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...












1












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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?










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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.












  • 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












$begingroup$


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?










share|cite|improve this question











$endgroup$



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




    $begingroup$
    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








1





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








edited Jan 16 at 16:17







ntua_math

















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 $c=kt$.
    $endgroup$
    – Derek Holt
    Jan 16 at 16:14










  • $begingroup$
    You are right. I changed it. @DerekHolt
    $endgroup$
    – ntua_math
    Jan 16 at 16:18
















  • 1




    $begingroup$
    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












1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $t$.






share|cite|improve this answer











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  • $begingroup$
    Thanks a lot!!! @mathcounterexamples.net
    $endgroup$
    – ntua_math
    Jan 16 at 17:00


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $t$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot!!! @mathcounterexamples.net
    $endgroup$
    – ntua_math
    Jan 16 at 17:00
















0












$begingroup$

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 $t$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot!!! @mathcounterexamples.net
    $endgroup$
    – ntua_math
    Jan 16 at 17:00














0












0








0





$begingroup$

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 $t$.






share|cite|improve this answer











$endgroup$



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 $t$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 16:29

























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


















  • $begingroup$
    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



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