Suppose that *A* is an arbitrary monoid having {H}_{1}^{\ast} and {H}_{1} as the {\mathcal{H}}^{\ast}- and ℋ- classes containing the identity element {1}_{A} of *A*. Moreover, let us assume that *β* and *γ* are morphisms from *A* into {H}_{1}^{\ast} and, for an element *u* in {H}_{1}, let {\lambda}_{u} be the inner automorphism of {H}_{1}^{\ast} defined by x\mapsto ux{u}^{-1} such that \gamma {\lambda}_{u}=\beta \gamma.

Now one can consider the set S={\mathbb{N}}^{0}\times {\mathbb{N}}^{0}\times A\times {\mathbb{N}}^{0}\times {\mathbb{N}}^{0} into a semigroup with a multiplication

where d=max(p,{n}^{\prime}) and {\beta}^{0}, {\gamma}^{0} are interpreted as the identity map of *A*, and also {u}^{0} is interpreted as the identity {1}_{A} of *A*. In [8], Yu Shung and Li-Min Wang showed that *S* is a monoid with the identity (0,0,{1}_{A},0,0). In fact, this new monoid S={\mathbb{N}}^{0}\times {\mathbb{N}}^{0}\times A\times {\mathbb{N}}^{0}\times {\mathbb{N}}^{0} is denoted by {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) and called *generalized Bruck-Reilly* ∗-*extension* of *A* determined by the morphisms *β*, *γ* and the element *u*.

The following lemmas were established in [8].

**Lemma 2** *If* (m,n,v,p,q)\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u), *then* (m,n,v,p,q) *is an idempotent if and only if* m=q, n=p *and* *v* *is idempotent*.

**Lemma 3** *If* (m,n,v,p,q)\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u), *then* (m,n,v,p,q) *has an inverse*

({m}^{\prime},{n}^{\prime},{v}^{\prime},{p}^{\prime},{q}^{\prime})\in S

*if and only if* {v}^{\prime} *is an inverse of* *v* *in* *A* *while* {m}^{\prime}=q, {n}^{\prime}=p, {p}^{\prime}=n *and* {q}^{\prime}=m.

Then we have an immediate consequence as in the following.

**Corollary 2** *Let* *A* *be a monoid*. *Then* {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) *is regular if and only if* *A* *is regular*.

In this section, we mainly characterize the properties coregularity and strongly *π*-inverse over the generalized Bruck-Reilly ∗-extensions of monoids. More specifically, for a given monoid *A*, we determine the maximal submonoid of {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u), which can be held coregularity if *A* satisfies particular properties.

Our first observation is the following.

**Lemma 4** *The set* \mathcal{L}:=\{(m,n,v,n,m)|v\in A,m,n\in {\mathbb{N}}^{0}\} *is a submonoid of* {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u).

*Proof* By considering the multiplication in (4), the proof can be seen easily. □

It turns out that all coregular elements in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) belong to the submonoid ℒ .

**Lemma 5** *Let* (m,n,v,p,q)\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u). *If* (m,n,v,p,q) *is coregular then* m=q *and* n=p.

*Proof* Let (m,n,v,p,q)\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) be a coregular element. Then there exists an element ({m}^{\prime},{n}^{\prime},{v}^{\prime},{p}^{\prime},{q}^{\prime})\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) such that

Let q={m}^{\prime} and {q}^{\prime}=m. Then we have

for some w\in A, where z=max(p,{n}^{\prime}) and {z}^{\prime}=max({p}^{\prime}-{n}^{\prime}+z,n). This implies that

n=n+{n}^{\prime}-p-{p}^{\prime}+{z}^{\prime}

(5)

and

By (6), we have n={z}^{\prime}. Applying this in (5), we get n+{n}^{\prime}-p-{p}^{\prime}+n=n and thus n+{n}^{\prime}=p+{p}^{\prime}. Further, we have

for some {w}^{\prime}\in A, where Z=max({p}^{\prime},n) and {Z}^{\prime}=max(p-n+Z,{n}^{\prime}). This implies that m={m}^{\prime}, q={q}^{\prime},

n={n}^{\prime}+n-{p}^{\prime}-p+{Z}^{\prime}

(7)

and

p={p}^{\prime}-{n}^{\prime}+{Z}^{\prime}.

(8)

By writing the equality (8) in (7), we get {n}^{\prime}={p}^{\prime}. Together with n+{n}^{\prime}=p+{p}^{\prime}, we obtain n=p. By assuming q={m}^{\prime}, we also get m=q.

Now let (x,y,t,z,w)({x}^{\prime},{y}^{\prime},{t}^{\prime},{z}^{\prime},{w}^{\prime})=({x}^{\u2033},{y}^{\u2033},{t}^{\u2033},{z}^{\u2033},{w}^{\u2033}). Then it is easy to verify that {x}^{\u2033}\ge {x}^{\prime},x and {y}^{\u2033}\ge {y}^{\prime},y. If w\ne {x}^{\prime}, we can easily see that {x}^{\u2033}>{x}^{\prime},x or {y}^{\u2033}>{y}^{\prime},y. This shows that (m,n,a,p,q)({m}^{\prime},{n}^{\prime},{a}^{\prime},{p}^{\prime},{q}^{\prime})(m,n,a,p,q)\ne (m,n,a,p,q) if q\ne {m}^{\prime} or {q}^{\prime}\ne m. Hence, q\ne {m}^{\prime} or {q}^{\prime}\ne m is not possible. □

Then we have the following result.

**Theorem 3** *Let* *A* *be a monoid*. *Then the submonoid* ℒ *of* {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) *is coregular if and only if* *A* *is coregular*.

*Proof* Suppose that \mathcal{L}=\{(m,n,v,n,m)|v\in A,m,n\in {\mathbb{N}}^{0}\}\le {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) is coregular. For each ({m}^{\prime},0,v,0,{m}^{\prime}) in ℒ, there exists an element ({m}^{\prime},{n}^{\prime},{v}^{\prime},{n}^{\prime},{m}^{\prime})\in \mathcal{L} such that

\begin{array}{rcl}({m}^{\prime},0,v,0,{m}^{\prime})({m}^{\prime},{n}^{\prime},{v}^{\prime},{n}^{\prime},{m}^{\prime})({m}^{\prime},0,v,0,{m}^{\prime})& =& ({m}^{\prime},{n}^{\prime},\left(v{\beta}^{{n}^{\prime}}\right){v}^{\prime}\left(v{\beta}^{{n}^{\prime}}\right),{n}^{\prime},{m}^{\prime})\\ =& ({m}^{\prime},0,v,0,{m}^{\prime})\end{array}

(9)

and

\begin{array}{rcl}({m}^{\prime},{n}^{\prime},{v}^{\prime},{n}^{\prime},{m}^{\prime})({m}^{\prime},0,v,0,{m}^{\prime})({m}^{\prime},{n}^{\prime},{v}^{\prime},{n}^{\prime},{m}^{\prime})& =& ({m}^{\prime},{n}^{\prime},{v}^{\prime}\left(v{\beta}^{{n}^{\prime}}\right){v}^{\prime},{n}^{\prime},{m}^{\prime})\\ =& ({m}^{\prime},0,v,0,{m}^{\prime}).\end{array}

(10)

By (9) and (10), we obtain {n}^{\prime}=0, and hence v{v}^{\prime}v=v and {v}^{\prime}v{v}^{\prime}=v. So, *A* is coregular.

Conversely, let *A* be a coregular monoid and (m,n,v,n,m)\in \mathcal{L}. Then there exists an element {v}^{\prime}\in A with v{v}^{\prime}v=v and {v}^{\prime}v{v}^{\prime}=v. Therefore, for (m,n,{v}^{\prime},n,m)\in \mathcal{L}, we get

Hence, \mathcal{L}\le {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) is a coregular monoid, as desired. □

In the final theorem, we consider strongly *π*-inverse property.

**Theorem 4** {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) *is strongly* *π*-*inverse if and only if* *A* *is regular and the idempotents in* *A* *commute*.

*Proof* We will follow the same format as in the proof of Theorem 2. So, let us suppose that {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) is a strongly *π*-inverse monoid, and let a\in A. Then, for (0,0,a,1,0)\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u), there is an element r\in \mathbb{N} with {(0,0,a,1,0)}^{r}\in Reg{\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u). It is easily seen that {(0,0,a,1,0)}^{r}=(0,0,a{(a\beta )}^{r-1},r,0). Moreover, there is an element {a}^{\prime}\in A such that (0,r,{a}^{\prime},0,0) is an inverse of (0,0,a{(a\beta )}^{r-1},r,0) by Lemma 3. From here, we have

\begin{array}{rcl}(0,0,a{(a\beta )}^{r-1},r,0)& =& (0,0,a{(a\beta )}^{r-1},r,0)(0,r,{a}^{\prime},0,0)(0,0,a{(a\beta )}^{r-1},r,0)\\ =& (0,0,a{(a\beta )}^{r-1}{a}^{\prime}a{(a\beta )}^{r-1},r,0).\end{array}

This actually shows that

a{(a\beta )}^{r-1}=a{(a\beta )}^{r-1}{a}^{\prime}a{(a\beta )}^{r-1}.

(11)

At the same time, since *aβ* is in the {\mathcal{H}}^{\ast}-class of the {1}_{A}, there exists an inverse element {({(a\beta )}^{r-1})}^{-1}. Thus, by (11), we get a=a{(a\beta )}^{r-1}{a}^{\prime}a, in other words, a\in RegA. Hence, *A* is regular. Now, let us show that the elements in E(A) are commutative to conclude the necessity part of the proof. To do that, consider any two elements {v}_{1} and {v}_{2} in E(A). Thus, (0,0,{v}_{1},0,0),(0,0,{v}_{2},0,0)\in E({\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)) (by Lemma 2) and we have

\begin{array}{rcl}(0,0,{v}_{1}{v}_{2},0,0)& =& (0,0,{v}_{1},0,0)(0,0,{v}_{2},0,0)\\ =& (0,0,{v}_{2},0,0)(0,0,{v}_{1},0,0)=(0,0,{v}_{2}{v}_{1},0,0).\end{array}

So, {v}_{1}{v}_{2}={v}_{2}{v}_{1}, as required.

Conversely, let us suppose that *A* is regular. Then {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u) is regular, where *π*-regular by Corollary 2. Now we need to show that the elements in E({\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)) commute. To do that, let us take (m,n,e,n,m),({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})\in E({\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)), and thus e{e}^{\prime}={e}^{\prime}e by Lemma 2. Now, by considering the multiplication (m,n,e,n,m)({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime}) as defined in (4), we have the following cases.

Case (i): If m={m}^{\prime}, then we get

(m,n,e,n,m)({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})=(m,{d}^{\prime},\left(e{\beta}^{{d}^{\prime}-n}\right)\left({e}^{\prime}{\beta}^{{d}^{\prime}-{n}^{\prime}}\right),{d}^{\prime},{m}^{\prime})

and

({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})(m,n,e,n,m)=({m}^{\prime},{d}^{\prime},\left({e}^{\prime}{\beta}^{{d}^{\prime}-{n}^{\prime}}\right)\left(e{\beta}^{{d}^{\prime}-n}\right),{d}^{\prime},{m}^{\prime}),

respectively, where {d}^{\prime}=max(n,{n}^{\prime}). Since e,{e}^{\prime}\in E(A), we deduce that both {e}^{\prime}{\beta}^{{d}^{\prime}-{n}^{\prime}} and e{\beta}^{{d}^{\prime}-n} are the elements of E(A), in other words,

\left({e}^{\prime}{\beta}^{{d}^{\prime}-{n}^{\prime}}\right)\left(e{\beta}^{{d}^{\prime}-n}\right)=\left(e{\beta}^{{d}^{\prime}-n}\right)\left({e}^{\prime}{\beta}^{{d}^{\prime}-{n}^{\prime}}\right).

Thus, (m,n,e,n,m)({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})=({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})(m,n,e,n,m).

Case (ii): If m<{m}^{\prime} or m>{m}^{\prime}, then we get

or

respectively. Since (({u}^{-n}(e\gamma ){u}^{n}){\gamma}^{{m}^{\prime}-m-1}){\beta}^{{n}^{\prime}},(({u}^{-{n}^{\prime}}({e}^{\prime}\gamma ){u}^{{n}^{\prime}}){\gamma}^{m-{m}^{\prime}-1}){\beta}^{n}\in E(A), we clearly obtain (m,n,e,n,m)({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})=({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})(m,n,e,n,m).

Hence, the result. □