Inverse of a product in semigroupExistence of universal enveloping inverse semigroup (similar to “Grothendieck group”)Idempotent separating congruence of an inverse semigroup.Is the universal inverse semigroup of a commutative semigroup an embedding?inverse on topological semigroupSemigroup isomorphismExample of an inverse semigroupFinitely generated Clifford semigroupEvery inverse semigroup is a groupProperties of regular semigroupExample of a locally inverse semigroup which isn't a generalized inverse semigroup
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Inverse of a product in semigroup
Existence of universal enveloping inverse semigroup (similar to “Grothendieck group”)Idempotent separating congruence of an inverse semigroup.Is the universal inverse semigroup of a commutative semigroup an embedding?inverse on topological semigroupSemigroup isomorphismExample of an inverse semigroupFinitely generated Clifford semigroupEvery inverse semigroup is a groupProperties of regular semigroupExample of a locally inverse semigroup which isn't a generalized inverse semigroup
$begingroup$
Disclaimer. The following might end up being a stupid question
Let $S$ be a regular semigroup i.e for every $sin S$ we can write $s=sas$ for some $ain S$ (called an inverse of $s$).
Do we know how to compute explicitly an inverse of a product? That is given $s=sas$ and $t=tbt$, can we tell what an inverse of $st$ would be?
Some attempts go as follows
$$st(btsa)st = s(tbt)(sas)t = sstt =st Leftarrow s,tin E(S) $$
We could also do
$$ st(bca)st = (sas)t(bca)s(tbt) = s(astb)c(astb)t = s(astb)t = (sas)(tbt) = st $$
We've assumed $cin V(astb)$, which is provided by regularity, and the arithmetic checks out, but I'm not entirely convinced by this.
abstract-algebra semigroups
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
$endgroup$
|
show 3 more comments
$begingroup$
Disclaimer. The following might end up being a stupid question
Let $S$ be a regular semigroup i.e for every $sin S$ we can write $s=sas$ for some $ain S$ (called an inverse of $s$).
Do we know how to compute explicitly an inverse of a product? That is given $s=sas$ and $t=tbt$, can we tell what an inverse of $st$ would be?
Some attempts go as follows
$$st(btsa)st = s(tbt)(sas)t = sstt =st Leftarrow s,tin E(S) $$
We could also do
$$ st(bca)st = (sas)t(bca)s(tbt) = s(astb)c(astb)t = s(astb)t = (sas)(tbt) = st $$
We've assumed $cin V(astb)$, which is provided by regularity, and the arithmetic checks out, but I'm not entirely convinced by this.
abstract-algebra semigroups
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
$endgroup$
$begingroup$
In the first one, why should $s$ and $t$ be in $E(S)$ (assuming that means the set of idempotents of $S$)? Or are you pointing out (correctly) that that's only valid for $s$ and $t$ in $E(S)$? (We know that $sa, as, tb, bt in E(S)$, but not $s$ or $t$). But the second proof is indeed correct. Any particular reason you're not convinced by it?
$endgroup$
– M. Vinay
Mar 22 at 12:19
$begingroup$
@M.Vinay In the first one if we assumed $s,t$ were idempotents, then we'd have our inverse. Notice that in that case, there is no magical $c$. For the second one I just tried to fit something between $st(ldots)st$ and it checked out, but it now relies on some new $c$. I get the feeling like I'm cheating, because the goal is to find an explicit inverse, however, $c$ merely invokes regularity, we don't know what it looks like.
$endgroup$
– Alvin Lepik
Mar 22 at 12:23
1
$begingroup$
Well the reverse is true. In the first case you've assumed special properties for $s$ and $t$, which may not hold [to give a trivial and extreme example: a group is a regular semigroup, but the only idempotent element is the identity]. In the second case, you know that such a $c$ exists because of regularity, so there's no cheating there. In fact, we should somewhat expect that the generalised inverse of $st$ might depend on some completely new element other than $s$ and $t$ — i.e., we should expect the involvement of some such $c$.
$endgroup$
– M. Vinay
Mar 22 at 12:27
$begingroup$
Sorry, I was thinking about this again, and I see that I'd lost track of the true meaning of the problem when we discussed this two days back. You're right, since we don't know what $c$ is, knowing a generalised inverse of $st$ in terms of the unknown $c$ is as good [bad] as saying "Let $d$ be the generalised inverse of $st$".
$endgroup$
– M. Vinay
Mar 24 at 13:05
1
$begingroup$
One lead I'm investigating right now is using reflexive generalised inverses of $s$ and $t$. That is, $q$ such $sqs = s$ and $qsq = q$ (and similarly one for $t$). If $a$ is a generalised inverse of $s$, then $q = asa$ is a reflexive generalised inverse of $a$.
$endgroup$
– M. Vinay
Mar 24 at 14:17
|
show 3 more comments
$begingroup$
Disclaimer. The following might end up being a stupid question
Let $S$ be a regular semigroup i.e for every $sin S$ we can write $s=sas$ for some $ain S$ (called an inverse of $s$).
Do we know how to compute explicitly an inverse of a product? That is given $s=sas$ and $t=tbt$, can we tell what an inverse of $st$ would be?
Some attempts go as follows
$$st(btsa)st = s(tbt)(sas)t = sstt =st Leftarrow s,tin E(S) $$
We could also do
$$ st(bca)st = (sas)t(bca)s(tbt) = s(astb)c(astb)t = s(astb)t = (sas)(tbt) = st $$
We've assumed $cin V(astb)$, which is provided by regularity, and the arithmetic checks out, but I'm not entirely convinced by this.
abstract-algebra semigroups
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
$endgroup$
Disclaimer. The following might end up being a stupid question
Let $S$ be a regular semigroup i.e for every $sin S$ we can write $s=sas$ for some $ain S$ (called an inverse of $s$).
Do we know how to compute explicitly an inverse of a product? That is given $s=sas$ and $t=tbt$, can we tell what an inverse of $st$ would be?
Some attempts go as follows
$$st(btsa)st = s(tbt)(sas)t = sstt =st Leftarrow s,tin E(S) $$
We could also do
$$ st(bca)st = (sas)t(bca)s(tbt) = s(astb)c(astb)t = s(astb)t = (sas)(tbt) = st $$
We've assumed $cin V(astb)$, which is provided by regularity, and the arithmetic checks out, but I'm not entirely convinced by this.
abstract-algebra semigroups
abstract-algebra semigroups
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
asked Mar 22 at 10:53
Alvin LepikAlvin Lepik
2,8261924
2,8261924
$begingroup$
In the first one, why should $s$ and $t$ be in $E(S)$ (assuming that means the set of idempotents of $S$)? Or are you pointing out (correctly) that that's only valid for $s$ and $t$ in $E(S)$? (We know that $sa, as, tb, bt in E(S)$, but not $s$ or $t$). But the second proof is indeed correct. Any particular reason you're not convinced by it?
$endgroup$
– M. Vinay
Mar 22 at 12:19
$begingroup$
@M.Vinay In the first one if we assumed $s,t$ were idempotents, then we'd have our inverse. Notice that in that case, there is no magical $c$. For the second one I just tried to fit something between $st(ldots)st$ and it checked out, but it now relies on some new $c$. I get the feeling like I'm cheating, because the goal is to find an explicit inverse, however, $c$ merely invokes regularity, we don't know what it looks like.
$endgroup$
– Alvin Lepik
Mar 22 at 12:23
1
$begingroup$
Well the reverse is true. In the first case you've assumed special properties for $s$ and $t$, which may not hold [to give a trivial and extreme example: a group is a regular semigroup, but the only idempotent element is the identity]. In the second case, you know that such a $c$ exists because of regularity, so there's no cheating there. In fact, we should somewhat expect that the generalised inverse of $st$ might depend on some completely new element other than $s$ and $t$ — i.e., we should expect the involvement of some such $c$.
$endgroup$
– M. Vinay
Mar 22 at 12:27
$begingroup$
Sorry, I was thinking about this again, and I see that I'd lost track of the true meaning of the problem when we discussed this two days back. You're right, since we don't know what $c$ is, knowing a generalised inverse of $st$ in terms of the unknown $c$ is as good [bad] as saying "Let $d$ be the generalised inverse of $st$".
$endgroup$
– M. Vinay
Mar 24 at 13:05
1
$begingroup$
One lead I'm investigating right now is using reflexive generalised inverses of $s$ and $t$. That is, $q$ such $sqs = s$ and $qsq = q$ (and similarly one for $t$). If $a$ is a generalised inverse of $s$, then $q = asa$ is a reflexive generalised inverse of $a$.
$endgroup$
– M. Vinay
Mar 24 at 14:17
|
show 3 more comments
$begingroup$
In the first one, why should $s$ and $t$ be in $E(S)$ (assuming that means the set of idempotents of $S$)? Or are you pointing out (correctly) that that's only valid for $s$ and $t$ in $E(S)$? (We know that $sa, as, tb, bt in E(S)$, but not $s$ or $t$). But the second proof is indeed correct. Any particular reason you're not convinced by it?
$endgroup$
– M. Vinay
Mar 22 at 12:19
$begingroup$
@M.Vinay In the first one if we assumed $s,t$ were idempotents, then we'd have our inverse. Notice that in that case, there is no magical $c$. For the second one I just tried to fit something between $st(ldots)st$ and it checked out, but it now relies on some new $c$. I get the feeling like I'm cheating, because the goal is to find an explicit inverse, however, $c$ merely invokes regularity, we don't know what it looks like.
$endgroup$
– Alvin Lepik
Mar 22 at 12:23
1
$begingroup$
Well the reverse is true. In the first case you've assumed special properties for $s$ and $t$, which may not hold [to give a trivial and extreme example: a group is a regular semigroup, but the only idempotent element is the identity]. In the second case, you know that such a $c$ exists because of regularity, so there's no cheating there. In fact, we should somewhat expect that the generalised inverse of $st$ might depend on some completely new element other than $s$ and $t$ — i.e., we should expect the involvement of some such $c$.
$endgroup$
– M. Vinay
Mar 22 at 12:27
$begingroup$
Sorry, I was thinking about this again, and I see that I'd lost track of the true meaning of the problem when we discussed this two days back. You're right, since we don't know what $c$ is, knowing a generalised inverse of $st$ in terms of the unknown $c$ is as good [bad] as saying "Let $d$ be the generalised inverse of $st$".
$endgroup$
– M. Vinay
Mar 24 at 13:05
1
$begingroup$
One lead I'm investigating right now is using reflexive generalised inverses of $s$ and $t$. That is, $q$ such $sqs = s$ and $qsq = q$ (and similarly one for $t$). If $a$ is a generalised inverse of $s$, then $q = asa$ is a reflexive generalised inverse of $a$.
$endgroup$
– M. Vinay
Mar 24 at 14:17
$begingroup$
In the first one, why should $s$ and $t$ be in $E(S)$ (assuming that means the set of idempotents of $S$)? Or are you pointing out (correctly) that that's only valid for $s$ and $t$ in $E(S)$? (We know that $sa, as, tb, bt in E(S)$, but not $s$ or $t$). But the second proof is indeed correct. Any particular reason you're not convinced by it?
$endgroup$
– M. Vinay
Mar 22 at 12:19
$begingroup$
In the first one, why should $s$ and $t$ be in $E(S)$ (assuming that means the set of idempotents of $S$)? Or are you pointing out (correctly) that that's only valid for $s$ and $t$ in $E(S)$? (We know that $sa, as, tb, bt in E(S)$, but not $s$ or $t$). But the second proof is indeed correct. Any particular reason you're not convinced by it?
$endgroup$
– M. Vinay
Mar 22 at 12:19
$begingroup$
@M.Vinay In the first one if we assumed $s,t$ were idempotents, then we'd have our inverse. Notice that in that case, there is no magical $c$. For the second one I just tried to fit something between $st(ldots)st$ and it checked out, but it now relies on some new $c$. I get the feeling like I'm cheating, because the goal is to find an explicit inverse, however, $c$ merely invokes regularity, we don't know what it looks like.
$endgroup$
– Alvin Lepik
Mar 22 at 12:23
$begingroup$
@M.Vinay In the first one if we assumed $s,t$ were idempotents, then we'd have our inverse. Notice that in that case, there is no magical $c$. For the second one I just tried to fit something between $st(ldots)st$ and it checked out, but it now relies on some new $c$. I get the feeling like I'm cheating, because the goal is to find an explicit inverse, however, $c$ merely invokes regularity, we don't know what it looks like.
$endgroup$
– Alvin Lepik
Mar 22 at 12:23
1
1
$begingroup$
Well the reverse is true. In the first case you've assumed special properties for $s$ and $t$, which may not hold [to give a trivial and extreme example: a group is a regular semigroup, but the only idempotent element is the identity]. In the second case, you know that such a $c$ exists because of regularity, so there's no cheating there. In fact, we should somewhat expect that the generalised inverse of $st$ might depend on some completely new element other than $s$ and $t$ — i.e., we should expect the involvement of some such $c$.
$endgroup$
– M. Vinay
Mar 22 at 12:27
$begingroup$
Well the reverse is true. In the first case you've assumed special properties for $s$ and $t$, which may not hold [to give a trivial and extreme example: a group is a regular semigroup, but the only idempotent element is the identity]. In the second case, you know that such a $c$ exists because of regularity, so there's no cheating there. In fact, we should somewhat expect that the generalised inverse of $st$ might depend on some completely new element other than $s$ and $t$ — i.e., we should expect the involvement of some such $c$.
$endgroup$
– M. Vinay
Mar 22 at 12:27
$begingroup$
Sorry, I was thinking about this again, and I see that I'd lost track of the true meaning of the problem when we discussed this two days back. You're right, since we don't know what $c$ is, knowing a generalised inverse of $st$ in terms of the unknown $c$ is as good [bad] as saying "Let $d$ be the generalised inverse of $st$".
$endgroup$
– M. Vinay
Mar 24 at 13:05
$begingroup$
Sorry, I was thinking about this again, and I see that I'd lost track of the true meaning of the problem when we discussed this two days back. You're right, since we don't know what $c$ is, knowing a generalised inverse of $st$ in terms of the unknown $c$ is as good [bad] as saying "Let $d$ be the generalised inverse of $st$".
$endgroup$
– M. Vinay
Mar 24 at 13:05
1
1
$begingroup$
One lead I'm investigating right now is using reflexive generalised inverses of $s$ and $t$. That is, $q$ such $sqs = s$ and $qsq = q$ (and similarly one for $t$). If $a$ is a generalised inverse of $s$, then $q = asa$ is a reflexive generalised inverse of $a$.
$endgroup$
– M. Vinay
Mar 24 at 14:17
$begingroup$
One lead I'm investigating right now is using reflexive generalised inverses of $s$ and $t$. That is, $q$ such $sqs = s$ and $qsq = q$ (and similarly one for $t$). If $a$ is a generalised inverse of $s$, then $q = asa$ is a reflexive generalised inverse of $a$.
$endgroup$
– M. Vinay
Mar 24 at 14:17
|
show 3 more comments
0
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$begingroup$
In the first one, why should $s$ and $t$ be in $E(S)$ (assuming that means the set of idempotents of $S$)? Or are you pointing out (correctly) that that's only valid for $s$ and $t$ in $E(S)$? (We know that $sa, as, tb, bt in E(S)$, but not $s$ or $t$). But the second proof is indeed correct. Any particular reason you're not convinced by it?
$endgroup$
– M. Vinay
Mar 22 at 12:19
$begingroup$
@M.Vinay In the first one if we assumed $s,t$ were idempotents, then we'd have our inverse. Notice that in that case, there is no magical $c$. For the second one I just tried to fit something between $st(ldots)st$ and it checked out, but it now relies on some new $c$. I get the feeling like I'm cheating, because the goal is to find an explicit inverse, however, $c$ merely invokes regularity, we don't know what it looks like.
$endgroup$
– Alvin Lepik
Mar 22 at 12:23
1
$begingroup$
Well the reverse is true. In the first case you've assumed special properties for $s$ and $t$, which may not hold [to give a trivial and extreme example: a group is a regular semigroup, but the only idempotent element is the identity]. In the second case, you know that such a $c$ exists because of regularity, so there's no cheating there. In fact, we should somewhat expect that the generalised inverse of $st$ might depend on some completely new element other than $s$ and $t$ — i.e., we should expect the involvement of some such $c$.
$endgroup$
– M. Vinay
Mar 22 at 12:27
$begingroup$
Sorry, I was thinking about this again, and I see that I'd lost track of the true meaning of the problem when we discussed this two days back. You're right, since we don't know what $c$ is, knowing a generalised inverse of $st$ in terms of the unknown $c$ is as good [bad] as saying "Let $d$ be the generalised inverse of $st$".
$endgroup$
– M. Vinay
Mar 24 at 13:05
1
$begingroup$
One lead I'm investigating right now is using reflexive generalised inverses of $s$ and $t$. That is, $q$ such $sqs = s$ and $qsq = q$ (and similarly one for $t$). If $a$ is a generalised inverse of $s$, then $q = asa$ is a reflexive generalised inverse of $a$.
$endgroup$
– M. Vinay
Mar 24 at 14:17