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 , 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 .
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
By (6), we have . Applying this in (5), we get and thus . Further, we have
for some , where and . This implies that , ,
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
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
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
respectively, where . Since , we deduce that both and are the elements of , in other words,
Case (ii): If or , then we get
respectively. Since , we clearly obtain .
Hence, the result. □