Suppose that A is an arbitrary monoid having and as the - and ℋ- classes containing the identity element of A. Moreover, let us assume that β and γ are morphisms from A into and, for an element u in , let be the inner automorphism of defined by such that .
Now one can consider the set into a semigroup with a multiplication
where and , are interpreted as the identity map of A, and also is interpreted as the identity of A. In [8], Yu Shung and Li-Min Wang showed that S is a monoid with the identity . In fact, this new monoid is denoted by 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 , then is an idempotent if and only if , and v is idempotent.
Lemma 3 If , then has an inverse
if and only if is an inverse of v in A while , , and .
Then we have an immediate consequence as in the following.
Corollary 2 Let A be a monoid. Then 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 , which can be held coregularity if A satisfies particular properties.
Our first observation is the following.
Lemma 4 The set is a submonoid of .
Proof By considering the multiplication in (4), the proof can be seen easily. □
It turns out that all coregular elements in belong to the submonoid ℒ .
Lemma 5 Let . If is coregular then and .
Proof Let be a coregular element. Then there exists an element such that
Let and . Then we have
for some , where and . This implies that
(5)
and
By (6), we have . Applying this in (5), we get and thus . Further, we have
for some , where and . This implies that , ,
(7)
and
By writing the equality (8) in (7), we get . Together with , we obtain . By assuming , we also get .
Now let . Then it is easy to verify that and . If , we can easily see that or . This shows that if or . Hence, or is not possible. □
Then we have the following result.
Theorem 3 Let A be a monoid. Then the submonoid ℒ of is coregular if and only if A is coregular.
Proof Suppose that is coregular. For each in ℒ, there exists an element such that
(9)
and
(10)
By (9) and (10), we obtain , and hence and . So, A is coregular.
Conversely, let A be a coregular monoid and . Then there exists an element with and . Therefore, for , we get
Hence, is a coregular monoid, as desired. □
In the final theorem, we consider strongly π-inverse property.
Theorem 4 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 is a strongly π-inverse monoid, and let . Then, for , there is an element with . It is easily seen that . Moreover, there is an element such that is an inverse of by Lemma 3. From here, we have
This actually shows that
(11)
At the same time, since aβ is in the -class of the , there exists an inverse element . Thus, by (11), we get , in other words, . Hence, A is regular. Now, let us show that the elements in are commutative to conclude the necessity part of the proof. To do that, consider any two elements and in . Thus, (by Lemma 2) and we have
So, , as required.
Conversely, let us suppose that A is regular. Then is regular, where π-regular by Corollary 2. Now we need to show that the elements in commute. To do that, let us take , and thus by Lemma 2. Now, by considering the multiplication as defined in (4), we have the following cases.
Case (i): If , then we get
and
respectively, where . Since , we deduce that both and are the elements of , in other words,
Thus, .
Case (ii): If or , then we get
or
respectively. Since , we clearly obtain .
Hence, the result. □