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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2024; 64(1): 15-30

Published online March 31, 2024

### Generalized Inverses and Solutions to Equations in Rings with Involution

Yue Sui and Junchao Wei

Department of Mathematics, Yangzhou University,Yangzhou, 225002, P. R. China
e-mail : suiyue052@126.com and jcweiyz@126.com

Received: May 4, 2021; Revised: June 23, 2022; Accepted: July 25, 2022

In this paper, we focus on partial isometry elements and strongly EP elements or a ring. We construct characterizing equations such that an element which is both group invertible and MP-invertible, is a partial isometry element, or is strongly EP, exactly when these equations have a solution in a given set. In particular, an element aR#R is a partial isometry element if and only if the equation x = x(a)*a has at least one solution in {a, a#, a, a*, (a#)*, (a)*}. An element aR#R is a strongly EP element if and only if the equation (a)*xa = xaa has at least one solution in {a, a#, a, a*, (a#)*, (a)*}. These characterizations extend many well-known results.

Keywords: EP element, Normal EP element, Strongly EP element, Partial isometry

### 1. Introduction

Throughout this paper, R denotes an associative ring with 1. We write E(R) and J(R) to denote the set of all idempotents and the Jacobson radical of R, respectively.

An element aR is said to be group invertible if there exists a#R such that

aa#a=a,a#aa#=a#,aa#=a#a.

The element a# is called the group inverse of a, which is uniquely determined by the above equations [1, 9].

An involution *:aa* in a ring R is an anti-isomorphism of degree 2, that is,

(a*)*=a,(a+b)*=a*+b*,(ab)*=b*a*.

An element a in R is called normal if aa*=a*a.

An element a in R is called the Moore-Penrose inverse (MP-inverse) of a [6, 10], if

aaa=a,aaa=a,(aa)*=aa,(aa)*=aa.

If such a exists, then it is unique [6]. Denote by R# and R the set of group invertible elements of R and the set of all MP-invertible elements of R, respectively.

An element a is said to be EP if aR#R and satisfies a#=a [3, 6]. We denote by REP the set of all EP elements of R. Note that if aR is normal, then aREP, see [6]. An element aR is called normal EP if a is normal and aR. Denote by RNEP the set of all normal EP elements of R.

An element a is called a partial isometry if a=a* and a is called a strongly EP element if aREP is a partial isometry. We denote the sets of all partial isometry elements and strongly EP elements of R by RPI and RSEP, respectively.

In [7], D. Mosić and D. S. Djordjević presented some characterizations of EP elements in rings with involution. In addition, some equivalent conditions for the element a in a ring with involution to be a partial isometry are given. Recent researches on EP elements in rings with involution have produced some interesting results, see [4, 9, 12]. The necessary and sufficient conditions for the existence of a common solution and the general common solution of the equation axb=c (a,b are regular elements) were given for rings with involution in [2]. In [13, 15], a new kind of characterizations of generalized inverse elements has been studied by means of the solution of constructed equations recently.

Motivated by these articles above, this paper is intended to provide, by using certain equations admitting solutions in a definite set, further sufficient and necessary conditions for an element in a ring with involution to be an EP element, partial isometry, normal EP element, and strongly EP element. This is a new way to study generalized inverses in rings.

### 2. EP elements

Lemma 2.1.[5, 7] Let aR#R. If a*aa#(1-aa)=0, then aREP.

Proof. Pre-multiplying the equality a*aa#(1-aa)=0 by (a)*, we have aa#(1-aa)=0. That is aa#=aa. Hence aREP by [7, Theorem 1.6] or [5].

Lemma 2.2. Let aR#R. Then aRPI if and only if a*a=aa.

Proof. ⇒ The equality obviously holds since a*=a.

⇐ Post-multiplying a*a=aa by a, one has a*aa=aaa. Applying the involution to the last equality, we have aa2=aa(a)*, it follows that a2=a(a)*. Post-multiplying the equality by a*, we get a2a*=a2a. Pre-multiplying a2a*=a2a by a#, one has aa*=aa. Thus aRPI by [7, Theorem 2.1].

In order to prove the theorems given in this paper more clearly, we briefly review the following existing conclusions:

Lemma 2.3.[11, Lemma 2.2] Let aR#. Then (a#)*R=a*R and R(a#)*=Ra*.

Lemma 2.4.[11, Lemma 2.3] Let aR. Then

(1) aR=aaR=aa*R and Ra=Raa=Ra*a.

(2) a*R=aR=a*aR=aaR and Ra*=Ra=Raa*=Raa.}

Lemma 2.5.[12, Theorem 3.9] Let aR. Then the following are equivalent:

(1) aREP;

(2) aR# and aRa*R;

(3) aR# and RaRa*;

(4) aR# and a*RaR;

(5) aR# and Ra*Ra.}

Lemma 2.6.[14, Lemma 2.1] Let aR#R. Then the following conditions are satisfied:

(1) a*R=a*a2R=a*aa#R=(a#)*R;

(2) Ra=Ra#=Raa*a#=Ra*a=Ra*a*a=Raa*a;

(3) (a#)*aaR=(a#)*a#aR=(a#)*a#a*R;

(4) a#R=aR and Ra*=Ra.}

In [14, Theorem 2.4], the authors proved that an element aR#R can be an EP element if and only if the equation axa#+axa*=xaa+a*ax has at least one solution in the set χa={a,a#,a,a*,(a#)*,(a)*}.

Recall that an element a is said to be EP if aR#R and satisfies a#=a. Thus, we can modify the above existing theorem in [14] and construct the following equation, with the help of which we can explore a new kind of characterization of EP elements:

axa+axa*=xaa#+a*ax.

Theorem 2.7. Let aR#R. Then aREP if and only if equation (2.1) has at least one solution in χa={a,a#,a,a*,(a#)*,(a)*}.

Proof. ⇒ Obviously, x=a is a solution because a=a#.

⇐ (1) If x=a is a solution, then a2a+a2a*=a2a#+a*a2=a+a*a2. By Lemma 2.6, we have

a*R=a*a2R=(a2a+a2a*-a)RaR.

Therefore aREP by Lemma 2.5.

(2) If x=a# is a solution, then aa#a+aa#a*=a#aa#+a*aa#=a#+a*aa#. By Lemma 2.6, we obtain that

a*R=a*aa#R=(aa#a+aa#a*-a#)RaR.

The fact that aREP follows from Lemma 2.5.

(3) If x=a is a solution, then aaa+aaa*=aaa#+a*aa=aaa#+a*. Pre-multiplying it by aa#, we obtain aaa+aaa*=a#+a#aa*. By Lemma 2.6, we have

Ra#=R(aaa+aaa*-a#aa*)Ra+Ra*=Ra.

Since Ra#=Ra,Ra=Ra* by Lemma 2.6, we get RaRa*. From Lemma 2.5, aREP.

(4) If x=a* is a solution, then aa*a+aa*a*=a*aa#+a*aa*. Post-multiplying it by (1-aa), we have a*aa#(1-aa)=0. By Lemma 2.1, we get aREP.

(5) If x=(a#)* is a solution, then a(a#)*a+a(a#)*a*=(a#)*aa#+a*a(a#)*. Post-multiplying it by (1-aa), we get (a#)*aa#(1-aa)=0. Pre-multiplying the last equation by (a2)*, we get a*aa#(1-aa)=0. Therefore aREP by Lemma 2.1.

(6) If x=(a)* is a solution, then a(a)*a+a(a)*a*=(a)*aa#+a*a(a)*. Taking involution of the above equality, we obtain that

(a)*aa*+aaa*=(a#)*a*a+aa*a.

Ra=Raa*a=R((a)*aa*+aaa*-(a#)*a*a)Ra*+Ra=Ra=Ra*.

From Lemma 2.5, aREP.

Multipying the equation (2.1) on the right by a, we obtain:

axaa+axa*a=ax+a*axa.

Theorem 2.8. Let aR#R. Then aREP if and only if the equation (2.2) has at least one solution in χa.

Proof. ⇒Obviously, x=a is a solution.

⇐ (1) If x=a is a solution, then a2aa+a2a*a=a2+a*a3. It is immediate that a2a*a=a*a3. By Lemma 2.4, we obtain that

a*R=a*a3R=a2a*aRaR.

Hence aREP by Lemma 2.5.

(2) If x=a# is a solution, then aa#aa+aa#a*a=aa#+a*aa#a. That is aa#a*a=a*a. Post-multiplying it by a, we obtain that aa#a*a2=a*a2.

By Lemma 2.6, we get

a*R=a*a2R=aa#a*a2RaR.

Therefore, aREP by Lemma 2.5.

(3) If x=a is a solution, then aaaa+aaa*a=aa+a*aaa=aa+a*a. Post-multiplying it by a, we have aaaa2+aaa*a2=a+a*a2. We thus get

a*R=a*a2R=(aaaa2+aaa*a2-a)RaR

by Lemma 2.6.

And then it follows from Lemma 2.5 that aREP.

(4) If x=a* is a solution, then aa*aa+aa*a*a=aa*+a*aa*a. We conclude from Lemma 2.4 that

Ra*=Raa*=R(aa*aa+aa*a*a-a*aa*a)Ra.

Hence aREP by Lemma 2.5.

(5) If x=(a#)* is a solution, then a(a#)*aa+a(a#)*a*a=a(a#)*+a*a(a#)*a. Pre-multiplying it by a, we get (a#)*aa+(a#)*a*a=(a#)*+aa*a(a#)*a. Then from Lemma 2.3, we obtain that

Ra*=R(a#)*=R((a#)*aa+(a#)*a*a-aa*a(a#)*a)Ra,

which yields aREP by Lemma 2.5.

(6) If x=(a)* is a solution, then a(a)*aa+a(a)*a*a=a(a)*+a*a(a)*a. That is a2=a*a(a)*a. Post-multiplying it by a#, we obtain that a=a*a(a)*aa#. Then

aR=a*a(a)*aa#Ra*R.

Therefore, aREP by Lemma 2.5.

Further, we revised the equation (2.2) as follows:

axaa+xaa*a=ax+a*axa.

Theorem 2.9. Let aR#R. Then aREP if and only if the equation (2.3) has at least one solution in χa.

Proof.x=a is a solution since aa=aa.

⇐ (1) If x=a is a solution, then a2aa+a2a*a=a2+a*a3. It is immediate from the proof of Theorem 2.8(1) that aREP.

(2) If x=a# is a solution, then aa#aa+a#aa*a=aa#+a*aa#a. Then aREP by the proof of Theorem 2.8(2) since aa#=a#a.

(3) If x=a is a solution, then aaaa+aaa*a=aa+a*aaa=aa+a*a. That is aaaa=aa. Applying the involution, one has aa=aa2a. By Lemma 2.4 and Lemma 2.5, we have aREP.

(4) If x=a* is a solution, then aa*aa+a*aa*a=aa*+a*aa*a. That is aa*aa=aa*. From Lemma 2.4, we obtain that

Ra*=Raa*=Raa*aaRa.

By Lemma 2.5, aREP.

(5) If x=(a#)* is a solution, then a(a#)*aa+(a#)*aa*a=a(a#)*+a*a(a#)*a. Pre-multiplying it by a, we have (a#)*aa+a(a#)*aa*a=(a#)*+aa*a(a#)*a. By Lemma 2.3, we have

Ra*=R(a#)*=R((a#)*aa+a(a#)*aa*a-aa*a(a#)*a)Ra,

which gives aREP by Lemma 2.5.

(6) If x=(a)* is a solution, then a(a)*aa+(a)*aa*a=a(a)*+a*a(a)*a. That is (a)*aa*a=a*a(a)*a.

Pre-multiplying it by (1-aa), we have

(1-aa)a*a(a)*a=0.

Hence 0=(1-aa)a*a(aaa)*a=(1-aa)a*a(a)*aa2.

Multiplying the last equality by a# on the right, we obtain that 0=(1-aa)a*a(a)*aa=(1-aa)a*a(a)*. Post-multiply it by a* and then we have

(1-aa)a*a2a=0.

Post-multiplying it by aa#a, we get (1-aa)a*=0, which implies a=a2a. Consequently, aREP.

Theorem 2.10. Let aR#R. Then aREP if and only if the equality axa=a* has at least one solution.

Proof. ⇒ Since aREP, we have aa=aa. Hence x=aa*a is a solution of the equation axa=a*.

⇐ Assume that axa=a* have a solution x0. Then Ra*=Rax0aRa, it follows that aREP by Lemma 2.5.

Theorem 2.11. Let aR#R. Then aREP if and only if the equation axa=aa#-aa has at least one solution.

Proof. ⇒ Since aREP, we have aa#-aa=0. Then x=aa#-aa is a solution of the equation axa=aa#-aa.

⇐ Assume that axa=aa#-aa has a solution. Then, by [2, Theorem 2.1], we get

aa#aa=aa(aa#aa)aa.

Pre-multiplying (2.4) by a, we obtain that

a-a2a=a(aa#-aa)aa=a-a2aaa.

That is a2a=a2aaa.

On the other hand, post-multiply (2.4) by a, we have

aa(aa#aa)aa2=0.

Pre-multiplying (2.5) by a and then post-multiplying the last equation by a#, we obtain that

a(aa#-aa)aa=0.

That is a=a2aaa. Obviously, we can deduce that a=a2a. Consequently, aREP.

Let aR. Write a0={xR|ax=0}. Clearly, a0 is a right ideal of R, which is called the right annihilator of a. Similarly, we can define  0a. Then, we have the following theorem.

Theorem 2.12. Let aR#R. Then aREP if and only if a0=(a)0.

Proof. ⇒ Since aREP, we have a#=a. Therefore (a#)o=(a)0. Note that a0=(a#)0. Then a0=(a)0.

⇐ Assume that a0=(a)0. Note that 1-aa(a#)0. Then 1-aa(a)0, which implies a=aaa. Hence RaRa, one obtains aREP by Lemma 2.4 and Lemma 2.5.

### 3. Partial Isometry Elements

Recall that an element cR is semi-idempotent if c-c2J(R). Using the semi-idempotent elements of R, we have the following theorem.

Theorem 3.1. Let aR. Then aRPI if and only if the following two conditions hold:

(1) aa* is a semi-idempotent;

(2) a-a*R#.}

Proof. The equality a*=a implies that

aa*-aa*aa*=0J(R)anda-a*=0E(R)R#.

On the contrary, assume that aa* is semi-idempotent and a-a*R#. Take aa*-aa*aa*=xJ(R). Then, by the proof of [14, Theorem 2.6], one has a-a*=a(a)*axJ(R). Set z=(a-a*)# because a-a*R#. Then a-a*=(a-a*)z(a-a*). Thus, we get

(a-a*)(1-z(a-a*))=(a-a*)-(a-a*)z(a-a*)=0.

Since z(a-a*)J(R), we obtain that 1-z(a-a*) is invertible. Hence a-a*=0, so aRPI.

Theorem 3.2. Let aR. Then the following conditions are equivalent:

(1) a(a)*E(R);

(2) (a)*aE(R);

(3) aRPI;

(4) a(a)* is a semi-idempotent and a*-aE(R);

(5) aa-a(a)*E(R);

(6) aa-(a)*aE(R);

(7) (a)*a is a semi-idempotent and a*-aE(R).

Proof. (1)⇒(2) From the assumption, we know that a(a)*=a(a)*a(a)*. Pre-multiplying it by a and then post-multiplying the last equality by a, we get (a)*a=(a)*a(a)*a.

(2)⇒(3) From (2), we obtain that (a)*a=(a)*a(a)*a. Post-multiplying it by aa*, we get aa=(a)*a. Multiply the equality by a on the right and then we get a=(a)*. Applying involution to a=(a)*, we get a*=a. Consequently, aRPI.

(3)⇒(4) Since aRPI, a*=a. Then we know that a*-a=0E(R) and a(a)*-a(a)*a(a)*=0J(R). It is immediate that a(a)* is a semi-idempotent and a*-aE(R).

(4)⇒(5) Write x=aa-a(a)*. Then x-x2=a(a)*-a(a)*a(a)*J(R) by hypothesis. Clearly, a(x-x2)a*a=a-(a)*. Note that a*-aE(R). Then a-(a)*E(R), this gives a(x-x2)a*aJ(R)E(R), so a(x-x2)a*a=0. It follows that a=(a)*. Hence aa-a(a)*E(R).

(5)⇒(6) From (5), we know that a(a)*=a(a)*a(a)*. Pre-multiplying it by a and then multiplying the last equality by a on the right, we obtain that

(a)*a=(a)*a(a)*a.

Hence aa-(a)*a-(aa-(a)*a)(aa-(a)*a)=(a)*a-(a)*a(a)*a=0. Consequently, aa-(a)*aE(R).

(6)⇒(7) From (6), we obtain that (a)*a=(a)*a(a)*a. So (a)*a is an idempotent. By (2)⇒(3), we get a*=a, which gives a*-a=0E(R).

(7)⇒(1) Similar to (4)⇒(5), we obtain that a*=a. Then

a(a)*a(a)*=aaa(a)*=a(a)*.

Therefore, a(a)*E(R).

Lemma 3.3. Let aR. Then aRPI if and only if a=a(a)*a.

Proof. ⇒ Since aRPI, a*=a. And then we can easily get

a(a)*a=aaa=a.

⇐ From the assumption, we know that a=a(a)*a. Pre-multiplying it by a and then post-multiplying the last equality by a, we get a=(a)*. Taking involution of the above equality, we obtain a*=a. So aRPI.

Lemma 3.4. Let aR#R. Then aRPI if and only if a2=a(a)*.

Proof. ⇒ We know that a*=a according to the assumption. Then we can get the following equality.

a(a)*=a(a*)*=a2.

⇐ From the assumption, we know that a2=a(a)*. Post-multiplying it by a*, we have a2a*=a2a. Hence aRPI by [8, Theorem 2.1].

Lemma 3.5. Let aR#R. Then aRPI if and only if the following equation has at least one solution in χa:

x=x(a)*a.

Proof. ⇒ Obviously, a* is a solution.

⇐ (1) If x=a is a solution, then a=a(a)*a. Post-multiplying it by a, we have a2=a(a)*. By Lemma 3.4, aRPI.

(2) If x=a# is a solution, then a#=a#(a)*a is a solution. Pre-multiplying it by a2, we get a=a(a)*a. By the proof of (1), we know aRPI.

(3) If x=a is a solution, then a=a(a)*a. Hence aRPI by Lemma 3.3.

(4) If x=a* is a solution, then a*=a*(a)*a=a. It is immediate that aRPI.

(5) If x=(a#)* is a solution, then (a#)*=(a#)*(a)*a. Post-multiplying it by a, we get (a#)*a=(a#)*(a)*. Applying involution to the above equality, we have a*a#=aa#. Then we deduce that aRPI by [8, Theorem 2.1].

(6) If x=(a)* is a solution, then (a)*=(a)*(a)*a. Firstly, multiply the equality on the right by a, apply involution to the latest equation, and then we get a*a=aa. By Lemma 2.2, aRPI.

Similarly, we have the following theorem.

Theorem 3.6. Let aR#R. Then aRPI if and only if the following equation has at least one solution in χa:

x=(a)*ax.

Using the symmetricity, we have the following corollary.

Corollary 3.7. Let aR#R. Then aRPI if and only if the following equation has at least one solution in χa:

x=xa(a)*.

### 4. Normal EP Elements

Lemma 4.1. Let aR and xR.

(1) If a(a)*ax=0, then ax=0;

(2) If xa(a)*a=0, then xa=0.

Proof. (1) Pre-multiplying the equality a(a)*ax=0 by a*a, we immediately have ax=0.

(2) Similarly, we can prove (2).

Lemma 4.2. Let aRR# and xR. If a(a)*ax=0, then ax=0.

Proof. Since a(a)*=a2a(a)*, pre-multiplying the equality a(a)*ax=0 by a#, one has (a)*ax=0. Pre-multiplying the last equality by a*, one obtains ax=0.

Lemma 4.3. Let aRR# and xR. If x(a#)*(a)*a=0, then x(a#)*=0.

Proof. Post-multiplying x(a#)*(a)*a=0 by aa*, we have x(a#)*aa=0. Note that (a#)*aa=(a#)*. Thus, x(a#)*=0.

Lemma 4.4. Let aRR# and xR. If (a)*(a)*ax=0, then ax=0.

Proof. Pre-multiplying (a)*(a)*ax=0 by a#a*, we have a#(a)*ax=0. Pre-multiplying the last equation by a2, we obtain that a(a)*ax=0. By Lemma 4.2, ax=0.

Lemma 4.5.[14, Lemma 2.3] Let aR#R. Then aREP if and only if one of the following conditions holds:

(1)RaRa;

(2)RaRa;

(4)aRaR;

(6)aRaR;

(3)Ra#Ra*;

(5)Ra#Ra.

Lemma 4.6.[14, Lemma 2.11] Let aR#R. Then aRNEP if and only if (a)*a=a(a)*.

Theorem 4.7. Let aR#R. Then aRNEP if and only if the following equation has at least one solution in χa:

xa(a)*=x(a)*a.

Proof. ⇒ By [11, Corollary 2.8], we know that x=a is a solution.

⇐ (1) If x=a is a solution, then aa(a)*=a(a)*a. That is (a)*=a(a)*a. This infers that Ra=R(a)*=Ra(a)*aRa=Ra* by [11, Lemma 2.1]. It follows from Lemma 2.5 that aREP. Moreover, post-multiplying (a)*=a(a)*a by a, we get (a)*a=a(a)*. Take involution of the last equation. It follows a*a=aa*. By [11, Lemma 2.7], a is normal. Hence aRNEP.

(2) If x=a# is a solution, then a#a(a)*=a#(a)*a. Note that (a#)0=a0. Then we get aa(a)*=a(a)*a. Hence aRNEP by (1).

(3) If x=a is a solution, then aa(a)*=a(a)*a. Note that (a)*=(a)*aa. Then a(a)*a(1-aa)=0. By Lemma 4.1, we have a(1-aa)=0. Then, Ra=RaaaRa. Thus, by Lemma 4.5, we obtain aREP and aa=aa. On the other hand,

a(a)*=aaa(a)*=aaa(a)*=aa(a)*a=(a)*a,

which shows aRNEP by Lemma 4.6.

(4) If x=a* is a solution, then a*a(a)*=a*(a)*a=aaa=a. Similar to the proof of (1), aRNEP.

(5) If x=(a#)* is a solution, then (a#)*a(a)*=(a#)*(a)*a. Applying involution to it, then a(a)*a#=(a)*aa#. We get a(a)*a=(a)* because  0a=0(a#). Similar to the proof of (1), aRNEP.

(6) If x=(a)* is a solution, then (a)*a(a)*=(a)*(a)*a. Taking involution of the equality, we obtain that

a(a)*a=(a)*aa.

Similar to the proof of (3), aRNEP.

Theorem 4.8. Let aR#R. Then aRNEP if and only if the following equation has at least one solution in χa:

x(a)*a=(a)*ax.

Proof. ⇒ Since aRNEP, x=a is a solution.

⇐ (1) If x=a is a solution, then a(a)*a=(a)*aa. Post-multiplying it by (1-aa), we have a(a)*a(1-aa)=0. By Lemma 4.2, a(1-aa)=0.

Hence aREP. It follows that

(a)*a=aa(a)*a=aa(a)*a=a(a)*aa=a(a)*.

Therefore, aRNEP according to Lemma 4.6.

(2) If x=a# is a solution, then a#(a)*a=(a)*aa#. Post-multiplying it by (1-aa), we get

a#(a)*a(1-aa)=0.

By Lemma 4.2 and the proof of (1), aREP.

Post-multiplying a#(a)*a=(a)*aa# by a, we have a(a)*=a#(a)*=(a)*a. Hence aRNEP by Lemma 4.6.

(3) If x=a is a solution, then a(a)*a=(a)*aa. Pre-multiplying it by (1-aa), we have

(1-aa)a(a)*a=0.

By Lemma 4.1, (1-aa)a=0. Hence aREP. Then x=a# is a solution. By (2), aRNEP.

(4) If x=a* is a solution, then a*(a)*a=(a)*aa*. That is a=(a)*aa*. Similar to the proof of (1) in Theorem 4.7, we have aRNEP.

(5) If x=(a#)* is a solution, then (a#)*(a)*a=(a)*a(a#)*. Pre-multiplying it by (1-aa), we get

(1-aa)(a#)*(a)*a=0.

By Lemma 4.3, (1-aa)(a#)*=0. This gives a#=a#aa. Therefore, aREP. Multiplying (a#)*(a)*a=(a)*a(a#)* on the left by a*, we obtain that

(a)*a=a(a#)*=a(a)*,

which implies aRNEP by Lemma 4.6.

(6) If x=(a)* is a solution, then (a)*(a)*a=(a)*a(a)*. Post-multiplying it by (1-aa), we have (a)*(a)*a(1-aa)=0. By Lemma 4.4, we obtain that a(1-aa)=0, which yields aREP. Therefore, x=(a#)* is a solution. By (5), aRNEP.

### 5. Strongly EP elements

Theorem 5.1. Let aR#R. Then aRSEP if and only if the following equation has at least one solution in χa:

x=(a#)*xa#.

Proof. ⇒ Note that a#=a=a* since aRSEP. Hence x=a* is a solution.

⇐ (1) If x=a is a solution, then a=(a#)*aa#. Post-multiplying the equality by a, one gets a2=(a#)*a. Hence

a2a=(a#)*aa=(a#)*,

which leads to a#=aaa*. Then we have Ra#=Raaa*Ra*. Thus, aREP by Lemma 4.5. Then, we get a=a*, which implies aRPI. Hence aRSEP.

(2) If x=a# is a solution, then a#=(a#)*a#a#. Multiplying the equality by a2 from the right, one obtains a=(a#)*aa#. By the proof of (1), we get aRSEP.

(3) If x=a is a solution, then a=(a#)*aa#. Multiplying the equality by aa from the right, we have

aaa=(a#)*aa#aa=(a#)*aa#=a.

Then, Ra=RaaaRa. Thus, by Lemma 4.5, we obtain aREP, which gives x=a=a#. By (2), aRSEP.

(4) If x=a* is a solution, then a*=(a#)*a*a#. We can deduce that

a*aa=(a#)*a*a#aa=(a#)*a*a#=a*.

Applying the involution to the equality, one has a=aa2. Thus, we get aR=aa2RaR, which implies aREP by Lemma 4.5. Then, we find that

a*=(a#)*a*a#=(a)*a*a#=aaa#=a#,

which gives aRSEP.

(5) If x=(a#)* is a solution, then (a#)*=(a#)*(a#)*a#. Hence we deduce that

a*=a*a*(a#)*=a*a*(a#)*(a#)*a#=(a#)*a*a#.

By (4), we get aRSEP.

(6) If x=(a)* is a solution, then (a)*=(a#)*(a)*a#. Applying involution to the equality, we have a=(a#)*aa#. By (3), aRSEP.

Theorem 5.2. Let aR#R. Then aRSEP if and only if the following equation has at least one solution in χa:

xaa=x(a)*a.

Proof. ⇒ Obviously x=a is a solution since a*=a=a#.

⇐ (1) If x=a is a solution, then a=aaa=a(a)*a. Hence

Ra=Ra(a)*aRa.

By Lemma 4.5, aREP. Post-multiplying a=a(a)*a by a, we get a2=a(a)*. Thus a*=a by Lemma 3.4, which implies aRSEP.

(2) If x=a# is a solution, then a#=a#aa=a#(a)*a. Pre-multiplying the equality by a2, we have a=a(a)*a. By (1), aRSEP.

(3) If x=a is a solution, then aaa=a(a)*a. Note that (a)0=(a*)0. Then we get a*aa=a. Therefore aRSEP by [7, Theorem 2.3].

(4) If x=a* is a solution, then a*aa=a*(a)*a=a. This gives a(1-aa)=0, so aREP. Post-multiplying a*aa=a by a, we get a*a=aa. Then we obtain that aRSEP by Lemma 2.2.

(5) If x=(a#)* is a solution, then (a#)*aa=(a#)*(a)*a. Taking involution of the equality, we deduce that

aaa#=(a)*aa#.

This implies aaa=(a)*aa=(a)* because  0(a#)=  0a. Hence a*aa=a, which infers aRSEP by [7, Theorem 2.3].

(6) If x=(a)* is a solution, then (a)*=(a)*aa=(a)*(a)*a. Post-multiplying the equality (a)*=(a)*(a)*a by (1-aa), we have (a)*(1-aa)=0 which implies a=aaa. Hence aREP. Post-multiplying (a)*=(a)*(a)*a by a, we get (a)*a=(a)*(a)*. Applying involution to the equality, we obtain that a*a=aa. Therefore, aRSEP according to Lemma 2.2.

Theorem 5.3. Let aR#R. Then aRSEP if and only if the following equation has at least one solution in χa:

(a)*xa=xaa.

Proof.x=a is a solution since a=a*=a#.

⇐ (1) If x=a is a solution, then (a)*aa=aaa=a. Hence Ra=R(a)*aaRa. By Lemma 4.5, aREP. Post-multiplying (a)*aa=a by a, we obtain that (a)*a=a2. Per-multiplying the equation by a*, we get aa2=a*a2. Post-multiplying the last equation by a#a, we obtain a=a*. Therefore aRSEP.

(2) If x=a# is a solution, then (a)*a#a=a#aa=a#. Observe that

Ra#=R(a)*a#aRa.

This implies that aREP by Lemma 4.5. Then, we can obtain (a)*aa#=a#. Since  0(a)=  0(a#), we get (a)*aa=a. That is (a)*=a. Therefore, aRSEP.

(3) If x=a is a solution, then (a)*aa=aaa. Taking involution of the equality, we have (a)*(a)*a=aa(a)*. Pre-multiplying the last equality by (1-aa), we have

(1-aa)(a)*(a)*a=0.

Post-multiplying by aa*, we get (1-aa)(a)*aa=0, it is immediate that

(1-aa)(a)*=(1-aa)(a)*aa=(1-aa)(a)*aa2aa#a=(1-aa)(a)*aaa#a=0.

Hence aREP. On the other hand, pre-multiply (a)*aa=aaa by a*, and we obtain that aa=a*a, which implies aRSEP by Lemma 2.2.

(4) If x=a* is a solution, then (a)*a*a=a*aa. Taking involution of the equality, we get (a)*aa=aa2. Hence we obtain that

Ra=Ra#a2=Ra#aaa2Raa2=R(a)*aaRa.

Therefore aREP by Lemma 4.5. Post-multiplying (a)*aa=aa2 by a, we have (a)*a=a2. Then we deduce that a*=a by the proof of (1). Therefore aRSEP.

(5) If x=(a#)* is a solution, then (a)*(a#)*a=(a#)*aa. Applying involution to it, we get (a)*a#a=aaa#. Post-multiplying the last equality by aa, we obtain that (a)*a#a=a. Then, by [11, Lemma 2.1] we know that

aR=(a)*a#aR(a)*R=aR.

By Lemma 4.5, aREP. Post-multiplying (a)*a#a=a by a3, we get (a)*a=a2. From Lemma 3.4, we deduce that a*=a, which implies aRSEP.

(6) If x=(a)* is a solution, then (a)*(a)*a=(a)*aa. By Theorem 5.2 (6), aRSEP.

Theorem 5.4. Let aR#R. Then aRSEP if and only if the following equation has at least one solution in χa:

ax(a)*=xaa.

Proof.x=a* is a solution since a=a*=a#.

⇐ (1) If x=a is a solution, then aa(a)*=a2a. Taking involution of the equality, we have aaa=aaa*. By [14, Theorem 2.15] (3), we deduce that aRSEP.

(2) If x=a# is a solution, then aa#(a)*=a#aa. Pre-multiplying the equality by a, we get a#(a)*=aa. Taking involution of the last equality and then post-multiplying the obtained equality by a, we obtain that a(a#)*a=a. It is evident that

aR=a(a#)*aRaR,

which shows aREP by Lemma 4.5. Furthermore, post-multiply a#(a)*=aa by a* and thus we get a=a*. Hence aRSEP.

(3) If x=a is a solution, then aa(a)*=aaa=a. Similar to the proof of Theorem 5.1 (3), we deduce that aRSEP.

(4) If x=a* is a solution, then aa*(a)*=a*aa=a*. Similar to the proof of Theorem 5.3 (1), we get aRSEP.

(5) If x=(a#)* is a solution, then a(a#)*(a)*=(a#)*aa. Taking involution of the equality, we get aa#(a)*=aaa#=a#. Similar to the proof of Theorem 5.1 (2), we have aRSEP.

(6) If x=(a)* is a solution, then a(a)*(a)*=(a)*aa. Similar to the proof of Theorem 5.3 (3), we know that aRSEP.

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