Given a cycle $c in S_n $ with $ ord(c) = s $ and $ s = kt $, prove that $c^k$ is a product of $k$ cycles of...
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I came across this question in a recent exam.
Given that $ ord(c) = s $, we assume that $c^s = c^{kt} = (id) implies (c^{k})^t = (id)$.
That means that $c^k$ is a cycle of order $t$.
Can you please help me on the next step?
abstract-algebra group-theory permutations permutation-cycles
asked Jan 16 at 15:17
ntua_mathntua_math
85
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closed as off-topic by Shaun, Alexander Gruber♦ Jan 17 at 3:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I came across this question in a recent exam.
Given that $ ord(c) = s $, we assume that $c^s = c^{kt} = (id) implies (c^{k})^t = (id)$.
That means that $c^k$ is a cycle of order $t$.
Can you please help me on the next step?
abstract-algebra group-theory permutations permutation-cycles
asked Jan 16 at 15:17
ntua_mathntua_math
85
$endgroup$
closed as off-topic by Shaun, Alexander Gruber♦ Jan 17 at 3:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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There appears to be a typo in the subject line. You probably want $s=kt$, not $c=kt$.
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– Derek Holt
Jan 16 at 16:14
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You are right. I changed it. @DerekHolt
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– ntua_math
Jan 16 at 16:18
add a comment |
$begingroup$
I came across this question in a recent exam.
Given that $ ord(c) = s $, we assume that $c^s = c^{kt} = (id) implies (c^{k})^t = (id)$.
That means that $c^k$ is a cycle of order $t$.
Can you please help me on the next step?
abstract-algebra group-theory permutations permutation-cycles
asked Jan 16 at 15:17
ntua_mathntua_math
85
$endgroup$
I came across this question in a recent exam.
Given that $ ord(c) = s $, we assume that $c^s = c^{kt} = (id) implies (c^{k})^t = (id)$.
That means that $c^k$ is a cycle of order $t$.
Can you please help me on the next step?
abstract-algebra group-theory permutations permutation-cycles
abstract-algebra group-theory permutations permutation-cycles
asked Jan 16 at 15:17
ntua_mathntua_math
85
asked Jan 16 at 15:17
ntua_mathntua_math
85
edited Jan 16 at 16:17
ntua_math
asked Jan 16 at 15:17
ntua_mathntua_math
85
asked Jan 16 at 15:17
ntua_mathntua_math
85
asked Jan 16 at 15:17
ntua_mathntua_math
85
85
closed as off-topic by Shaun, Alexander Gruber♦ Jan 17 at 3:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shaun, Alexander Gruber♦ Jan 17 at 3:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
There appears to be a typo in the subject line. You probably want $s=kt$, not $c=kt$.
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– Derek Holt
Jan 16 at 16:14
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You are right. I changed it. @DerekHolt
$endgroup$
– ntua_math
Jan 16 at 16:18
add a comment |
1
$begingroup$
There appears to be a typo in the subject line. You probably want $s=kt$, not $c=kt$.
$endgroup$
– Derek Holt
Jan 16 at 16:14
$begingroup$
You are right. I changed it. @DerekHolt
$endgroup$
– ntua_math
Jan 16 at 16:18
1
1
$begingroup$
There appears to be a typo in the subject line. You probably want $s=kt$, not $c=kt$.
$endgroup$
– Derek Holt
Jan 16 at 16:14
$begingroup$
There appears to be a typo in the subject line. You probably want $s=kt$, not $c=kt$.
$endgroup$
– Derek Holt
Jan 16 at 16:14
$begingroup$
You are right. I changed it. @DerekHolt
$endgroup$
– ntua_math
Jan 16 at 16:18
$begingroup$
You are right. I changed it. @DerekHolt
$endgroup$
– ntua_math
Jan 16 at 16:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Power of a cycle may not be a cycle
Your sentence That means that $c^k$ is a cycle of order $t$. is not true.
Take $S_4$ as an example. $c = (1 2 3 4)$ is a cycle of order 4. However, $c^2 = (1 3) (2 4)$ is not a cycle. It is a product of two cycles each one of order 2.
It can be proven that for a cycle $c$ of order $k$, $c^i$ is a cycle if and only if $i, k$ are coprime integers.
Coming back to your case
Using what I mentioned above, if $ord(c) = s$ and $s=kt$, $c^k$ is not a cycle if $1 < k <s$.
If $c = (a_0 dots a_k a_{k+1} dots a_{2k} dots a_{(t-1)k} dots a_{tk-1})$ is a cycle or order $s=kt$, then you'll be able to prove that $c^k$ is the product of the cycles
$$c= (a_0 a_k dots a_{(t-1)k})(a_1 a_{k+1} dots a_{(t-1)k+1}) dots (a_{k-1} a_{2k-1} dots a_{tk-1})$$
each one being of order $t$.
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
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$begingroup$
Thanks a lot!!! @mathcounterexamples.net
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– ntua_math
Jan 16 at 17:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Power of a cycle may not be a cycle
Your sentence That means that $c^k$ is a cycle of order $t$. is not true.
Take $S_4$ as an example. $c = (1 2 3 4)$ is a cycle of order 4. However, $c^2 = (1 3) (2 4)$ is not a cycle. It is a product of two cycles each one of order 2.
It can be proven that for a cycle $c$ of order $k$, $c^i$ is a cycle if and only if $i, k$ are coprime integers.
Coming back to your case
Using what I mentioned above, if $ord(c) = s$ and $s=kt$, $c^k$ is not a cycle if $1 < k <s$.
If $c = (a_0 dots a_k a_{k+1} dots a_{2k} dots a_{(t-1)k} dots a_{tk-1})$ is a cycle or order $s=kt$, then you'll be able to prove that $c^k$ is the product of the cycles
$$c= (a_0 a_k dots a_{(t-1)k})(a_1 a_{k+1} dots a_{(t-1)k+1}) dots (a_{k-1} a_{2k-1} dots a_{tk-1})$$
each one being of order $t$.
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Thanks a lot!!! @mathcounterexamples.net
$endgroup$
– ntua_math
Jan 16 at 17:00
add a comment |
$begingroup$
Power of a cycle may not be a cycle
Your sentence That means that $c^k$ is a cycle of order $t$. is not true.
Take $S_4$ as an example. $c = (1 2 3 4)$ is a cycle of order 4. However, $c^2 = (1 3) (2 4)$ is not a cycle. It is a product of two cycles each one of order 2.
It can be proven that for a cycle $c$ of order $k$, $c^i$ is a cycle if and only if $i, k$ are coprime integers.
Coming back to your case
Using what I mentioned above, if $ord(c) = s$ and $s=kt$, $c^k$ is not a cycle if $1 < k <s$.
If $c = (a_0 dots a_k a_{k+1} dots a_{2k} dots a_{(t-1)k} dots a_{tk-1})$ is a cycle or order $s=kt$, then you'll be able to prove that $c^k$ is the product of the cycles
$$c= (a_0 a_k dots a_{(t-1)k})(a_1 a_{k+1} dots a_{(t-1)k+1}) dots (a_{k-1} a_{2k-1} dots a_{tk-1})$$
each one being of order $t$.
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Thanks a lot!!! @mathcounterexamples.net
$endgroup$
– ntua_math
Jan 16 at 17:00
add a comment |
$begingroup$
Power of a cycle may not be a cycle
Your sentence That means that $c^k$ is a cycle of order $t$. is not true.
Take $S_4$ as an example. $c = (1 2 3 4)$ is a cycle of order 4. However, $c^2 = (1 3) (2 4)$ is not a cycle. It is a product of two cycles each one of order 2.
It can be proven that for a cycle $c$ of order $k$, $c^i$ is a cycle if and only if $i, k$ are coprime integers.
Coming back to your case
Using what I mentioned above, if $ord(c) = s$ and $s=kt$, $c^k$ is not a cycle if $1 < k <s$.
If $c = (a_0 dots a_k a_{k+1} dots a_{2k} dots a_{(t-1)k} dots a_{tk-1})$ is a cycle or order $s=kt$, then you'll be able to prove that $c^k$ is the product of the cycles
$$c= (a_0 a_k dots a_{(t-1)k})(a_1 a_{k+1} dots a_{(t-1)k+1}) dots (a_{k-1} a_{2k-1} dots a_{tk-1})$$
each one being of order $t$.
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
Power of a cycle may not be a cycle
Your sentence That means that $c^k$ is a cycle of order $t$. is not true.
Take $S_4$ as an example. $c = (1 2 3 4)$ is a cycle of order 4. However, $c^2 = (1 3) (2 4)$ is not a cycle. It is a product of two cycles each one of order 2.
It can be proven that for a cycle $c$ of order $k$, $c^i$ is a cycle if and only if $i, k$ are coprime integers.
Coming back to your case
Using what I mentioned above, if $ord(c) = s$ and $s=kt$, $c^k$ is not a cycle if $1 < k <s$.
If $c = (a_0 dots a_k a_{k+1} dots a_{2k} dots a_{(t-1)k} dots a_{tk-1})$ is a cycle or order $s=kt$, then you'll be able to prove that $c^k$ is the product of the cycles
$$c= (a_0 a_k dots a_{(t-1)k})(a_1 a_{k+1} dots a_{(t-1)k+1}) dots (a_{k-1} a_{2k-1} dots a_{tk-1})$$
each one being of order $t$.
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Jan 16 at 16:29
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 16:20
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
$begingroup$
Thanks a lot!!! @mathcounterexamples.net
$endgroup$
– ntua_math
Jan 16 at 17:00
add a comment |
$begingroup$
Thanks a lot!!! @mathcounterexamples.net
$endgroup$
– ntua_math
Jan 16 at 17:00
$begingroup$
Thanks a lot!!! @mathcounterexamples.net
$endgroup$
– ntua_math
Jan 16 at 17:00
$begingroup$
Thanks a lot!!! @mathcounterexamples.net
$endgroup$
– ntua_math
Jan 16 at 17:00
add a comment |
1
$begingroup$
There appears to be a typo in the subject line. You probably want $s=kt$, not $c=kt$.
$endgroup$
– Derek Holt
Jan 16 at 16:14
$begingroup$
You are right. I changed it. @DerekHolt
$endgroup$
– ntua_math
Jan 16 at 16:18