|\right|c}ab.$$

Proof. (1) We first check that $a^{\left|\right|b}a{a}^{\left|\right|c}$ satisfies these three equations. Actually, we have

$$a^{\left|\right|b}a{a}^{\left|\right|c}a{a}^{\left|\right|b}a{a}^{\left|\right|c}=a^{\left|\right|b}a{a}^{\left|\right|b}a{a}^{\left|\right|c}=a^{\left|\right|b}a{a}^{\left|\right|c},$$

$$ba{a}^{\left|\right|b}a{a}^{\left|\right|c}=ba{a}^{\left|\right|c}\mathrm{and}{a}^{\left|\right|b}a{a}^{\left|\right|c}a=a^{\left|\right|b}a.$$

If y also satisfies these equations, supposing that $a^{\left|\right|b}=sb$ for some $s\in S$, then

$$y=yay=a^{\left|\right|b}ay=sbay=sba{a}^{\left|\right|c}=a^{\left|\right|b}a{a}^{\left|\right|c}.$$

(2) and (3) can be proved similarly. □

Let $A\in {\mathbb{C}}^{n\times n}$ with $\mathrm{ind}\left(A\right)=m$. Taking $b=A^m$ and $c=A^{\ast }$ in Proposition 16, we recover the characterizations of the DMP inverse ([18], Theorem 2.2), dual DMP inverse and CMP inverse ([19], Theorem 2.1).

Particularly, $a^{\left|\right|b}a{a}^{\left|\right|c}$, $a^{\left|\right|c}a{a}^{\left|\right|b}$ and $a^{\left|\right|c}a{a}^{\left|\right|b}a{a}^{\left|\right|c}$ can be expressed as the $(\_,\_)$-inverses of a.

Proposition 17.

Let S be any semigroup and $a,b,c\in S$. If a is invertible along b and c such that $a{a}^{\left|\right|c}a=a$, then

(1)

$a^{\left|\right|b}a{a}^{\left|\right|c}$ is the $(b,ba{a}^{\left|\right|c})$-inverse of a;

(2)

$a^{\left|\right|(c,c)}a{a}^{\left|\right|(b,b)}$ is the $(a^{\left|\right|c}ab,b)$-inverse of a;

(3)

$a^{\left|\right|(c,c)}a{a}^{\left|\right|(b,b)}a{a}^{\left|\right|c}$ is the $(a^{\left|\right|c}ab,ba{a}^{\left|\right|c})$-inverse of a.

Proof. (1) It is obvious that $a^{\left|\right|b}a{a}^{\left|\right|c}\in bS\cap ba{a}^{\left|\right|c}S$. Meanwhile, we have

$$a^{\left|\right|b}a{a}^{\left|\right|c}ab=a^{\left|\right|b}ab=b,$$

$$ba{a}^{\left|\right|c}a{a}^{\left|\right|b}a{a}^{\left|\right|c}=ba{a}^{\left|\right|b}a{a}^{\left|\right|c}=ba{a}^{\left|\right|c}.$$

So $a^{\left|\right|b}a{a}^{\left|\right|c}$ is the $(b,ba{a}^{\left|\right|c})$-inverse of a.

(2) and (3) can be proved in a similar way. □

Let $A\in {\mathbb{C}}^{n\times n}$ with $\mathrm{ind}\left(A\right)=m$. Taking $b=A^m$ and $c=A^{\ast }$ in Proposition 17, we have $A^{D,†}=A^{\left|\right|(A^m,A^m{A}^{†})}=A^{\left|\right|(A^D,A^m{A}^{†})}$, which are Theorem 3.2 in [20] and Theorem 3.6 in [21].

Corollary 3.

Let $A\in {\mathbb{C}}^{n\times n}$ with $\mathrm{ind}\left(A\right)=m$. Then $A^{c†}$ is the $(A^{†}{A}^m,A^m{A}^{†})$-inverse of A.