Submonoid generated by a subset
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
asked Jan 5 at 22:27
AlexAlex
254
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
asked Jan 5 at 22:27
AlexAlex
254
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
– jgon
Jan 5 at 22:30
2
$1$ 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.
– Derek Holt
Jan 5 at 22:51
add a comment |
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
asked Jan 5 at 22:27
AlexAlex
254
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
asked Jan 5 at 22:27
AlexAlex
254
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 22:27
AlexAlex
254
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 5 at 22:32
Alex
asked Jan 5 at 22:27
AlexAlex
254
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 5 at 22:27
AlexAlex
254
asked Jan 5 at 22:27
AlexAlex
254
254
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
– jgon
Jan 5 at 22:30
2
$1$ 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.
– Derek Holt
Jan 5 at 22:51
add a comment |
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
– jgon
Jan 5 at 22:30
2
$1$ 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.
– Derek Holt
Jan 5 at 22:51
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
– jgon
Jan 5 at 22:30
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
– jgon
Jan 5 at 22:30
2
2
$1$ 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.
– Derek Holt
Jan 5 at 22:51
$1$ 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.
– Derek Holt
Jan 5 at 22:51
add a comment |
1 Answer
1
active
oldest
votes
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 } $.
answered 2 days ago
NawajNawaj
283
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Alex is a new contributor. Be nice, and check out our Code of Conduct.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063260%2fsubmonoid-generated-by-a-subset%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 } $.
answered 2 days ago
NawajNawaj
283
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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 } $.
answered 2 days ago
NawajNawaj
283
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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 } $.
answered 2 days ago
NawajNawaj
283
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 } $.
answered 2 days ago
NawajNawaj
283
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
NawajNawaj
283
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
NawajNawaj
283
answered 2 days ago
NawajNawaj
283
283
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nawaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Alex is a new contributor. Be nice, and check out our Code of Conduct.
Alex is a new contributor. Be nice, and check out our Code of Conduct.
Alex is a new contributor. Be nice, and check out our Code of Conduct.
Alex is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063260%2fsubmonoid-generated-by-a-subset%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I'm pretty sure they intend $Q$ to be $Bbb{Q}$, the rational numbers.
– jgon
Jan 5 at 22:30
2
$1$ 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.
– Derek Holt
Jan 5 at 22:51