Submonoid generated by a subset












0












$begingroup$


Definition:




If $S$ is a subset of monoid $[M;∗]$, the submonoid generated by $S$, $⟨S⟩$, is defined by:




  • The identity of M belongs to $⟨S⟩$; and $a ∈ S ⇒ a ∈ ⟨S⟩$

  • $a, b ∈ ⟨S⟩ ⇒ a ∗ b ∈ ⟨S⟩$




Exercise:




Let $S_1$ = {$1$}, $S_2$ = {$frac{1}{2}$} and $S_3$ = {$-1$} be subsets of $Bbb{Q}$. Describe $⟨S_i⟩$ ($i = 1, 2, 3$) when $Bbb{Q}$ is a monoid with multiplication.




I'm not quite sure how to solve this. For example with the first subset I get that



{$1$, $id_Q$} ∈ $⟨S_1⟩$ ($id_Q$ being the identity of $Bbb{Q}$)



but that's about as far as I can get. And as applying the definition to the different subsets is definitely not the solution can someone explain me how to proceed or tell me what I'm doing wrong here?
Thank you!






abstract-algebra group-theory monoid







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$endgroup$












  • $begingroup$
    I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
    $endgroup$
    – jgon
    Jan 5 at 22:30






  • 2




    $begingroup$
    $1$ is the identity of ${mathbb Q}$, so $langle S_1 rangle = {1}$, $langle S_3rangle = { pm 1 }$, and I'll leave $langle S_2 rangle$ to you.
    $endgroup$
    – Derek Holt
    Jan 5 at 22:51


















0












$begingroup$


Definition:




If $S$ is a subset of monoid $[M;∗]$, the submonoid generated by $S$, $⟨S⟩$, is defined by:




  • The identity of M belongs to $⟨S⟩$; and $a ∈ S ⇒ a ∈ ⟨S⟩$

  • $a, b ∈ ⟨S⟩ ⇒ a ∗ b ∈ ⟨S⟩$




Exercise:




Let $S_1$ = {$1$}, $S_2$ = {$frac{1}{2}$} and $S_3$ = {$-1$} be subsets of $Bbb{Q}$. Describe $⟨S_i⟩$ ($i = 1, 2, 3$) when $Bbb{Q}$ is a monoid with multiplication.




I'm not quite sure how to solve this. For example with the first subset I get that



{$1$, $id_Q$} ∈ $⟨S_1⟩$ ($id_Q$ being the identity of $Bbb{Q}$)



but that's about as far as I can get. And as applying the definition to the different subsets is definitely not the solution can someone explain me how to proceed or tell me what I'm doing wrong here?
Thank you!






abstract-algebra group-theory monoid







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
    $endgroup$
    – jgon
    Jan 5 at 22:30






  • 2




    $begingroup$
    $1$ is the identity of ${mathbb Q}$, so $langle S_1 rangle = {1}$, $langle S_3rangle = { pm 1 }$, and I'll leave $langle S_2 rangle$ to you.
    $endgroup$
    – Derek Holt
    Jan 5 at 22:51
















0












0








0





$begingroup$


Definition:




If $S$ is a subset of monoid $[M;∗]$, the submonoid generated by $S$, $⟨S⟩$, is defined by:




  • The identity of M belongs to $⟨S⟩$; and $a ∈ S ⇒ a ∈ ⟨S⟩$

  • $a, b ∈ ⟨S⟩ ⇒ a ∗ b ∈ ⟨S⟩$




Exercise:




Let $S_1$ = {$1$}, $S_2$ = {$frac{1}{2}$} and $S_3$ = {$-1$} be subsets of $Bbb{Q}$. Describe $⟨S_i⟩$ ($i = 1, 2, 3$) when $Bbb{Q}$ is a monoid with multiplication.




I'm not quite sure how to solve this. For example with the first subset I get that



{$1$, $id_Q$} ∈ $⟨S_1⟩$ ($id_Q$ being the identity of $Bbb{Q}$)



but that's about as far as I can get. And as applying the definition to the different subsets is definitely not the solution can someone explain me how to proceed or tell me what I'm doing wrong here?
Thank you!






abstract-algebra group-theory monoid







share|cite|improve this question











$endgroup$




Definition:




If $S$ is a subset of monoid $[M;∗]$, the submonoid generated by $S$, $⟨S⟩$, is defined by:




  • The identity of M belongs to $⟨S⟩$; and $a ∈ S ⇒ a ∈ ⟨S⟩$

  • $a, b ∈ ⟨S⟩ ⇒ a ∗ b ∈ ⟨S⟩$




Exercise:




Let $S_1$ = {$1$}, $S_2$ = {$frac{1}{2}$} and $S_3$ = {$-1$} be subsets of $Bbb{Q}$. Describe $⟨S_i⟩$ ($i = 1, 2, 3$) when $Bbb{Q}$ is a monoid with multiplication.




I'm not quite sure how to solve this. For example with the first subset I get that



{$1$, $id_Q$} ∈ $⟨S_1⟩$ ($id_Q$ being the identity of $Bbb{Q}$)



but that's about as far as I can get. And as applying the definition to the different subsets is definitely not the solution can someone explain me how to proceed or tell me what I'm doing wrong here?
Thank you!






abstract-algebra group-theory monoid




abstract-algebra group-theory monoid






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 5 at 22:32







Alex

















asked Jan 5 at 22:27









AlexAlex

304




304












  • $begingroup$
    I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
    $endgroup$
    – jgon
    Jan 5 at 22:30






  • 2




    $begingroup$
    $1$ is the identity of ${mathbb Q}$, so $langle S_1 rangle = {1}$, $langle S_3rangle = { pm 1 }$, and I'll leave $langle S_2 rangle$ to you.
    $endgroup$
    – Derek Holt
    Jan 5 at 22:51




















  • $begingroup$
    I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
    $endgroup$
    – jgon
    Jan 5 at 22:30






  • 2




    $begingroup$
    $1$ is the identity of ${mathbb Q}$, so $langle S_1 rangle = {1}$, $langle S_3rangle = { pm 1 }$, and I'll leave $langle S_2 rangle$ to you.
    $endgroup$
    – Derek Holt
    Jan 5 at 22:51


















$begingroup$
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
$endgroup$
– jgon
Jan 5 at 22:30




$begingroup$
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
$endgroup$
– jgon
Jan 5 at 22:30




2




2




$begingroup$
$1$ is the identity of ${mathbb Q}$, so $langle S_1 rangle = {1}$, $langle S_3rangle = { pm 1 }$, and I'll leave $langle S_2 rangle$ to you.
$endgroup$
– Derek Holt
Jan 5 at 22:51






$begingroup$
$1$ is the identity of ${mathbb Q}$, so $langle S_1 rangle = {1}$, $langle S_3rangle = { pm 1 }$, and I'll leave $langle S_2 rangle$ to you.
$endgroup$
– Derek Holt
Jan 5 at 22:51












1 Answer
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1












$begingroup$

Under multiplication, identity element of $ mathbb{Q} $ is $1$. So, $ 1 in langle S_i rangle $ for any $i$. Furthermore, we have to make sure that the group is closed under the binary operation, multiplication in this case. Consider $ langle S_1 rangle $. $1$ is already there and certainly, $ langle S_1 rangle = { 1 } $. On the other hand, $ langle S_3 rangle$ requires we add $ 1$ to the set. Then, we are done as product of $ 1 $ and $ - 1$ (and no matter how many times or how you operate these two) is always $ pm 1 $. So, $ langle S_3 rangle = { pm 1 } $. Finally, for $ langle S_3 rangle $, you can see that it will have $ 1 $ and $ frac{1}{2} $ at least. However, $frac{1}{2}cdot frac{1}{2} = frac{1}{4}$ so we have to make sure it's in the set. By similar argument $ frac{1}{2^n}$ such that $n geq 0 $ should be in the set. So, we have $ langle S_3 rangle = { frac{1}{2^n} mid n geq 0 } $.






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    $begingroup$

    Under multiplication, identity element of $ mathbb{Q} $ is $1$. So, $ 1 in langle S_i rangle $ for any $i$. Furthermore, we have to make sure that the group is closed under the binary operation, multiplication in this case. Consider $ langle S_1 rangle $. $1$ is already there and certainly, $ langle S_1 rangle = { 1 } $. On the other hand, $ langle S_3 rangle$ requires we add $ 1$ to the set. Then, we are done as product of $ 1 $ and $ - 1$ (and no matter how many times or how you operate these two) is always $ pm 1 $. So, $ langle S_3 rangle = { pm 1 } $. Finally, for $ langle S_3 rangle $, you can see that it will have $ 1 $ and $ frac{1}{2} $ at least. However, $frac{1}{2}cdot frac{1}{2} = frac{1}{4}$ so we have to make sure it's in the set. By similar argument $ frac{1}{2^n}$ such that $n geq 0 $ should be in the set. So, we have $ langle S_3 rangle = { frac{1}{2^n} mid n geq 0 } $.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Under multiplication, identity element of $ mathbb{Q} $ is $1$. So, $ 1 in langle S_i rangle $ for any $i$. Furthermore, we have to make sure that the group is closed under the binary operation, multiplication in this case. Consider $ langle S_1 rangle $. $1$ is already there and certainly, $ langle S_1 rangle = { 1 } $. On the other hand, $ langle S_3 rangle$ requires we add $ 1$ to the set. Then, we are done as product of $ 1 $ and $ - 1$ (and no matter how many times or how you operate these two) is always $ pm 1 $. So, $ langle S_3 rangle = { pm 1 } $. Finally, for $ langle S_3 rangle $, you can see that it will have $ 1 $ and $ frac{1}{2} $ at least. However, $frac{1}{2}cdot frac{1}{2} = frac{1}{4}$ so we have to make sure it's in the set. By similar argument $ frac{1}{2^n}$ such that $n geq 0 $ should be in the set. So, we have $ langle S_3 rangle = { frac{1}{2^n} mid n geq 0 } $.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Under multiplication, identity element of $ mathbb{Q} $ is $1$. So, $ 1 in langle S_i rangle $ for any $i$. Furthermore, we have to make sure that the group is closed under the binary operation, multiplication in this case. Consider $ langle S_1 rangle $. $1$ is already there and certainly, $ langle S_1 rangle = { 1 } $. On the other hand, $ langle S_3 rangle$ requires we add $ 1$ to the set. Then, we are done as product of $ 1 $ and $ - 1$ (and no matter how many times or how you operate these two) is always $ pm 1 $. So, $ langle S_3 rangle = { pm 1 } $. Finally, for $ langle S_3 rangle $, you can see that it will have $ 1 $ and $ frac{1}{2} $ at least. However, $frac{1}{2}cdot frac{1}{2} = frac{1}{4}$ so we have to make sure it's in the set. By similar argument $ frac{1}{2^n}$ such that $n geq 0 $ should be in the set. So, we have $ langle S_3 rangle = { frac{1}{2^n} mid n geq 0 } $.






        share|cite|improve this answer









        $endgroup$



        Under multiplication, identity element of $ mathbb{Q} $ is $1$. So, $ 1 in langle S_i rangle $ for any $i$. Furthermore, we have to make sure that the group is closed under the binary operation, multiplication in this case. Consider $ langle S_1 rangle $. $1$ is already there and certainly, $ langle S_1 rangle = { 1 } $. On the other hand, $ langle S_3 rangle$ requires we add $ 1$ to the set. Then, we are done as product of $ 1 $ and $ - 1$ (and no matter how many times or how you operate these two) is always $ pm 1 $. So, $ langle S_3 rangle = { pm 1 } $. Finally, for $ langle S_3 rangle $, you can see that it will have $ 1 $ and $ frac{1}{2} $ at least. However, $frac{1}{2}cdot frac{1}{2} = frac{1}{4}$ so we have to make sure it's in the set. By similar argument $ frac{1}{2^n}$ such that $n geq 0 $ should be in the set. So, we have $ langle S_3 rangle = { frac{1}{2^n} mid n geq 0 } $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 9:56









        NawajNawaj

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