How to compute the powers of $tau$?
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When working through the Symmetric and Alternating Groups section of my Abstract Algebra textbook, I came across Theorem 7.25 that sates:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Then I saw the following example:
The permutation $tau = (12)(34)(567)$ is a product of disjoint cycles of lengths 2, 2, and 3. The least common multiple of 2, 2, and 3 is 6. Theorem 7.25 tells us that tau has order 6. You can verify this directly by computing the powers of $tau$.
$tau = (12)(34)(567), tau^2 = (576), tau^3 = (12)(34),$
$tau^4 = (567), tau^5 = (12)(34)(576), tau^6 = (1).$
But I am not sure how the book is calculating $tau^2, tau^3, ... , tau^6$. I know the product of permutations is a composition of functions, but I must be missing something or doing something wrong to get $tau^2$. Any help would be appreciated!
abstract-algebra permutations
asked Dec 7 '18 at 18:32
AMN52AMN52
326
$endgroup$
add a comment |
$begingroup$
When working through the Symmetric and Alternating Groups section of my Abstract Algebra textbook, I came across Theorem 7.25 that sates:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Then I saw the following example:
The permutation $tau = (12)(34)(567)$ is a product of disjoint cycles of lengths 2, 2, and 3. The least common multiple of 2, 2, and 3 is 6. Theorem 7.25 tells us that tau has order 6. You can verify this directly by computing the powers of $tau$.
$tau = (12)(34)(567), tau^2 = (576), tau^3 = (12)(34),$
$tau^4 = (567), tau^5 = (12)(34)(576), tau^6 = (1).$
But I am not sure how the book is calculating $tau^2, tau^3, ... , tau^6$. I know the product of permutations is a composition of functions, but I must be missing something or doing something wrong to get $tau^2$. Any help would be appreciated!
abstract-algebra permutations
asked Dec 7 '18 at 18:32
AMN52AMN52
326
$endgroup$
1
$begingroup$
$tau^2=(5,7,6)$
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 18:34
add a comment |
$begingroup$
When working through the Symmetric and Alternating Groups section of my Abstract Algebra textbook, I came across Theorem 7.25 that sates:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Then I saw the following example:
The permutation $tau = (12)(34)(567)$ is a product of disjoint cycles of lengths 2, 2, and 3. The least common multiple of 2, 2, and 3 is 6. Theorem 7.25 tells us that tau has order 6. You can verify this directly by computing the powers of $tau$.
$tau = (12)(34)(567), tau^2 = (576), tau^3 = (12)(34),$
$tau^4 = (567), tau^5 = (12)(34)(576), tau^6 = (1).$
But I am not sure how the book is calculating $tau^2, tau^3, ... , tau^6$. I know the product of permutations is a composition of functions, but I must be missing something or doing something wrong to get $tau^2$. Any help would be appreciated!
abstract-algebra permutations
asked Dec 7 '18 at 18:32
AMN52AMN52
326
$endgroup$
When working through the Symmetric and Alternating Groups section of my Abstract Algebra textbook, I came across Theorem 7.25 that sates:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Then I saw the following example:
The permutation $tau = (12)(34)(567)$ is a product of disjoint cycles of lengths 2, 2, and 3. The least common multiple of 2, 2, and 3 is 6. Theorem 7.25 tells us that tau has order 6. You can verify this directly by computing the powers of $tau$.
$tau = (12)(34)(567), tau^2 = (576), tau^3 = (12)(34),$
$tau^4 = (567), tau^5 = (12)(34)(576), tau^6 = (1).$
But I am not sure how the book is calculating $tau^2, tau^3, ... , tau^6$. I know the product of permutations is a composition of functions, but I must be missing something or doing something wrong to get $tau^2$. Any help would be appreciated!
abstract-algebra permutations
abstract-algebra permutations
asked Dec 7 '18 at 18:32
AMN52AMN52
326
asked Dec 7 '18 at 18:32
AMN52AMN52
326
edited Dec 7 '18 at 22:25
AMN52
asked Dec 7 '18 at 18:32
AMN52AMN52
326
asked Dec 7 '18 at 18:32
AMN52AMN52
326
asked Dec 7 '18 at 18:32
AMN52AMN52
326
326
1
$begingroup$
$tau^2=(5,7,6)$
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 18:34
add a comment |
1
$begingroup$
$tau^2=(5,7,6)$
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 18:34
1
1
$begingroup$
$tau^2=(5,7,6)$
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 18:34
$begingroup$
$tau^2=(5,7,6)$
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 18:34
add a comment |
1 Answer
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$begingroup$
Recall that disjoint cycles commute, so $tau^n=(12)^n(34)^n(567)^n$. Can you conclude from here? Additional hint: $(12)^n=(12)$ if $n$ is odd, $(12)^n=mathrm{Id}$ otherwise.
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
$endgroup$
1
$begingroup$
Thank you for the help, that clarified things for me!
$endgroup$
– AMN52
Dec 7 '18 at 20:08
add a comment |
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1 Answer
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$begingroup$
Recall that disjoint cycles commute, so $tau^n=(12)^n(34)^n(567)^n$. Can you conclude from here? Additional hint: $(12)^n=(12)$ if $n$ is odd, $(12)^n=mathrm{Id}$ otherwise.
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
$endgroup$
1
$begingroup$
Thank you for the help, that clarified things for me!
$endgroup$
– AMN52
Dec 7 '18 at 20:08
add a comment |
$begingroup$
Recall that disjoint cycles commute, so $tau^n=(12)^n(34)^n(567)^n$. Can you conclude from here? Additional hint: $(12)^n=(12)$ if $n$ is odd, $(12)^n=mathrm{Id}$ otherwise.
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
$endgroup$
1
$begingroup$
Thank you for the help, that clarified things for me!
$endgroup$
– AMN52
Dec 7 '18 at 20:08
add a comment |
$begingroup$
Recall that disjoint cycles commute, so $tau^n=(12)^n(34)^n(567)^n$. Can you conclude from here? Additional hint: $(12)^n=(12)$ if $n$ is odd, $(12)^n=mathrm{Id}$ otherwise.
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
$endgroup$
Recall that disjoint cycles commute, so $tau^n=(12)^n(34)^n(567)^n$. Can you conclude from here? Additional hint: $(12)^n=(12)$ if $n$ is odd, $(12)^n=mathrm{Id}$ otherwise.
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
answered Dec 7 '18 at 18:50
FedericoFederico
4,984514
4,984514
1
$begingroup$
Thank you for the help, that clarified things for me!
$endgroup$
– AMN52
Dec 7 '18 at 20:08
add a comment |
1
$begingroup$
Thank you for the help, that clarified things for me!
$endgroup$
– AMN52
Dec 7 '18 at 20:08
1
1
$begingroup$
Thank you for the help, that clarified things for me!
$endgroup$
– AMN52
Dec 7 '18 at 20:08
$begingroup$
Thank you for the help, that clarified things for me!
$endgroup$
– AMN52
Dec 7 '18 at 20:08
add a comment |
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$begingroup$
$tau^2=(5,7,6)$
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 18:34