Show that $alpha = (1 2 3)(2 3 4)(5 6 7)(7 8 9 10)$ has order $10$ in $S_n$ $(ngeq10)$.
So I have done problem #13 in section 7.5 of t book Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford (ISBN: 978-1-1115696-2-4). multiple times over now, but I still get the order of $alpha$ is 6. Here is some of my work:
$alpha = (12)(34)(5678910)$
$alpha^2 = (579)(6810)$
$alpha^3 = (12)(34)(58)(69)(710)$
$alpha^6 = alpha^3 • alpha^3 = [(12)(34)(58)(69)(710)] • [(12)(34)(58)(69)(710)] = (1)$
Even by Theorem 7.25 I am getting that the order of $alpha$ is 6. Theorem 7.25 states:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Is there something I am doing incorrectly, or is the book wrong?
P.S. I tried to post images of my work and the book question but “I need at least 10 reputation to post images” or so the error message said.
abstract-algebra permutations least-common-multiple
asked Dec 7 at 22:22
AMN52
326
|
show 1 more comment
So I have done problem #13 in section 7.5 of t book Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford (ISBN: 978-1-1115696-2-4). multiple times over now, but I still get the order of $alpha$ is 6. Here is some of my work:
$alpha = (12)(34)(5678910)$
$alpha^2 = (579)(6810)$
$alpha^3 = (12)(34)(58)(69)(710)$
$alpha^6 = alpha^3 • alpha^3 = [(12)(34)(58)(69)(710)] • [(12)(34)(58)(69)(710)] = (1)$
Even by Theorem 7.25 I am getting that the order of $alpha$ is 6. Theorem 7.25 states:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Is there something I am doing incorrectly, or is the book wrong?
P.S. I tried to post images of my work and the book question but “I need at least 10 reputation to post images” or so the error message said.
abstract-algebra permutations least-common-multiple
asked Dec 7 at 22:22
AMN52
326
Those aren't disjoint cycles in the title. What does your decomposition into disjoint cycles look like?
– jmerry
Dec 7 at 22:32
The theorem says "disjoint".
– Tobias Kildetoft
Dec 7 at 22:33
Which book are you referring to?
– Shaun
Dec 7 at 22:36
Please edit the question to include which book you mean.
– Shaun
Dec 7 at 22:41
The book is Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford with ISBN: 978-1-1115696-2-4.
– AMN52
Dec 7 at 22:44
|
show 1 more comment
So I have done problem #13 in section 7.5 of t book Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford (ISBN: 978-1-1115696-2-4). multiple times over now, but I still get the order of $alpha$ is 6. Here is some of my work:
$alpha = (12)(34)(5678910)$
$alpha^2 = (579)(6810)$
$alpha^3 = (12)(34)(58)(69)(710)$
$alpha^6 = alpha^3 • alpha^3 = [(12)(34)(58)(69)(710)] • [(12)(34)(58)(69)(710)] = (1)$
Even by Theorem 7.25 I am getting that the order of $alpha$ is 6. Theorem 7.25 states:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Is there something I am doing incorrectly, or is the book wrong?
P.S. I tried to post images of my work and the book question but “I need at least 10 reputation to post images” or so the error message said.
abstract-algebra permutations least-common-multiple
asked Dec 7 at 22:22
AMN52
326
So I have done problem #13 in section 7.5 of t book Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford (ISBN: 978-1-1115696-2-4). multiple times over now, but I still get the order of $alpha$ is 6. Here is some of my work:
$alpha = (12)(34)(5678910)$
$alpha^2 = (579)(6810)$
$alpha^3 = (12)(34)(58)(69)(710)$
$alpha^6 = alpha^3 • alpha^3 = [(12)(34)(58)(69)(710)] • [(12)(34)(58)(69)(710)] = (1)$
Even by Theorem 7.25 I am getting that the order of $alpha$ is 6. Theorem 7.25 states:
The order of a permutation $tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $tau$.
Is there something I am doing incorrectly, or is the book wrong?
P.S. I tried to post images of my work and the book question but “I need at least 10 reputation to post images” or so the error message said.
abstract-algebra permutations least-common-multiple
abstract-algebra permutations least-common-multiple
asked Dec 7 at 22:22
AMN52
326
asked Dec 7 at 22:22
AMN52
326
edited Dec 7 at 22:50
asked Dec 7 at 22:22
AMN52
326
asked Dec 7 at 22:22
AMN52
326
asked Dec 7 at 22:22
AMN52
326
326
Those aren't disjoint cycles in the title. What does your decomposition into disjoint cycles look like?
– jmerry
Dec 7 at 22:32
The theorem says "disjoint".
– Tobias Kildetoft
Dec 7 at 22:33
Which book are you referring to?
– Shaun
Dec 7 at 22:36
Please edit the question to include which book you mean.
– Shaun
Dec 7 at 22:41
The book is Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford with ISBN: 978-1-1115696-2-4.
– AMN52
Dec 7 at 22:44
|
show 1 more comment
Those aren't disjoint cycles in the title. What does your decomposition into disjoint cycles look like?
– jmerry
Dec 7 at 22:32
The theorem says "disjoint".
– Tobias Kildetoft
Dec 7 at 22:33
Which book are you referring to?
– Shaun
Dec 7 at 22:36
Please edit the question to include which book you mean.
– Shaun
Dec 7 at 22:41
The book is Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford with ISBN: 978-1-1115696-2-4.
– AMN52
Dec 7 at 22:44
Those aren't disjoint cycles in the title. What does your decomposition into disjoint cycles look like?
– jmerry
Dec 7 at 22:32
Those aren't disjoint cycles in the title. What does your decomposition into disjoint cycles look like?
– jmerry
Dec 7 at 22:32
The theorem says "disjoint".
– Tobias Kildetoft
Dec 7 at 22:33
The theorem says "disjoint".
– Tobias Kildetoft
Dec 7 at 22:33
Which book are you referring to?
– Shaun
Dec 7 at 22:36
Which book are you referring to?
– Shaun
Dec 7 at 22:36
Please edit the question to include which book you mean.
– Shaun
Dec 7 at 22:41
Please edit the question to include which book you mean.
– Shaun
Dec 7 at 22:41
The book is Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford with ISBN: 978-1-1115696-2-4.
– AMN52
Dec 7 at 22:44
The book is Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford with ISBN: 978-1-1115696-2-4.
– AMN52
Dec 7 at 22:44
|
show 1 more comment
1 Answer
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You're right. The decomposition into disjoint cycles yields
$$(1, 2, 3)(2, 3 ,4)(5, 6, 7)(7 ,8, 9, 10)=(1,2)(3,4) (5,6,7,8,9,10).$$
answered Dec 7 at 22:38
Bernard
118k638112
add a comment |
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1 Answer
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1 Answer
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You're right. The decomposition into disjoint cycles yields
$$(1, 2, 3)(2, 3 ,4)(5, 6, 7)(7 ,8, 9, 10)=(1,2)(3,4) (5,6,7,8,9,10).$$
answered Dec 7 at 22:38
Bernard
118k638112
add a comment |
You're right. The decomposition into disjoint cycles yields
$$(1, 2, 3)(2, 3 ,4)(5, 6, 7)(7 ,8, 9, 10)=(1,2)(3,4) (5,6,7,8,9,10).$$
answered Dec 7 at 22:38
Bernard
118k638112
add a comment |
You're right. The decomposition into disjoint cycles yields
$$(1, 2, 3)(2, 3 ,4)(5, 6, 7)(7 ,8, 9, 10)=(1,2)(3,4) (5,6,7,8,9,10).$$
answered Dec 7 at 22:38
Bernard
118k638112
You're right. The decomposition into disjoint cycles yields
$$(1, 2, 3)(2, 3 ,4)(5, 6, 7)(7 ,8, 9, 10)=(1,2)(3,4) (5,6,7,8,9,10).$$
answered Dec 7 at 22:38
Bernard
118k638112
answered Dec 7 at 22:38
Bernard
118k638112
answered Dec 7 at 22:38
Bernard
118k638112
answered Dec 7 at 22:38
Bernard
118k638112
118k638112
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Those aren't disjoint cycles in the title. What does your decomposition into disjoint cycles look like?
– jmerry
Dec 7 at 22:32
The theorem says "disjoint".
– Tobias Kildetoft
Dec 7 at 22:33
Which book are you referring to?
– Shaun
Dec 7 at 22:36
Please edit the question to include which book you mean.
– Shaun
Dec 7 at 22:41
The book is Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford with ISBN: 978-1-1115696-2-4.
– AMN52
Dec 7 at 22:44