Inner Automorphism using Transpositions Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why is conjugation by an odd permutation in $S_n$ not an inner automorphism on $A_n$?Finding the centralizer of a permutationOn the Commutativity of Cycles in Permutation GroupsProof that disjoint cycles commuteProving $Z(S_n)=1$ for $n geq 3$.The cycle decomposition of $sigma$ contains an $r$-cycleTransposition definition permutationWhy is it trivial to see that $binomn2 = fracn!(n-2m)!m!2^m$ has no solutions for $m > 3$?Commutativity of cycles in decomposition of commutative permutationsPermutations that commute with a given transposition
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Inner Automorphism using Transpositions
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why is conjugation by an odd permutation in $S_n$ not an inner automorphism on $A_n$?Finding the centralizer of a permutationOn the Commutativity of Cycles in Permutation GroupsProof that disjoint cycles commuteProving $Z(S_n)=1$ for $n geq 3$.The cycle decomposition of $sigma$ contains an $r$-cycleTransposition definition permutationWhy is it trivial to see that $binomn2 = fracn!(n-2m)!m!2^m$ has no solutions for $m > 3$?Commutativity of cycles in decomposition of commutative permutationsPermutations that commute with a given transposition
$begingroup$
Test question that I got destroyed by. Thought about it all weekend and came up with an explanation. Today, scores back, and prof gave a very different explanation from the one I had in mind. But I'm wondering if my explanation could also work? I'm pretty sure the question is about a special case of inner automorphism, though we haven't covered this topic yet so neither his proof nor mine makes explicit reference to the topic. Here's the question:
Let $n geq 3$ be an integer. Suppose that $sigma in S_n$ sends $a$ to $b$, where $a$ and $b$ are distinct elements of $1, 2, ..., n$. With justification, tell whether there exists a transposition $tau in S_n$ such that $tau sigma tau neq sigma$.
His explanation:
Let $tau = (b c)$ with $c in 1, 2, ..., n setminus a, b $. Then $tau sigma tau (a) = c$ whereas, by assumption in the question stem, $sigma(a) = b$.
(Note that if there is an error above, it is certainly an error I made taking notes and not my prof's error!) This proof is definitely very direct. But I find it unsatisfying because it doesn't give me a lot of insight into when $tau sigma tau = sigma$ and when not. So my own explanation focuses on finding a criterion for when $tau sigma tau = sigma$, and then showing that any $sigma in S_n, n geq 3$ will sometimes fail that criterion:
I claim that $tau sigma tau = sigma$ if and only if $sigma$ and $tau$ commute.
($Rightarrow$) If $tau sigma tau = sigma$, then $sigma tau = tau^-1 sigma = tau sigma$, where the last equality is because $tau$ is a transposition. Therefore $tau$ and $sigma$ commute.
($Leftarrow$) If $tau$ and $sigma$ commute, then $tau sigma tau = tau tau sigma = sigma$, where again the last equality is because $tau$ is a transposition.
However, for any $sigma in S_n, n geq 3$, there exists some $tau$ that is neither disjoint to $sigma$ nor equal to $sigma$ (if $sigma$ is a transposition), and therefore there is some $tau$ that does not commute with $sigma$, meaning that $tau sigma tau neq sigma$.
I think the two explanations are actually equivalent because in prof's proof, both $tau$ and $sigma$ move $b$, so they are not disjoint. And the criteria that $c neq a$ means that $tau neq sigma$, if $sigma$ is a transposition. I also think mine is correct because $tau sigma tau$ is an inner automorphism when $tau$ is a transposition, and for any inner automorphism, $A B A^-1 = B$ iff A and B commute...right?
Thank you in advance!
abstract-algebra permutations automorphism-group
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
$endgroup$
add a comment |
$begingroup$
Test question that I got destroyed by. Thought about it all weekend and came up with an explanation. Today, scores back, and prof gave a very different explanation from the one I had in mind. But I'm wondering if my explanation could also work? I'm pretty sure the question is about a special case of inner automorphism, though we haven't covered this topic yet so neither his proof nor mine makes explicit reference to the topic. Here's the question:
Let $n geq 3$ be an integer. Suppose that $sigma in S_n$ sends $a$ to $b$, where $a$ and $b$ are distinct elements of $1, 2, ..., n$. With justification, tell whether there exists a transposition $tau in S_n$ such that $tau sigma tau neq sigma$.
His explanation:
Let $tau = (b c)$ with $c in 1, 2, ..., n setminus a, b $. Then $tau sigma tau (a) = c$ whereas, by assumption in the question stem, $sigma(a) = b$.
(Note that if there is an error above, it is certainly an error I made taking notes and not my prof's error!) This proof is definitely very direct. But I find it unsatisfying because it doesn't give me a lot of insight into when $tau sigma tau = sigma$ and when not. So my own explanation focuses on finding a criterion for when $tau sigma tau = sigma$, and then showing that any $sigma in S_n, n geq 3$ will sometimes fail that criterion:
I claim that $tau sigma tau = sigma$ if and only if $sigma$ and $tau$ commute.
($Rightarrow$) If $tau sigma tau = sigma$, then $sigma tau = tau^-1 sigma = tau sigma$, where the last equality is because $tau$ is a transposition. Therefore $tau$ and $sigma$ commute.
($Leftarrow$) If $tau$ and $sigma$ commute, then $tau sigma tau = tau tau sigma = sigma$, where again the last equality is because $tau$ is a transposition.
However, for any $sigma in S_n, n geq 3$, there exists some $tau$ that is neither disjoint to $sigma$ nor equal to $sigma$ (if $sigma$ is a transposition), and therefore there is some $tau$ that does not commute with $sigma$, meaning that $tau sigma tau neq sigma$.
I think the two explanations are actually equivalent because in prof's proof, both $tau$ and $sigma$ move $b$, so they are not disjoint. And the criteria that $c neq a$ means that $tau neq sigma$, if $sigma$ is a transposition. I also think mine is correct because $tau sigma tau$ is an inner automorphism when $tau$ is a transposition, and for any inner automorphism, $A B A^-1 = B$ iff A and B commute...right?
Thank you in advance!
abstract-algebra permutations automorphism-group
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
$endgroup$
1
$begingroup$
Your "if and only if" claim definitely looks correct to me. What is the reasoning behind the "disjoint and not equal" part? Is that related to a special property of transpositions? It's not true in general that this stops permutations from commuting. For example, (12) commutes with (12)(34) even though they are not disjoint and not equal.
$endgroup$
– CJD
Apr 2 at 3:14
$begingroup$
Ah good point! My "disjoint and not equal" part is probably just incorrect. I was thinking of "if two cycles are disjoint then they commute," but I incorrectly extended that idea by replacing "cycles" with "permutations in general", and by making it an "iff" statement, when it isn't. So then it is possibly accurate to say that if $tau$ and $sigma$ do not commute, then $tau sigma tau neq sigma$. But I need a better justification for why there exists $tau$ that does not commute with $sigma$? Or is that still not quite right?
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:28
1
$begingroup$
I agree with your comment, that if you can find a $tau$ that does not commute with $sigma$, then you are done. Do you know that $S_n$ has trivial center when $n geq 3$? And do you know that $S_n$ is generated by transpositions? If you know both those things, you could try using proof by contradiction: Assume that $sigma$ commutes with every transposition...
$endgroup$
– CJD
Apr 2 at 3:33
1
$begingroup$
Great points. For this class, we do not "know" that $S_n$ has trivial center when $n geq 3$ yet. By that I mean that it isn't something we've shown in class yet, so it wouldn't have been "fair game" to just state that on this test. (And in fact, the next part of the problem actually asked, "What can you conclude about the center of $S_n$ based on your answer to the preceding question?") HOWEVER, your point about $S_n$ being generated by transpositions is extremely interesting, and I will think more about that. I think that might be the missing piece. Thanks so much!
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:41
add a comment |
$begingroup$
Test question that I got destroyed by. Thought about it all weekend and came up with an explanation. Today, scores back, and prof gave a very different explanation from the one I had in mind. But I'm wondering if my explanation could also work? I'm pretty sure the question is about a special case of inner automorphism, though we haven't covered this topic yet so neither his proof nor mine makes explicit reference to the topic. Here's the question:
Let $n geq 3$ be an integer. Suppose that $sigma in S_n$ sends $a$ to $b$, where $a$ and $b$ are distinct elements of $1, 2, ..., n$. With justification, tell whether there exists a transposition $tau in S_n$ such that $tau sigma tau neq sigma$.
His explanation:
Let $tau = (b c)$ with $c in 1, 2, ..., n setminus a, b $. Then $tau sigma tau (a) = c$ whereas, by assumption in the question stem, $sigma(a) = b$.
(Note that if there is an error above, it is certainly an error I made taking notes and not my prof's error!) This proof is definitely very direct. But I find it unsatisfying because it doesn't give me a lot of insight into when $tau sigma tau = sigma$ and when not. So my own explanation focuses on finding a criterion for when $tau sigma tau = sigma$, and then showing that any $sigma in S_n, n geq 3$ will sometimes fail that criterion:
I claim that $tau sigma tau = sigma$ if and only if $sigma$ and $tau$ commute.
($Rightarrow$) If $tau sigma tau = sigma$, then $sigma tau = tau^-1 sigma = tau sigma$, where the last equality is because $tau$ is a transposition. Therefore $tau$ and $sigma$ commute.
($Leftarrow$) If $tau$ and $sigma$ commute, then $tau sigma tau = tau tau sigma = sigma$, where again the last equality is because $tau$ is a transposition.
However, for any $sigma in S_n, n geq 3$, there exists some $tau$ that is neither disjoint to $sigma$ nor equal to $sigma$ (if $sigma$ is a transposition), and therefore there is some $tau$ that does not commute with $sigma$, meaning that $tau sigma tau neq sigma$.
I think the two explanations are actually equivalent because in prof's proof, both $tau$ and $sigma$ move $b$, so they are not disjoint. And the criteria that $c neq a$ means that $tau neq sigma$, if $sigma$ is a transposition. I also think mine is correct because $tau sigma tau$ is an inner automorphism when $tau$ is a transposition, and for any inner automorphism, $A B A^-1 = B$ iff A and B commute...right?
Thank you in advance!
abstract-algebra permutations automorphism-group
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
$endgroup$
Test question that I got destroyed by. Thought about it all weekend and came up with an explanation. Today, scores back, and prof gave a very different explanation from the one I had in mind. But I'm wondering if my explanation could also work? I'm pretty sure the question is about a special case of inner automorphism, though we haven't covered this topic yet so neither his proof nor mine makes explicit reference to the topic. Here's the question:
Let $n geq 3$ be an integer. Suppose that $sigma in S_n$ sends $a$ to $b$, where $a$ and $b$ are distinct elements of $1, 2, ..., n$. With justification, tell whether there exists a transposition $tau in S_n$ such that $tau sigma tau neq sigma$.
His explanation:
Let $tau = (b c)$ with $c in 1, 2, ..., n setminus a, b $. Then $tau sigma tau (a) = c$ whereas, by assumption in the question stem, $sigma(a) = b$.
(Note that if there is an error above, it is certainly an error I made taking notes and not my prof's error!) This proof is definitely very direct. But I find it unsatisfying because it doesn't give me a lot of insight into when $tau sigma tau = sigma$ and when not. So my own explanation focuses on finding a criterion for when $tau sigma tau = sigma$, and then showing that any $sigma in S_n, n geq 3$ will sometimes fail that criterion:
I claim that $tau sigma tau = sigma$ if and only if $sigma$ and $tau$ commute.
($Rightarrow$) If $tau sigma tau = sigma$, then $sigma tau = tau^-1 sigma = tau sigma$, where the last equality is because $tau$ is a transposition. Therefore $tau$ and $sigma$ commute.
($Leftarrow$) If $tau$ and $sigma$ commute, then $tau sigma tau = tau tau sigma = sigma$, where again the last equality is because $tau$ is a transposition.
However, for any $sigma in S_n, n geq 3$, there exists some $tau$ that is neither disjoint to $sigma$ nor equal to $sigma$ (if $sigma$ is a transposition), and therefore there is some $tau$ that does not commute with $sigma$, meaning that $tau sigma tau neq sigma$.
I think the two explanations are actually equivalent because in prof's proof, both $tau$ and $sigma$ move $b$, so they are not disjoint. And the criteria that $c neq a$ means that $tau neq sigma$, if $sigma$ is a transposition. I also think mine is correct because $tau sigma tau$ is an inner automorphism when $tau$ is a transposition, and for any inner automorphism, $A B A^-1 = B$ iff A and B commute...right?
Thank you in advance!
abstract-algebra permutations automorphism-group
abstract-algebra permutations automorphism-group
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
edited Apr 2 at 2:47
1Teaches2Learn
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
asked Apr 2 at 2:24
1Teaches2Learn1Teaches2Learn
17528
17528
1
$begingroup$
Your "if and only if" claim definitely looks correct to me. What is the reasoning behind the "disjoint and not equal" part? Is that related to a special property of transpositions? It's not true in general that this stops permutations from commuting. For example, (12) commutes with (12)(34) even though they are not disjoint and not equal.
$endgroup$
– CJD
Apr 2 at 3:14
$begingroup$
Ah good point! My "disjoint and not equal" part is probably just incorrect. I was thinking of "if two cycles are disjoint then they commute," but I incorrectly extended that idea by replacing "cycles" with "permutations in general", and by making it an "iff" statement, when it isn't. So then it is possibly accurate to say that if $tau$ and $sigma$ do not commute, then $tau sigma tau neq sigma$. But I need a better justification for why there exists $tau$ that does not commute with $sigma$? Or is that still not quite right?
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:28
1
$begingroup$
I agree with your comment, that if you can find a $tau$ that does not commute with $sigma$, then you are done. Do you know that $S_n$ has trivial center when $n geq 3$? And do you know that $S_n$ is generated by transpositions? If you know both those things, you could try using proof by contradiction: Assume that $sigma$ commutes with every transposition...
$endgroup$
– CJD
Apr 2 at 3:33
1
$begingroup$
Great points. For this class, we do not "know" that $S_n$ has trivial center when $n geq 3$ yet. By that I mean that it isn't something we've shown in class yet, so it wouldn't have been "fair game" to just state that on this test. (And in fact, the next part of the problem actually asked, "What can you conclude about the center of $S_n$ based on your answer to the preceding question?") HOWEVER, your point about $S_n$ being generated by transpositions is extremely interesting, and I will think more about that. I think that might be the missing piece. Thanks so much!
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:41
add a comment |
1
$begingroup$
Your "if and only if" claim definitely looks correct to me. What is the reasoning behind the "disjoint and not equal" part? Is that related to a special property of transpositions? It's not true in general that this stops permutations from commuting. For example, (12) commutes with (12)(34) even though they are not disjoint and not equal.
$endgroup$
– CJD
Apr 2 at 3:14
$begingroup$
Ah good point! My "disjoint and not equal" part is probably just incorrect. I was thinking of "if two cycles are disjoint then they commute," but I incorrectly extended that idea by replacing "cycles" with "permutations in general", and by making it an "iff" statement, when it isn't. So then it is possibly accurate to say that if $tau$ and $sigma$ do not commute, then $tau sigma tau neq sigma$. But I need a better justification for why there exists $tau$ that does not commute with $sigma$? Or is that still not quite right?
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:28
1
$begingroup$
I agree with your comment, that if you can find a $tau$ that does not commute with $sigma$, then you are done. Do you know that $S_n$ has trivial center when $n geq 3$? And do you know that $S_n$ is generated by transpositions? If you know both those things, you could try using proof by contradiction: Assume that $sigma$ commutes with every transposition...
$endgroup$
– CJD
Apr 2 at 3:33
1
$begingroup$
Great points. For this class, we do not "know" that $S_n$ has trivial center when $n geq 3$ yet. By that I mean that it isn't something we've shown in class yet, so it wouldn't have been "fair game" to just state that on this test. (And in fact, the next part of the problem actually asked, "What can you conclude about the center of $S_n$ based on your answer to the preceding question?") HOWEVER, your point about $S_n$ being generated by transpositions is extremely interesting, and I will think more about that. I think that might be the missing piece. Thanks so much!
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:41
1
1
$begingroup$
Your "if and only if" claim definitely looks correct to me. What is the reasoning behind the "disjoint and not equal" part? Is that related to a special property of transpositions? It's not true in general that this stops permutations from commuting. For example, (12) commutes with (12)(34) even though they are not disjoint and not equal.
$endgroup$
– CJD
Apr 2 at 3:14
$begingroup$
Your "if and only if" claim definitely looks correct to me. What is the reasoning behind the "disjoint and not equal" part? Is that related to a special property of transpositions? It's not true in general that this stops permutations from commuting. For example, (12) commutes with (12)(34) even though they are not disjoint and not equal.
$endgroup$
– CJD
Apr 2 at 3:14
$begingroup$
Ah good point! My "disjoint and not equal" part is probably just incorrect. I was thinking of "if two cycles are disjoint then they commute," but I incorrectly extended that idea by replacing "cycles" with "permutations in general", and by making it an "iff" statement, when it isn't. So then it is possibly accurate to say that if $tau$ and $sigma$ do not commute, then $tau sigma tau neq sigma$. But I need a better justification for why there exists $tau$ that does not commute with $sigma$? Or is that still not quite right?
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:28
$begingroup$
Ah good point! My "disjoint and not equal" part is probably just incorrect. I was thinking of "if two cycles are disjoint then they commute," but I incorrectly extended that idea by replacing "cycles" with "permutations in general", and by making it an "iff" statement, when it isn't. So then it is possibly accurate to say that if $tau$ and $sigma$ do not commute, then $tau sigma tau neq sigma$. But I need a better justification for why there exists $tau$ that does not commute with $sigma$? Or is that still not quite right?
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:28
1
1
$begingroup$
I agree with your comment, that if you can find a $tau$ that does not commute with $sigma$, then you are done. Do you know that $S_n$ has trivial center when $n geq 3$? And do you know that $S_n$ is generated by transpositions? If you know both those things, you could try using proof by contradiction: Assume that $sigma$ commutes with every transposition...
$endgroup$
– CJD
Apr 2 at 3:33
$begingroup$
I agree with your comment, that if you can find a $tau$ that does not commute with $sigma$, then you are done. Do you know that $S_n$ has trivial center when $n geq 3$? And do you know that $S_n$ is generated by transpositions? If you know both those things, you could try using proof by contradiction: Assume that $sigma$ commutes with every transposition...
$endgroup$
– CJD
Apr 2 at 3:33
1
1
$begingroup$
Great points. For this class, we do not "know" that $S_n$ has trivial center when $n geq 3$ yet. By that I mean that it isn't something we've shown in class yet, so it wouldn't have been "fair game" to just state that on this test. (And in fact, the next part of the problem actually asked, "What can you conclude about the center of $S_n$ based on your answer to the preceding question?") HOWEVER, your point about $S_n$ being generated by transpositions is extremely interesting, and I will think more about that. I think that might be the missing piece. Thanks so much!
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:41
$begingroup$
Great points. For this class, we do not "know" that $S_n$ has trivial center when $n geq 3$ yet. By that I mean that it isn't something we've shown in class yet, so it wouldn't have been "fair game" to just state that on this test. (And in fact, the next part of the problem actually asked, "What can you conclude about the center of $S_n$ based on your answer to the preceding question?") HOWEVER, your point about $S_n$ being generated by transpositions is extremely interesting, and I will think more about that. I think that might be the missing piece. Thanks so much!
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:41
add a comment |
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$begingroup$
Your "if and only if" claim definitely looks correct to me. What is the reasoning behind the "disjoint and not equal" part? Is that related to a special property of transpositions? It's not true in general that this stops permutations from commuting. For example, (12) commutes with (12)(34) even though they are not disjoint and not equal.
$endgroup$
– CJD
Apr 2 at 3:14
$begingroup$
Ah good point! My "disjoint and not equal" part is probably just incorrect. I was thinking of "if two cycles are disjoint then they commute," but I incorrectly extended that idea by replacing "cycles" with "permutations in general", and by making it an "iff" statement, when it isn't. So then it is possibly accurate to say that if $tau$ and $sigma$ do not commute, then $tau sigma tau neq sigma$. But I need a better justification for why there exists $tau$ that does not commute with $sigma$? Or is that still not quite right?
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:28
1
$begingroup$
I agree with your comment, that if you can find a $tau$ that does not commute with $sigma$, then you are done. Do you know that $S_n$ has trivial center when $n geq 3$? And do you know that $S_n$ is generated by transpositions? If you know both those things, you could try using proof by contradiction: Assume that $sigma$ commutes with every transposition...
$endgroup$
– CJD
Apr 2 at 3:33
1
$begingroup$
Great points. For this class, we do not "know" that $S_n$ has trivial center when $n geq 3$ yet. By that I mean that it isn't something we've shown in class yet, so it wouldn't have been "fair game" to just state that on this test. (And in fact, the next part of the problem actually asked, "What can you conclude about the center of $S_n$ based on your answer to the preceding question?") HOWEVER, your point about $S_n$ being generated by transpositions is extremely interesting, and I will think more about that. I think that might be the missing piece. Thanks so much!
$endgroup$
– 1Teaches2Learn
Apr 2 at 3:41