Subsemigroup of a finite semigroup
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Let $S$ be a finite semigroup and $T subseteq S$ which satisfy the following property:
For $x, y in T$, we have $x, y in langle z rangle$ for some $z in S$.
If $H subseteq S$ satisfy the above property and $T subseteq H$, then $H = T$.
Is $T$ forms a subsemigroup of $S$.
I have tried to find out the counter example of a semigroup $S$ and a subset $T$ of $S$ which satisfies these two properties but it does not form a subsemigroup of $S$. But I am unable to find this. After that, I have tried to prove $T$ forms a subsemigroup. It is sufficient to prove for any $a, b, c in T$, there exist $d in S$ such that $ab, c in langle d rangle.$ From the given hhypothesis, there exist $x, y, z in S$ such that $a,b in langle x rangle$, $a, c in langle y rangle$, $b,c in langle z rangle$. I am stuck here. Thanks for your any kind of help.
abstract-algebra group-theory semigroups
asked Nov 16 at 5:17
user120386
1,057819
|
show 1 more comment
up vote
4
down vote
favorite
Let $S$ be a finite semigroup and $T subseteq S$ which satisfy the following property:
For $x, y in T$, we have $x, y in langle z rangle$ for some $z in S$.
If $H subseteq S$ satisfy the above property and $T subseteq H$, then $H = T$.
Is $T$ forms a subsemigroup of $S$.
I have tried to find out the counter example of a semigroup $S$ and a subset $T$ of $S$ which satisfies these two properties but it does not form a subsemigroup of $S$. But I am unable to find this. After that, I have tried to prove $T$ forms a subsemigroup. It is sufficient to prove for any $a, b, c in T$, there exist $d in S$ such that $ab, c in langle d rangle.$ From the given hhypothesis, there exist $x, y, z in S$ such that $a,b in langle x rangle$, $a, c in langle y rangle$, $b,c in langle z rangle$. I am stuck here. Thanks for your any kind of help.
abstract-algebra group-theory semigroups
asked Nov 16 at 5:17
user120386
1,057819
What does $langle z rangle$ mean?
– Paul
Nov 16 at 10:07
1
For $ z in S$, $langle z rangle = { z^n ; : ; n in mathbb N}$
– user120386
Nov 16 at 10:24
1
You haven't used the second property. I don't know the full answer yet, but it's not hard to prove that $T$ is at least closed under powers: if $xin T$, then $langle xranglesubseteq T$. To verify this, take $H=Tcuplangle xrangle$ and verify it satisfies the given property and contains $T$, from which you deduce $H=T$.
– Arturo Magidin
Nov 16 at 19:37
I have verified that for any $x in T$, we have $langle x rangle subseteq T$ so that $H = T = T cup langle x rangle$. From this, how we conclude that $T$ satisfy the closure property.
– user120386
Nov 17 at 6:29
@user120386: As I indicated, “I don’t know the full answer yet.” I just pointed out that you hadn’t used the second property (from which this one follows), or the finiteness, for that matter. I’ve been thinking about it bit on and off, but haven’t quite gotten it; I did not mean to imply that this property by itself would solve the problem. But it is an example of using the second property of $T$.
– Arturo Magidin
Nov 17 at 23:53
|
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $S$ be a finite semigroup and $T subseteq S$ which satisfy the following property:
For $x, y in T$, we have $x, y in langle z rangle$ for some $z in S$.
If $H subseteq S$ satisfy the above property and $T subseteq H$, then $H = T$.
Is $T$ forms a subsemigroup of $S$.
I have tried to find out the counter example of a semigroup $S$ and a subset $T$ of $S$ which satisfies these two properties but it does not form a subsemigroup of $S$. But I am unable to find this. After that, I have tried to prove $T$ forms a subsemigroup. It is sufficient to prove for any $a, b, c in T$, there exist $d in S$ such that $ab, c in langle d rangle.$ From the given hhypothesis, there exist $x, y, z in S$ such that $a,b in langle x rangle$, $a, c in langle y rangle$, $b,c in langle z rangle$. I am stuck here. Thanks for your any kind of help.
abstract-algebra group-theory semigroups
asked Nov 16 at 5:17
user120386
1,057819
Let $S$ be a finite semigroup and $T subseteq S$ which satisfy the following property:
For $x, y in T$, we have $x, y in langle z rangle$ for some $z in S$.
If $H subseteq S$ satisfy the above property and $T subseteq H$, then $H = T$.
Is $T$ forms a subsemigroup of $S$.
I have tried to find out the counter example of a semigroup $S$ and a subset $T$ of $S$ which satisfies these two properties but it does not form a subsemigroup of $S$. But I am unable to find this. After that, I have tried to prove $T$ forms a subsemigroup. It is sufficient to prove for any $a, b, c in T$, there exist $d in S$ such that $ab, c in langle d rangle.$ From the given hhypothesis, there exist $x, y, z in S$ such that $a,b in langle x rangle$, $a, c in langle y rangle$, $b,c in langle z rangle$. I am stuck here. Thanks for your any kind of help.
abstract-algebra group-theory semigroups
abstract-algebra group-theory semigroups
asked Nov 16 at 5:17
user120386
1,057819
asked Nov 16 at 5:17
user120386
1,057819
asked Nov 16 at 5:17
user120386
1,057819
asked Nov 16 at 5:17
user120386
1,057819
asked Nov 16 at 5:17
user120386
1,057819
1,057819
What does $langle z rangle$ mean?
– Paul
Nov 16 at 10:07
1
For $ z in S$, $langle z rangle = { z^n ; : ; n in mathbb N}$
– user120386
Nov 16 at 10:24
1
You haven't used the second property. I don't know the full answer yet, but it's not hard to prove that $T$ is at least closed under powers: if $xin T$, then $langle xranglesubseteq T$. To verify this, take $H=Tcuplangle xrangle$ and verify it satisfies the given property and contains $T$, from which you deduce $H=T$.
– Arturo Magidin
Nov 16 at 19:37
I have verified that for any $x in T$, we have $langle x rangle subseteq T$ so that $H = T = T cup langle x rangle$. From this, how we conclude that $T$ satisfy the closure property.
– user120386
Nov 17 at 6:29
@user120386: As I indicated, “I don’t know the full answer yet.” I just pointed out that you hadn’t used the second property (from which this one follows), or the finiteness, for that matter. I’ve been thinking about it bit on and off, but haven’t quite gotten it; I did not mean to imply that this property by itself would solve the problem. But it is an example of using the second property of $T$.
– Arturo Magidin
Nov 17 at 23:53
|
show 1 more comment
What does $langle z rangle$ mean?
– Paul
Nov 16 at 10:07
1
For $ z in S$, $langle z rangle = { z^n ; : ; n in mathbb N}$
– user120386
Nov 16 at 10:24
1
You haven't used the second property. I don't know the full answer yet, but it's not hard to prove that $T$ is at least closed under powers: if $xin T$, then $langle xranglesubseteq T$. To verify this, take $H=Tcuplangle xrangle$ and verify it satisfies the given property and contains $T$, from which you deduce $H=T$.
– Arturo Magidin
Nov 16 at 19:37
I have verified that for any $x in T$, we have $langle x rangle subseteq T$ so that $H = T = T cup langle x rangle$. From this, how we conclude that $T$ satisfy the closure property.
– user120386
Nov 17 at 6:29
@user120386: As I indicated, “I don’t know the full answer yet.” I just pointed out that you hadn’t used the second property (from which this one follows), or the finiteness, for that matter. I’ve been thinking about it bit on and off, but haven’t quite gotten it; I did not mean to imply that this property by itself would solve the problem. But it is an example of using the second property of $T$.
– Arturo Magidin
Nov 17 at 23:53
What does $langle z rangle$ mean?
– Paul
Nov 16 at 10:07
What does $langle z rangle$ mean?
– Paul
Nov 16 at 10:07
1
1
For $ z in S$, $langle z rangle = { z^n ; : ; n in mathbb N}$
– user120386
Nov 16 at 10:24
For $ z in S$, $langle z rangle = { z^n ; : ; n in mathbb N}$
– user120386
Nov 16 at 10:24
1
1
You haven't used the second property. I don't know the full answer yet, but it's not hard to prove that $T$ is at least closed under powers: if $xin T$, then $langle xranglesubseteq T$. To verify this, take $H=Tcuplangle xrangle$ and verify it satisfies the given property and contains $T$, from which you deduce $H=T$.
– Arturo Magidin
Nov 16 at 19:37
You haven't used the second property. I don't know the full answer yet, but it's not hard to prove that $T$ is at least closed under powers: if $xin T$, then $langle xranglesubseteq T$. To verify this, take $H=Tcuplangle xrangle$ and verify it satisfies the given property and contains $T$, from which you deduce $H=T$.
– Arturo Magidin
Nov 16 at 19:37
I have verified that for any $x in T$, we have $langle x rangle subseteq T$ so that $H = T = T cup langle x rangle$. From this, how we conclude that $T$ satisfy the closure property.
– user120386
Nov 17 at 6:29
I have verified that for any $x in T$, we have $langle x rangle subseteq T$ so that $H = T = T cup langle x rangle$. From this, how we conclude that $T$ satisfy the closure property.
– user120386
Nov 17 at 6:29
@user120386: As I indicated, “I don’t know the full answer yet.” I just pointed out that you hadn’t used the second property (from which this one follows), or the finiteness, for that matter. I’ve been thinking about it bit on and off, but haven’t quite gotten it; I did not mean to imply that this property by itself would solve the problem. But it is an example of using the second property of $T$.
– Arturo Magidin
Nov 17 at 23:53
@user120386: As I indicated, “I don’t know the full answer yet.” I just pointed out that you hadn’t used the second property (from which this one follows), or the finiteness, for that matter. I’ve been thinking about it bit on and off, but haven’t quite gotten it; I did not mean to imply that this property by itself would solve the problem. But it is an example of using the second property of $T$.
– Arturo Magidin
Nov 17 at 23:53
|
show 1 more comment
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What does $langle z rangle$ mean?
– Paul
Nov 16 at 10:07
1
For $ z in S$, $langle z rangle = { z^n ; : ; n in mathbb N}$
– user120386
Nov 16 at 10:24
1
You haven't used the second property. I don't know the full answer yet, but it's not hard to prove that $T$ is at least closed under powers: if $xin T$, then $langle xranglesubseteq T$. To verify this, take $H=Tcuplangle xrangle$ and verify it satisfies the given property and contains $T$, from which you deduce $H=T$.
– Arturo Magidin
Nov 16 at 19:37
I have verified that for any $x in T$, we have $langle x rangle subseteq T$ so that $H = T = T cup langle x rangle$. From this, how we conclude that $T$ satisfy the closure property.
– user120386
Nov 17 at 6:29
@user120386: As I indicated, “I don’t know the full answer yet.” I just pointed out that you hadn’t used the second property (from which this one follows), or the finiteness, for that matter. I’ve been thinking about it bit on and off, but haven’t quite gotten it; I did not mean to imply that this property by itself would solve the problem. But it is an example of using the second property of $T$.
– Arturo Magidin
Nov 17 at 23:53