Mixing
Prove that a semigroup satisfying $a^pb^q=ba$ is commutative
$begingroup$
Let $(S, cdot)$ be a semigroup. There are natural numbers $p,q geq 2$ such that $a^pb^q=ba$ for all $a,b in S$. Prove that $S$ is commutative.
I wrote
$$begin{align}
a^{p+1}b^{q+1} &=b^{(q+1)p}a^{(p+1)q} \
&=b^{p}cdot(b^q)^p cdot (a^p)^qcdot a^q \
&=b^pcdot a^p cdot b^q cdot a^q \
&= b^pcdot b cdot a cdot a^q \
&=b^{p+1}a^{q+1}.
end{align}$$
From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.
Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
abstract-algebra group-theory semigroups binary-operations
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
asked Jan 14 at 19:49
sgcsgc
916
$endgroup$
add a comment |
$begingroup$
Let $(S, cdot)$ be a semigroup. There are natural numbers $p,q geq 2$ such that $a^pb^q=ba$ for all $a,b in S$. Prove that $S$ is commutative.
I wrote
$$begin{align}
a^{p+1}b^{q+1} &=b^{(q+1)p}a^{(p+1)q} \
&=b^{p}cdot(b^q)^p cdot (a^p)^qcdot a^q \
&=b^pcdot a^p cdot b^q cdot a^q \
&= b^pcdot b cdot a cdot a^q \
&=b^{p+1}a^{q+1}.
end{align}$$
From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.
Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
abstract-algebra group-theory semigroups binary-operations
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
asked Jan 14 at 19:49
sgcsgc
916
$endgroup$
1
$begingroup$
This paper has many commutativity results for semigroups; I did not check yet, but perhaps there is a proof.
$endgroup$
– Dietrich Burde
Jan 14 at 20:13
1
$begingroup$
How do you get that $a^p = (b^q)^p$ or that $b^q = (a^p)^q$? That is, I don't understand your first or third equalities...
$endgroup$
– Arturo Magidin
Jan 14 at 20:15
1
$begingroup$
@ArturoMagidin I didn't. I just used the given relation: $(b^q)^p cdot (a^p)^q = a^p cdot b^q$. In the given identity instead of $a$ I wrote $b^q$ and instead of $b$ I wrote $a^p$
$endgroup$
– sgc
Jan 14 at 20:19
1
$begingroup$
Oh, I see. Thanks.
$endgroup$
– Arturo Magidin
Jan 14 at 20:26
1
$begingroup$
@sgc is it an exercise? Do you have any hint? For example, if the semi group posses the cancelative property then you are done.
$endgroup$
– Dog_69
Jan 14 at 21:39
add a comment |
$begingroup$
Let $(S, cdot)$ be a semigroup. There are natural numbers $p,q geq 2$ such that $a^pb^q=ba$ for all $a,b in S$. Prove that $S$ is commutative.
I wrote
$$begin{align}
a^{p+1}b^{q+1} &=b^{(q+1)p}a^{(p+1)q} \
&=b^{p}cdot(b^q)^p cdot (a^p)^qcdot a^q \
&=b^pcdot a^p cdot b^q cdot a^q \
&= b^pcdot b cdot a cdot a^q \
&=b^{p+1}a^{q+1}.
end{align}$$
From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.
Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
abstract-algebra group-theory semigroups binary-operations
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
asked Jan 14 at 19:49
sgcsgc
916
$endgroup$
Let $(S, cdot)$ be a semigroup. There are natural numbers $p,q geq 2$ such that $a^pb^q=ba$ for all $a,b in S$. Prove that $S$ is commutative.
I wrote
$$begin{align}
a^{p+1}b^{q+1} &=b^{(q+1)p}a^{(p+1)q} \
&=b^{p}cdot(b^q)^p cdot (a^p)^qcdot a^q \
&=b^pcdot a^p cdot b^q cdot a^q \
&= b^pcdot b cdot a cdot a^q \
&=b^{p+1}a^{q+1}.
end{align}$$
From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.
Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
abstract-algebra group-theory semigroups binary-operations
abstract-algebra group-theory semigroups binary-operations
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
asked Jan 14 at 19:49
sgcsgc
916
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
asked Jan 14 at 19:49
sgcsgc
916
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
edited Jan 14 at 21:49
Eric Wofsey
186k14215342
186k14215342
asked Jan 14 at 19:49
sgcsgc
916
asked Jan 14 at 19:49
sgcsgc
916
asked Jan 14 at 19:49
sgcsgc
916
916
1
$begingroup$
This paper has many commutativity results for semigroups; I did not check yet, but perhaps there is a proof.
$endgroup$
– Dietrich Burde
Jan 14 at 20:13
1
$begingroup$
How do you get that $a^p = (b^q)^p$ or that $b^q = (a^p)^q$? That is, I don't understand your first or third equalities...
$endgroup$
– Arturo Magidin
Jan 14 at 20:15
1
$begingroup$
@ArturoMagidin I didn't. I just used the given relation: $(b^q)^p cdot (a^p)^q = a^p cdot b^q$. In the given identity instead of $a$ I wrote $b^q$ and instead of $b$ I wrote $a^p$
$endgroup$
– sgc
Jan 14 at 20:19
1
$begingroup$
Oh, I see. Thanks.
$endgroup$
– Arturo Magidin
Jan 14 at 20:26
1
$begingroup$
@sgc is it an exercise? Do you have any hint? For example, if the semi group posses the cancelative property then you are done.
$endgroup$
– Dog_69
Jan 14 at 21:39
add a comment |
1
$begingroup$
This paper has many commutativity results for semigroups; I did not check yet, but perhaps there is a proof.
$endgroup$
– Dietrich Burde
Jan 14 at 20:13
1
$begingroup$
How do you get that $a^p = (b^q)^p$ or that $b^q = (a^p)^q$? That is, I don't understand your first or third equalities...
$endgroup$
– Arturo Magidin
Jan 14 at 20:15
1
$begingroup$
@ArturoMagidin I didn't. I just used the given relation: $(b^q)^p cdot (a^p)^q = a^p cdot b^q$. In the given identity instead of $a$ I wrote $b^q$ and instead of $b$ I wrote $a^p$
$endgroup$
– sgc
Jan 14 at 20:19
1
$begingroup$
Oh, I see. Thanks.
$endgroup$
– Arturo Magidin
Jan 14 at 20:26
1
$begingroup$
@sgc is it an exercise? Do you have any hint? For example, if the semi group posses the cancelative property then you are done.
$endgroup$
– Dog_69
Jan 14 at 21:39
1
1
$begingroup$
This paper has many commutativity results for semigroups; I did not check yet, but perhaps there is a proof.
$endgroup$
– Dietrich Burde
Jan 14 at 20:13
$begingroup$
This paper has many commutativity results for semigroups; I did not check yet, but perhaps there is a proof.
$endgroup$
– Dietrich Burde
Jan 14 at 20:13
1
1
$begingroup$
How do you get that $a^p = (b^q)^p$ or that $b^q = (a^p)^q$? That is, I don't understand your first or third equalities...
$endgroup$
– Arturo Magidin
Jan 14 at 20:15
$begingroup$
How do you get that $a^p = (b^q)^p$ or that $b^q = (a^p)^q$? That is, I don't understand your first or third equalities...
$endgroup$
– Arturo Magidin
Jan 14 at 20:15
1
1
$begingroup$
@ArturoMagidin I didn't. I just used the given relation: $(b^q)^p cdot (a^p)^q = a^p cdot b^q$. In the given identity instead of $a$ I wrote $b^q$ and instead of $b$ I wrote $a^p$
$endgroup$
– sgc
Jan 14 at 20:19
$begingroup$
@ArturoMagidin I didn't. I just used the given relation: $(b^q)^p cdot (a^p)^q = a^p cdot b^q$. In the given identity instead of $a$ I wrote $b^q$ and instead of $b$ I wrote $a^p$
$endgroup$
– sgc
Jan 14 at 20:19
1
1
$begingroup$
Oh, I see. Thanks.
$endgroup$
– Arturo Magidin
Jan 14 at 20:26
$begingroup$
Oh, I see. Thanks.
$endgroup$
– Arturo Magidin
Jan 14 at 20:26
1
1
$begingroup$
@sgc is it an exercise? Do you have any hint? For example, if the semi group posses the cancelative property then you are done.
$endgroup$
– Dog_69
Jan 14 at 21:39
$begingroup$
@sgc is it an exercise? Do you have any hint? For example, if the semi group posses the cancelative property then you are done.
$endgroup$
– Dog_69
Jan 14 at 21:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $ain S$. As you have observed, $a^2=a^{p+q}$. In particular, this means that the powers of $a$ are eventually periodic, say with period $d$. We additionally know that $dmid p+q-2$. But we also have $$a^3=color{red}{a^2}cdot color{blue}{a}=color{blue}{a^p}color{red}{(a^2)^q}=a^{p+2q}$$ which means $dmid p+2q-3$. Thus $d$ divides $(p+2q-3)-(p+q-2)=q-1$. Similarly, writing $a^3=acdot a^2$ we find that $d$ divides $p-1$.
Thus we have shown that for any $ain S$, there exists $N$ such that for any $ngeq N$, $$a^n=a^{n+p-1}=a^{n+q-1}.$$
Now let $a,bin S$; we will show that $ab=ba$. Notice first that $$ba=a^pb^q=(b^q)^p(a^p)^q=b^{pq}a^{pq}.$$ Iterating this, we have $ba=b^{(pq)^k}a^{(pq)^k}$ for any $k$. In particular, let us choose $k$ such that $n=(pq)^k$ is large enough such that $a^n=a^{n+p-1}$ and $b^n=b^{n+q-1}$. Now observe that $$ba=b^na^n=(a^n)^p(b^n)^q=a^{np}b^{nq}.$$ But $a^{np}=a^n$, since $a^n=a^{n+p-1}$ and $np-n$ is divisible by $p-1$, and similarly $b^{nq}=b^n$. Thus $$ba=a^nb^n,$$ where $n=(pq)^k$ for some $k$. But swapping the roles of $a$ and $b$ from the start of the argument, we know $a^nb^n=ab$, and thus $$ba=ab.$$
edited Jan 15 at 9:08
Shaun
9,221113684
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
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
});
}
});
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%2f3073660%2fprove-that-a-semigroup-satisfying-apbq-ba-is-commutative%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
$begingroup$
Let $ain S$. As you have observed, $a^2=a^{p+q}$. In particular, this means that the powers of $a$ are eventually periodic, say with period $d$. We additionally know that $dmid p+q-2$. But we also have $$a^3=color{red}{a^2}cdot color{blue}{a}=color{blue}{a^p}color{red}{(a^2)^q}=a^{p+2q}$$ which means $dmid p+2q-3$. Thus $d$ divides $(p+2q-3)-(p+q-2)=q-1$. Similarly, writing $a^3=acdot a^2$ we find that $d$ divides $p-1$.
Thus we have shown that for any $ain S$, there exists $N$ such that for any $ngeq N$, $$a^n=a^{n+p-1}=a^{n+q-1}.$$
Now let $a,bin S$; we will show that $ab=ba$. Notice first that $$ba=a^pb^q=(b^q)^p(a^p)^q=b^{pq}a^{pq}.$$ Iterating this, we have $ba=b^{(pq)^k}a^{(pq)^k}$ for any $k$. In particular, let us choose $k$ such that $n=(pq)^k$ is large enough such that $a^n=a^{n+p-1}$ and $b^n=b^{n+q-1}$. Now observe that $$ba=b^na^n=(a^n)^p(b^n)^q=a^{np}b^{nq}.$$ But $a^{np}=a^n$, since $a^n=a^{n+p-1}$ and $np-n$ is divisible by $p-1$, and similarly $b^{nq}=b^n$. Thus $$ba=a^nb^n,$$ where $n=(pq)^k$ for some $k$. But swapping the roles of $a$ and $b$ from the start of the argument, we know $a^nb^n=ab$, and thus $$ba=ab.$$
edited Jan 15 at 9:08
Shaun
9,221113684
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
add a comment |
$begingroup$
Let $ain S$. As you have observed, $a^2=a^{p+q}$. In particular, this means that the powers of $a$ are eventually periodic, say with period $d$. We additionally know that $dmid p+q-2$. But we also have $$a^3=color{red}{a^2}cdot color{blue}{a}=color{blue}{a^p}color{red}{(a^2)^q}=a^{p+2q}$$ which means $dmid p+2q-3$. Thus $d$ divides $(p+2q-3)-(p+q-2)=q-1$. Similarly, writing $a^3=acdot a^2$ we find that $d$ divides $p-1$.
Thus we have shown that for any $ain S$, there exists $N$ such that for any $ngeq N$, $$a^n=a^{n+p-1}=a^{n+q-1}.$$
Now let $a,bin S$; we will show that $ab=ba$. Notice first that $$ba=a^pb^q=(b^q)^p(a^p)^q=b^{pq}a^{pq}.$$ Iterating this, we have $ba=b^{(pq)^k}a^{(pq)^k}$ for any $k$. In particular, let us choose $k$ such that $n=(pq)^k$ is large enough such that $a^n=a^{n+p-1}$ and $b^n=b^{n+q-1}$. Now observe that $$ba=b^na^n=(a^n)^p(b^n)^q=a^{np}b^{nq}.$$ But $a^{np}=a^n$, since $a^n=a^{n+p-1}$ and $np-n$ is divisible by $p-1$, and similarly $b^{nq}=b^n$. Thus $$ba=a^nb^n,$$ where $n=(pq)^k$ for some $k$. But swapping the roles of $a$ and $b$ from the start of the argument, we know $a^nb^n=ab$, and thus $$ba=ab.$$
edited Jan 15 at 9:08
Shaun
9,221113684
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
add a comment |
$begingroup$
Let $ain S$. As you have observed, $a^2=a^{p+q}$. In particular, this means that the powers of $a$ are eventually periodic, say with period $d$. We additionally know that $dmid p+q-2$. But we also have $$a^3=color{red}{a^2}cdot color{blue}{a}=color{blue}{a^p}color{red}{(a^2)^q}=a^{p+2q}$$ which means $dmid p+2q-3$. Thus $d$ divides $(p+2q-3)-(p+q-2)=q-1$. Similarly, writing $a^3=acdot a^2$ we find that $d$ divides $p-1$.
Thus we have shown that for any $ain S$, there exists $N$ such that for any $ngeq N$, $$a^n=a^{n+p-1}=a^{n+q-1}.$$
Now let $a,bin S$; we will show that $ab=ba$. Notice first that $$ba=a^pb^q=(b^q)^p(a^p)^q=b^{pq}a^{pq}.$$ Iterating this, we have $ba=b^{(pq)^k}a^{(pq)^k}$ for any $k$. In particular, let us choose $k$ such that $n=(pq)^k$ is large enough such that $a^n=a^{n+p-1}$ and $b^n=b^{n+q-1}$. Now observe that $$ba=b^na^n=(a^n)^p(b^n)^q=a^{np}b^{nq}.$$ But $a^{np}=a^n$, since $a^n=a^{n+p-1}$ and $np-n$ is divisible by $p-1$, and similarly $b^{nq}=b^n$. Thus $$ba=a^nb^n,$$ where $n=(pq)^k$ for some $k$. But swapping the roles of $a$ and $b$ from the start of the argument, we know $a^nb^n=ab$, and thus $$ba=ab.$$
edited Jan 15 at 9:08
Shaun
9,221113684
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
Let $ain S$. As you have observed, $a^2=a^{p+q}$. In particular, this means that the powers of $a$ are eventually periodic, say with period $d$. We additionally know that $dmid p+q-2$. But we also have $$a^3=color{red}{a^2}cdot color{blue}{a}=color{blue}{a^p}color{red}{(a^2)^q}=a^{p+2q}$$ which means $dmid p+2q-3$. Thus $d$ divides $(p+2q-3)-(p+q-2)=q-1$. Similarly, writing $a^3=acdot a^2$ we find that $d$ divides $p-1$.
Thus we have shown that for any $ain S$, there exists $N$ such that for any $ngeq N$, $$a^n=a^{n+p-1}=a^{n+q-1}.$$
Now let $a,bin S$; we will show that $ab=ba$. Notice first that $$ba=a^pb^q=(b^q)^p(a^p)^q=b^{pq}a^{pq}.$$ Iterating this, we have $ba=b^{(pq)^k}a^{(pq)^k}$ for any $k$. In particular, let us choose $k$ such that $n=(pq)^k$ is large enough such that $a^n=a^{n+p-1}$ and $b^n=b^{n+q-1}$. Now observe that $$ba=b^na^n=(a^n)^p(b^n)^q=a^{np}b^{nq}.$$ But $a^{np}=a^n$, since $a^n=a^{n+p-1}$ and $np-n$ is divisible by $p-1$, and similarly $b^{nq}=b^n$. Thus $$ba=a^nb^n,$$ where $n=(pq)^k$ for some $k$. But swapping the roles of $a$ and $b$ from the start of the argument, we know $a^nb^n=ab$, and thus $$ba=ab.$$
edited Jan 15 at 9:08
Shaun
9,221113684
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
edited Jan 15 at 9:08
Shaun
9,221113684
edited Jan 15 at 9:08
Shaun
9,221113684
edited Jan 15 at 9:08
Shaun
9,221113684
9,221113684
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
answered Jan 14 at 21:47
Eric WofseyEric Wofsey
186k14215342
186k14215342
add a comment |
add a comment |
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.
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%2f3073660%2fprove-that-a-semigroup-satisfying-apbq-ba-is-commutative%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
1
$begingroup$
This paper has many commutativity results for semigroups; I did not check yet, but perhaps there is a proof.
$endgroup$
– Dietrich Burde
Jan 14 at 20:13
1
$begingroup$
How do you get that $a^p = (b^q)^p$ or that $b^q = (a^p)^q$? That is, I don't understand your first or third equalities...
$endgroup$
– Arturo Magidin
Jan 14 at 20:15
1
$begingroup$
@ArturoMagidin I didn't. I just used the given relation: $(b^q)^p cdot (a^p)^q = a^p cdot b^q$. In the given identity instead of $a$ I wrote $b^q$ and instead of $b$ I wrote $a^p$
$endgroup$
– sgc
Jan 14 at 20:19
1
$begingroup$
Oh, I see. Thanks.
$endgroup$
– Arturo Magidin
Jan 14 at 20:26
1
$begingroup$
@sgc is it an exercise? Do you have any hint? For example, if the semi group posses the cancelative property then you are done.
$endgroup$
– Dog_69
Jan 14 at 21:39