Mixing
Show that a semigroup with $aS cup a = bS cup b$ and $Sa cup a = Sb cup b$ is a group
Clash Royale CLAN TAG#URR8PPP
up vote
10
down vote
favorite
Let $S$ be a semigroup such that for all $a,b in S$
$$
aS cup a = bS cup b quad textand quad
Sa cup a = Sb cup b.
$$
where $aS = as : s in S $ and similarly $Sa$. I want to show that $S$ is a group.
The above conditions imply
$$
a in bS
$$
for $a ne b$, hence $S setminusb subseteq bS$ for all $b in S$. If $S$ contains an idempotent $e$, then as $e in eS$ we have $eS = S = Se$ and $e$ is the identity. With an identity element it is easy to see that $aS = S$ and $S = Sa$ for every $a in S$, hence that $S$ is a group.
So, if I could show that $S$ contains an idempotent then it follows that $S$ is a group. But here I am stuck, so could anyone provide a hint how to proceed?
By the way, this is an exercise from J. Howie Fundamentals of Semigroup Theory (page 61).
abstract-algebra group-theory semigroups
edited yesterday
darij grinberg
9,19132959
asked Jul 31 at 12:48
StefanH
7,81941858
 |Â
show 7 more comments
up vote
10
down vote
favorite
Let $S$ be a semigroup such that for all $a,b in S$
$$
aS cup a = bS cup b quad textand quad
Sa cup a = Sb cup b.
$$
where $aS = as : s in S $ and similarly $Sa$. I want to show that $S$ is a group.
The above conditions imply
$$
a in bS
$$
for $a ne b$, hence $S setminusb subseteq bS$ for all $b in S$. If $S$ contains an idempotent $e$, then as $e in eS$ we have $eS = S = Se$ and $e$ is the identity. With an identity element it is easy to see that $aS = S$ and $S = Sa$ for every $a in S$, hence that $S$ is a group.
So, if I could show that $S$ contains an idempotent then it follows that $S$ is a group. But here I am stuck, so could anyone provide a hint how to proceed?
By the way, this is an exercise from J. Howie Fundamentals of Semigroup Theory (page 61).
abstract-algebra group-theory semigroups
edited yesterday
darij grinberg
9,19132959
asked Jul 31 at 12:48
StefanH
7,81941858
Why is this downvoted?
– greedoid
Jul 31 at 12:51
Before downvoting please leave a comment what I could improve (or add a comment to your downvote at least)... I dont know either why this was downvoted!
– StefanH
Jul 31 at 12:55
1
@DerekHolt That being said, if the purpose of actively downvoting is, as far as I can tell, to make this site a better place, then leaving a comment is the only way to teach the poster what they can do better. So downvoting without a comment serves little purpose except possibly scare new users away from the site (which, if I'm allowed to be cynical, might be a goal to some).
– Arthur
Jul 31 at 13:33
1
I don't know much about semigroups, but i had a quick look at the book. Lemma 2.2.1 seems to say that multiplying by $s$ on the right is a bijection of $S$, and similarly on the left. So, for $a in S$, there exists $e$ with $ae=a$, and then $ae^2=ae$, so $e^2=e$, and $e$ is an idempotent.
– Derek Holt
Jul 31 at 14:01
1
@DerekHolt Okay, yes by my assummption there is just one $mathcal L$-class! So this is fine, and also by this cancellability follows.
– StefanH
Jul 31 at 14:06
 |Â
show 7 more comments
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Let $S$ be a semigroup such that for all $a,b in S$
$$
aS cup a = bS cup b quad textand quad
Sa cup a = Sb cup b.
$$
where $aS = as : s in S $ and similarly $Sa$. I want to show that $S$ is a group.
The above conditions imply
$$
a in bS
$$
for $a ne b$, hence $S setminusb subseteq bS$ for all $b in S$. If $S$ contains an idempotent $e$, then as $e in eS$ we have $eS = S = Se$ and $e$ is the identity. With an identity element it is easy to see that $aS = S$ and $S = Sa$ for every $a in S$, hence that $S$ is a group.
So, if I could show that $S$ contains an idempotent then it follows that $S$ is a group. But here I am stuck, so could anyone provide a hint how to proceed?
By the way, this is an exercise from J. Howie Fundamentals of Semigroup Theory (page 61).
abstract-algebra group-theory semigroups
edited yesterday
darij grinberg
9,19132959
asked Jul 31 at 12:48
StefanH
7,81941858
Let $S$ be a semigroup such that for all $a,b in S$
$$
aS cup a = bS cup b quad textand quad
Sa cup a = Sb cup b.
$$
where $aS = as : s in S $ and similarly $Sa$. I want to show that $S$ is a group.
The above conditions imply
$$
a in bS
$$
for $a ne b$, hence $S setminusb subseteq bS$ for all $b in S$. If $S$ contains an idempotent $e$, then as $e in eS$ we have $eS = S = Se$ and $e$ is the identity. With an identity element it is easy to see that $aS = S$ and $S = Sa$ for every $a in S$, hence that $S$ is a group.
So, if I could show that $S$ contains an idempotent then it follows that $S$ is a group. But here I am stuck, so could anyone provide a hint how to proceed?
By the way, this is an exercise from J. Howie Fundamentals of Semigroup Theory (page 61).
abstract-algebra group-theory semigroups
edited yesterday
darij grinberg
9,19132959
asked Jul 31 at 12:48
StefanH
7,81941858
edited yesterday
darij grinberg
9,19132959
edited yesterday
darij grinberg
9,19132959
edited yesterday
darij grinberg
9,19132959
9,19132959
asked Jul 31 at 12:48
StefanH
7,81941858
asked Jul 31 at 12:48
StefanH
7,81941858
asked Jul 31 at 12:48
StefanH
7,81941858
7,81941858
Why is this downvoted?
– greedoid
Jul 31 at 12:51
Before downvoting please leave a comment what I could improve (or add a comment to your downvote at least)... I dont know either why this was downvoted!
– StefanH
Jul 31 at 12:55
1
@DerekHolt That being said, if the purpose of actively downvoting is, as far as I can tell, to make this site a better place, then leaving a comment is the only way to teach the poster what they can do better. So downvoting without a comment serves little purpose except possibly scare new users away from the site (which, if I'm allowed to be cynical, might be a goal to some).
– Arthur
Jul 31 at 13:33
1
I don't know much about semigroups, but i had a quick look at the book. Lemma 2.2.1 seems to say that multiplying by $s$ on the right is a bijection of $S$, and similarly on the left. So, for $a in S$, there exists $e$ with $ae=a$, and then $ae^2=ae$, so $e^2=e$, and $e$ is an idempotent.
– Derek Holt
Jul 31 at 14:01
1
@DerekHolt Okay, yes by my assummption there is just one $mathcal L$-class! So this is fine, and also by this cancellability follows.
– StefanH
Jul 31 at 14:06
 |Â
show 7 more comments
Why is this downvoted?
– greedoid
Jul 31 at 12:51
Before downvoting please leave a comment what I could improve (or add a comment to your downvote at least)... I dont know either why this was downvoted!
– StefanH
Jul 31 at 12:55
1
@DerekHolt That being said, if the purpose of actively downvoting is, as far as I can tell, to make this site a better place, then leaving a comment is the only way to teach the poster what they can do better. So downvoting without a comment serves little purpose except possibly scare new users away from the site (which, if I'm allowed to be cynical, might be a goal to some).
– Arthur
Jul 31 at 13:33
1
I don't know much about semigroups, but i had a quick look at the book. Lemma 2.2.1 seems to say that multiplying by $s$ on the right is a bijection of $S$, and similarly on the left. So, for $a in S$, there exists $e$ with $ae=a$, and then $ae^2=ae$, so $e^2=e$, and $e$ is an idempotent.
– Derek Holt
Jul 31 at 14:01
1
@DerekHolt Okay, yes by my assummption there is just one $mathcal L$-class! So this is fine, and also by this cancellability follows.
– StefanH
Jul 31 at 14:06
Why is this downvoted?
– greedoid
Jul 31 at 12:51
Why is this downvoted?
– greedoid
Jul 31 at 12:51
Before downvoting please leave a comment what I could improve (or add a comment to your downvote at least)... I dont know either why this was downvoted!
– StefanH
Jul 31 at 12:55
Before downvoting please leave a comment what I could improve (or add a comment to your downvote at least)... I dont know either why this was downvoted!
– StefanH
Jul 31 at 12:55
1
1
@DerekHolt That being said, if the purpose of actively downvoting is, as far as I can tell, to make this site a better place, then leaving a comment is the only way to teach the poster what they can do better. So downvoting without a comment serves little purpose except possibly scare new users away from the site (which, if I'm allowed to be cynical, might be a goal to some).
– Arthur
Jul 31 at 13:33
@DerekHolt That being said, if the purpose of actively downvoting is, as far as I can tell, to make this site a better place, then leaving a comment is the only way to teach the poster what they can do better. So downvoting without a comment serves little purpose except possibly scare new users away from the site (which, if I'm allowed to be cynical, might be a goal to some).
– Arthur
Jul 31 at 13:33
1
1
I don't know much about semigroups, but i had a quick look at the book. Lemma 2.2.1 seems to say that multiplying by $s$ on the right is a bijection of $S$, and similarly on the left. So, for $a in S$, there exists $e$ with $ae=a$, and then $ae^2=ae$, so $e^2=e$, and $e$ is an idempotent.
– Derek Holt
Jul 31 at 14:01
I don't know much about semigroups, but i had a quick look at the book. Lemma 2.2.1 seems to say that multiplying by $s$ on the right is a bijection of $S$, and similarly on the left. So, for $a in S$, there exists $e$ with $ae=a$, and then $ae^2=ae$, so $e^2=e$, and $e$ is an idempotent.
– Derek Holt
Jul 31 at 14:01
1
1
@DerekHolt Okay, yes by my assummption there is just one $mathcal L$-class! So this is fine, and also by this cancellability follows.
– StefanH
Jul 31 at 14:06
@DerekHolt Okay, yes by my assummption there is just one $mathcal L$-class! So this is fine, and also by this cancellability follows.
– StefanH
Jul 31 at 14:06
 |Â
show 7 more comments
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
As pointed out in the comments by DerekHolt, the key is Green's Lemma (as cited in the mentioned book on page 49). To formulate it, we start with an arbitrary semigroup $S$, and we define equivalence relations $mathcal R$ and $mathcal L$ on $S$ by setting
beginalign*
a mathcal R b & :Leftrightarrow a cup aS = b cup bS, \
a mathcal L b & :Leftrightarrow a cup Sa = b cup Sb
endalign*
for all $a, b in S$.
These are part of what is known as Green's relations.
Then we have:
(Green's Lemma) If $a ne b$ and $a mathcal R b$ with $as = b, a = bt$, then the maps
$$
x mapsto xs
quad mboxand quad
x mapsto xt
$$
are mutually inverse bijections between the $mathcal L$-classes of $a$ and $b$, which preserve $mathcal R$-classes. A similar (dual) relation holds if $amathcal L b$.
Proof: Its easy to see that $mathcal L$ is a left congruence and $mathcal R$ is a right congruence. Hence if $xmathcal L a$ then $xs mathcal L as$, so $xs mathcal L b$ because $b = as$. Similarly, right multiplication by $t$ maps the $mathcal L$-class of $b$ into the one of $a$. So the two maps between the $mathcal L$-classes are well-defined. Also, if $x = ua$ then
$$
xst = uast = ubt = ua = x
$$
and similarly $yts = y$ for all $y mathcal L b$; hence they are mutually inverse on the corresponding $mathcal L$-classes. Also, as $mathcal R$ is a right congruence, the images of $mathcal R$-related elements stay $mathcal R$-related, but further as $x = (xs)t$ for $x mathcal L a$ we have $x mathcal R xs$ and so the mapping stays in the same $mathcal R$-class for all elements $mathcal L$-equivalent to $a$. $square$
The assumption of the question gives that we just have a single $mathcal R$-class and a single $mathcal L$-class. So pick arbitrary $a ne b$ and $as = b$ and $a = bt$, then for every $x in S$ by the above we have
$$
xst = x
$$
giving that $e := st$ is idempotent (set $x := e$ above) and a right-identity. Further as the equations $ux = e$ and $xu = e$ are solvable for each $u in S$ it must be a group.
Remark 1: With further theory from the book it is shown that every non-empty intersection of an $mathcal L$-class and an $mathcal R$-class is a group.
Remark 2: In the paper Intersections of maximal ideals in semigroups by P. Grillet I found
An ideal $I$ is maximal iff $S - I$ is a $mathcal J$-class.
And the same proof works in case of right-ideals and $mathcal R$-classes and left-ideals and $mathcal L$-classes. So if $aS = S setminus a$, then as $aS$ is a right ideal it is maximal and $a$ would be a single $mathcal R$-class, which is excluded. Hence $aS = S$ for all $a in S$. Similar $Sa = S$ for all $a in S$ so that $S$ is a group. And the result follows without Green's lemma.
edited yesterday
darij grinberg
9,19132959
answered Jul 31 at 15:46
StefanH
7,81941858
add a comment |Â
up vote
1
down vote
Your question is indeed related to Green's relations. The hypothesis tells you that all the elements of your semigroup are $mathcalR$-equivalent and $mathcalL$-equivalent. Therefore, there are all $mathcalH$-equivalent. It now suffices to apply the following standard result (see for instance [1, Proposition 1.4]).
Let $H$ be an $mathcalH$-class of a semigroup. The following
conditions are equivalent:
- $H$ contains an idempotent,
- there exist $s$, $t in H$ such that $st in H$,
- $H$ is a group.
[1] P.A. Grillet, Semigroups, an introduction to the structure theory, Marcel Dekker, Inc., New York, 1995
answered yesterday
J.-E. Pin
16.6k21751
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
As pointed out in the comments by DerekHolt, the key is Green's Lemma (as cited in the mentioned book on page 49). To formulate it, we start with an arbitrary semigroup $S$, and we define equivalence relations $mathcal R$ and $mathcal L$ on $S$ by setting
beginalign*
a mathcal R b & :Leftrightarrow a cup aS = b cup bS, \
a mathcal L b & :Leftrightarrow a cup Sa = b cup Sb
endalign*
for all $a, b in S$.
These are part of what is known as Green's relations.
Then we have:
(Green's Lemma) If $a ne b$ and $a mathcal R b$ with $as = b, a = bt$, then the maps
$$
x mapsto xs
quad mboxand quad
x mapsto xt
$$
are mutually inverse bijections between the $mathcal L$-classes of $a$ and $b$, which preserve $mathcal R$-classes. A similar (dual) relation holds if $amathcal L b$.
Proof: Its easy to see that $mathcal L$ is a left congruence and $mathcal R$ is a right congruence. Hence if $xmathcal L a$ then $xs mathcal L as$, so $xs mathcal L b$ because $b = as$. Similarly, right multiplication by $t$ maps the $mathcal L$-class of $b$ into the one of $a$. So the two maps between the $mathcal L$-classes are well-defined. Also, if $x = ua$ then
$$
xst = uast = ubt = ua = x
$$
and similarly $yts = y$ for all $y mathcal L b$; hence they are mutually inverse on the corresponding $mathcal L$-classes. Also, as $mathcal R$ is a right congruence, the images of $mathcal R$-related elements stay $mathcal R$-related, but further as $x = (xs)t$ for $x mathcal L a$ we have $x mathcal R xs$ and so the mapping stays in the same $mathcal R$-class for all elements $mathcal L$-equivalent to $a$. $square$
The assumption of the question gives that we just have a single $mathcal R$-class and a single $mathcal L$-class. So pick arbitrary $a ne b$ and $as = b$ and $a = bt$, then for every $x in S$ by the above we have
$$
xst = x
$$
giving that $e := st$ is idempotent (set $x := e$ above) and a right-identity. Further as the equations $ux = e$ and $xu = e$ are solvable for each $u in S$ it must be a group.
Remark 1: With further theory from the book it is shown that every non-empty intersection of an $mathcal L$-class and an $mathcal R$-class is a group.
Remark 2: In the paper Intersections of maximal ideals in semigroups by P. Grillet I found
An ideal $I$ is maximal iff $S - I$ is a $mathcal J$-class.
And the same proof works in case of right-ideals and $mathcal R$-classes and left-ideals and $mathcal L$-classes. So if $aS = S setminus a$, then as $aS$ is a right ideal it is maximal and $a$ would be a single $mathcal R$-class, which is excluded. Hence $aS = S$ for all $a in S$. Similar $Sa = S$ for all $a in S$ so that $S$ is a group. And the result follows without Green's lemma.
edited yesterday
darij grinberg
9,19132959
answered Jul 31 at 15:46
StefanH
7,81941858
add a comment |Â
up vote
7
down vote
accepted
As pointed out in the comments by DerekHolt, the key is Green's Lemma (as cited in the mentioned book on page 49). To formulate it, we start with an arbitrary semigroup $S$, and we define equivalence relations $mathcal R$ and $mathcal L$ on $S$ by setting
beginalign*
a mathcal R b & :Leftrightarrow a cup aS = b cup bS, \
a mathcal L b & :Leftrightarrow a cup Sa = b cup Sb
endalign*
for all $a, b in S$.
These are part of what is known as Green's relations.
Then we have:
(Green's Lemma) If $a ne b$ and $a mathcal R b$ with $as = b, a = bt$, then the maps
$$
x mapsto xs
quad mboxand quad
x mapsto xt
$$
are mutually inverse bijections between the $mathcal L$-classes of $a$ and $b$, which preserve $mathcal R$-classes. A similar (dual) relation holds if $amathcal L b$.
Proof: Its easy to see that $mathcal L$ is a left congruence and $mathcal R$ is a right congruence. Hence if $xmathcal L a$ then $xs mathcal L as$, so $xs mathcal L b$ because $b = as$. Similarly, right multiplication by $t$ maps the $mathcal L$-class of $b$ into the one of $a$. So the two maps between the $mathcal L$-classes are well-defined. Also, if $x = ua$ then
$$
xst = uast = ubt = ua = x
$$
and similarly $yts = y$ for all $y mathcal L b$; hence they are mutually inverse on the corresponding $mathcal L$-classes. Also, as $mathcal R$ is a right congruence, the images of $mathcal R$-related elements stay $mathcal R$-related, but further as $x = (xs)t$ for $x mathcal L a$ we have $x mathcal R xs$ and so the mapping stays in the same $mathcal R$-class for all elements $mathcal L$-equivalent to $a$. $square$
The assumption of the question gives that we just have a single $mathcal R$-class and a single $mathcal L$-class. So pick arbitrary $a ne b$ and $as = b$ and $a = bt$, then for every $x in S$ by the above we have
$$
xst = x
$$
giving that $e := st$ is idempotent (set $x := e$ above) and a right-identity. Further as the equations $ux = e$ and $xu = e$ are solvable for each $u in S$ it must be a group.
Remark 1: With further theory from the book it is shown that every non-empty intersection of an $mathcal L$-class and an $mathcal R$-class is a group.
Remark 2: In the paper Intersections of maximal ideals in semigroups by P. Grillet I found
An ideal $I$ is maximal iff $S - I$ is a $mathcal J$-class.
And the same proof works in case of right-ideals and $mathcal R$-classes and left-ideals and $mathcal L$-classes. So if $aS = S setminus a$, then as $aS$ is a right ideal it is maximal and $a$ would be a single $mathcal R$-class, which is excluded. Hence $aS = S$ for all $a in S$. Similar $Sa = S$ for all $a in S$ so that $S$ is a group. And the result follows without Green's lemma.
edited yesterday
darij grinberg
9,19132959
answered Jul 31 at 15:46
StefanH
7,81941858
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
As pointed out in the comments by DerekHolt, the key is Green's Lemma (as cited in the mentioned book on page 49). To formulate it, we start with an arbitrary semigroup $S$, and we define equivalence relations $mathcal R$ and $mathcal L$ on $S$ by setting
beginalign*
a mathcal R b & :Leftrightarrow a cup aS = b cup bS, \
a mathcal L b & :Leftrightarrow a cup Sa = b cup Sb
endalign*
for all $a, b in S$.
These are part of what is known as Green's relations.
Then we have:
(Green's Lemma) If $a ne b$ and $a mathcal R b$ with $as = b, a = bt$, then the maps
$$
x mapsto xs
quad mboxand quad
x mapsto xt
$$
are mutually inverse bijections between the $mathcal L$-classes of $a$ and $b$, which preserve $mathcal R$-classes. A similar (dual) relation holds if $amathcal L b$.
Proof: Its easy to see that $mathcal L$ is a left congruence and $mathcal R$ is a right congruence. Hence if $xmathcal L a$ then $xs mathcal L as$, so $xs mathcal L b$ because $b = as$. Similarly, right multiplication by $t$ maps the $mathcal L$-class of $b$ into the one of $a$. So the two maps between the $mathcal L$-classes are well-defined. Also, if $x = ua$ then
$$
xst = uast = ubt = ua = x
$$
and similarly $yts = y$ for all $y mathcal L b$; hence they are mutually inverse on the corresponding $mathcal L$-classes. Also, as $mathcal R$ is a right congruence, the images of $mathcal R$-related elements stay $mathcal R$-related, but further as $x = (xs)t$ for $x mathcal L a$ we have $x mathcal R xs$ and so the mapping stays in the same $mathcal R$-class for all elements $mathcal L$-equivalent to $a$. $square$
The assumption of the question gives that we just have a single $mathcal R$-class and a single $mathcal L$-class. So pick arbitrary $a ne b$ and $as = b$ and $a = bt$, then for every $x in S$ by the above we have
$$
xst = x
$$
giving that $e := st$ is idempotent (set $x := e$ above) and a right-identity. Further as the equations $ux = e$ and $xu = e$ are solvable for each $u in S$ it must be a group.
Remark 1: With further theory from the book it is shown that every non-empty intersection of an $mathcal L$-class and an $mathcal R$-class is a group.
Remark 2: In the paper Intersections of maximal ideals in semigroups by P. Grillet I found
An ideal $I$ is maximal iff $S - I$ is a $mathcal J$-class.
And the same proof works in case of right-ideals and $mathcal R$-classes and left-ideals and $mathcal L$-classes. So if $aS = S setminus a$, then as $aS$ is a right ideal it is maximal and $a$ would be a single $mathcal R$-class, which is excluded. Hence $aS = S$ for all $a in S$. Similar $Sa = S$ for all $a in S$ so that $S$ is a group. And the result follows without Green's lemma.
edited yesterday
darij grinberg
9,19132959
answered Jul 31 at 15:46
StefanH
7,81941858
As pointed out in the comments by DerekHolt, the key is Green's Lemma (as cited in the mentioned book on page 49). To formulate it, we start with an arbitrary semigroup $S$, and we define equivalence relations $mathcal R$ and $mathcal L$ on $S$ by setting
beginalign*
a mathcal R b & :Leftrightarrow a cup aS = b cup bS, \
a mathcal L b & :Leftrightarrow a cup Sa = b cup Sb
endalign*
for all $a, b in S$.
These are part of what is known as Green's relations.
Then we have:
(Green's Lemma) If $a ne b$ and $a mathcal R b$ with $as = b, a = bt$, then the maps
$$
x mapsto xs
quad mboxand quad
x mapsto xt
$$
are mutually inverse bijections between the $mathcal L$-classes of $a$ and $b$, which preserve $mathcal R$-classes. A similar (dual) relation holds if $amathcal L b$.
Proof: Its easy to see that $mathcal L$ is a left congruence and $mathcal R$ is a right congruence. Hence if $xmathcal L a$ then $xs mathcal L as$, so $xs mathcal L b$ because $b = as$. Similarly, right multiplication by $t$ maps the $mathcal L$-class of $b$ into the one of $a$. So the two maps between the $mathcal L$-classes are well-defined. Also, if $x = ua$ then
$$
xst = uast = ubt = ua = x
$$
and similarly $yts = y$ for all $y mathcal L b$; hence they are mutually inverse on the corresponding $mathcal L$-classes. Also, as $mathcal R$ is a right congruence, the images of $mathcal R$-related elements stay $mathcal R$-related, but further as $x = (xs)t$ for $x mathcal L a$ we have $x mathcal R xs$ and so the mapping stays in the same $mathcal R$-class for all elements $mathcal L$-equivalent to $a$. $square$
The assumption of the question gives that we just have a single $mathcal R$-class and a single $mathcal L$-class. So pick arbitrary $a ne b$ and $as = b$ and $a = bt$, then for every $x in S$ by the above we have
$$
xst = x
$$
giving that $e := st$ is idempotent (set $x := e$ above) and a right-identity. Further as the equations $ux = e$ and $xu = e$ are solvable for each $u in S$ it must be a group.
Remark 1: With further theory from the book it is shown that every non-empty intersection of an $mathcal L$-class and an $mathcal R$-class is a group.
Remark 2: In the paper Intersections of maximal ideals in semigroups by P. Grillet I found
An ideal $I$ is maximal iff $S - I$ is a $mathcal J$-class.
And the same proof works in case of right-ideals and $mathcal R$-classes and left-ideals and $mathcal L$-classes. So if $aS = S setminus a$, then as $aS$ is a right ideal it is maximal and $a$ would be a single $mathcal R$-class, which is excluded. Hence $aS = S$ for all $a in S$. Similar $Sa = S$ for all $a in S$ so that $S$ is a group. And the result follows without Green's lemma.
edited yesterday
darij grinberg
9,19132959
answered Jul 31 at 15:46
StefanH
7,81941858
edited yesterday
darij grinberg
9,19132959
edited yesterday
darij grinberg
9,19132959
edited yesterday
darij grinberg
9,19132959
9,19132959
answered Jul 31 at 15:46
StefanH
7,81941858
answered Jul 31 at 15:46
StefanH
7,81941858
answered Jul 31 at 15:46
StefanH
7,81941858
7,81941858
add a comment |Â
add a comment |Â
up vote
1
down vote
Your question is indeed related to Green's relations. The hypothesis tells you that all the elements of your semigroup are $mathcalR$-equivalent and $mathcalL$-equivalent. Therefore, there are all $mathcalH$-equivalent. It now suffices to apply the following standard result (see for instance [1, Proposition 1.4]).
Let $H$ be an $mathcalH$-class of a semigroup. The following
conditions are equivalent:
- $H$ contains an idempotent,
- there exist $s$, $t in H$ such that $st in H$,
- $H$ is a group.
[1] P.A. Grillet, Semigroups, an introduction to the structure theory, Marcel Dekker, Inc., New York, 1995
answered yesterday
J.-E. Pin
16.6k21751
add a comment |Â
up vote
1
down vote
Your question is indeed related to Green's relations. The hypothesis tells you that all the elements of your semigroup are $mathcalR$-equivalent and $mathcalL$-equivalent. Therefore, there are all $mathcalH$-equivalent. It now suffices to apply the following standard result (see for instance [1, Proposition 1.4]).
Let $H$ be an $mathcalH$-class of a semigroup. The following
conditions are equivalent:
- $H$ contains an idempotent,
- there exist $s$, $t in H$ such that $st in H$,
- $H$ is a group.
[1] P.A. Grillet, Semigroups, an introduction to the structure theory, Marcel Dekker, Inc., New York, 1995
answered yesterday
J.-E. Pin
16.6k21751
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your question is indeed related to Green's relations. The hypothesis tells you that all the elements of your semigroup are $mathcalR$-equivalent and $mathcalL$-equivalent. Therefore, there are all $mathcalH$-equivalent. It now suffices to apply the following standard result (see for instance [1, Proposition 1.4]).
Let $H$ be an $mathcalH$-class of a semigroup. The following
conditions are equivalent:
- $H$ contains an idempotent,
- there exist $s$, $t in H$ such that $st in H$,
- $H$ is a group.
[1] P.A. Grillet, Semigroups, an introduction to the structure theory, Marcel Dekker, Inc., New York, 1995
answered yesterday
J.-E. Pin
16.6k21751
Your question is indeed related to Green's relations. The hypothesis tells you that all the elements of your semigroup are $mathcalR$-equivalent and $mathcalL$-equivalent. Therefore, there are all $mathcalH$-equivalent. It now suffices to apply the following standard result (see for instance [1, Proposition 1.4]).
Let $H$ be an $mathcalH$-class of a semigroup. The following
conditions are equivalent:
- $H$ contains an idempotent,
- there exist $s$, $t in H$ such that $st in H$,
- $H$ is a group.
[1] P.A. Grillet, Semigroups, an introduction to the structure theory, Marcel Dekker, Inc., New York, 1995
answered yesterday
J.-E. Pin
16.6k21751
answered yesterday
J.-E. Pin
16.6k21751
answered yesterday
J.-E. Pin
16.6k21751
answered yesterday
J.-E. Pin
16.6k21751
16.6k21751
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868013%2fshow-that-a-semigroup-with-as-cup-a-bs-cup-b-and-sa-cup-a%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Why is this downvoted?
– greedoid
Jul 31 at 12:51
Before downvoting please leave a comment what I could improve (or add a comment to your downvote at least)... I dont know either why this was downvoted!
– StefanH
Jul 31 at 12:55
1
@DerekHolt That being said, if the purpose of actively downvoting is, as far as I can tell, to make this site a better place, then leaving a comment is the only way to teach the poster what they can do better. So downvoting without a comment serves little purpose except possibly scare new users away from the site (which, if I'm allowed to be cynical, might be a goal to some).
– Arthur
Jul 31 at 13:33
1
I don't know much about semigroups, but i had a quick look at the book. Lemma 2.2.1 seems to say that multiplying by $s$ on the right is a bijection of $S$, and similarly on the left. So, for $a in S$, there exists $e$ with $ae=a$, and then $ae^2=ae$, so $e^2=e$, and $e$ is an idempotent.
– Derek Holt
Jul 31 at 14:01
1
@DerekHolt Okay, yes by my assummption there is just one $mathcal L$-class! So this is fine, and also by this cancellability follows.
– StefanH
Jul 31 at 14:06