Mixing
Power of a r-cycle
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Given $n ge 3$ and $2 le r le n$, let's define the permutation $rho in S_n$ (the symmetric group of degree $n$) by:
$rho(k)=k+1$, for $k in I_<:=lbrace 1,...,r-1 rbrace$
- $rho(r)=1$
$rho(k)=k$, for $kin I_>:=lbrace r+1,...,n rbrace$
I've thought to prove that $rho^r=iota_{S_n}$ -the identical permutation- in the following manner (reductio ad absurdum). Could you double check if the proof is correct, please?
Proof. Let's suppose that $rho^r ne iota_{S_n}$; then, $exists tilde k in I:=lbrace 1,...,n rbrace$, such that $rho^r(tilde k)=rho(rho^{r-1}(tilde k)) ne tilde k$. By the very same definition of $rho$, it follows that $rho^{r-1}(tilde k) in I_<^*:=lbrace 1,...,r-1 rbrace cup lbrace r rbrace$. Now, let's assume that $rho^j(tilde k) in I_<^*$ for some $0 le j le r-2$ (inductive hypothesis); then $rho^{j-1}(tilde k)=rho^{-1}(rho^j(tilde k)) in I_<^*$, because the restriction $rho_{|I_<^*}$ is a bijection of $I_<^*$ into itself; so, finally, $rho^j(tilde k) in I_<^*$ $forall j=0,dots,r-1$. This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$, which in its turn implies $tilde k=1$: contradiction, because by definition of $rho$, it is $rho^r(1)=1$.
abstract-algebra group-theory symmetric-groups
asked 2 days ago
Luca Carlotto
204
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Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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down vote
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Given $n ge 3$ and $2 le r le n$, let's define the permutation $rho in S_n$ (the symmetric group of degree $n$) by:
$rho(k)=k+1$, for $k in I_<:=lbrace 1,...,r-1 rbrace$
- $rho(r)=1$
$rho(k)=k$, for $kin I_>:=lbrace r+1,...,n rbrace$
I've thought to prove that $rho^r=iota_{S_n}$ -the identical permutation- in the following manner (reductio ad absurdum). Could you double check if the proof is correct, please?
Proof. Let's suppose that $rho^r ne iota_{S_n}$; then, $exists tilde k in I:=lbrace 1,...,n rbrace$, such that $rho^r(tilde k)=rho(rho^{r-1}(tilde k)) ne tilde k$. By the very same definition of $rho$, it follows that $rho^{r-1}(tilde k) in I_<^*:=lbrace 1,...,r-1 rbrace cup lbrace r rbrace$. Now, let's assume that $rho^j(tilde k) in I_<^*$ for some $0 le j le r-2$ (inductive hypothesis); then $rho^{j-1}(tilde k)=rho^{-1}(rho^j(tilde k)) in I_<^*$, because the restriction $rho_{|I_<^*}$ is a bijection of $I_<^*$ into itself; so, finally, $rho^j(tilde k) in I_<^*$ $forall j=0,dots,r-1$. This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$, which in its turn implies $tilde k=1$: contradiction, because by definition of $rho$, it is $rho^r(1)=1$.
abstract-algebra group-theory symmetric-groups
asked 2 days ago
Luca Carlotto
204
New contributor
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
"This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$," Why? If you have intermediate claims, it makes sense to delimit them (and their proofs) clearly, particularly when you are trying to formalize the proof.
– darij grinberg
2 days ago
@darij grinberg - You're right, that conclusion is undue, and frankly I'm stuck there. Any chance to conclude on that way?
– Luca Carlotto
2 days ago
A quick way is by proving (by induction) that $rho^i left(kright) in left{1,2,ldots,rright}$ and $rho^i left(kright) equiv k+i mod r$ for each $k in left{1,2,ldots,rright}$
– darij grinberg
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $n ge 3$ and $2 le r le n$, let's define the permutation $rho in S_n$ (the symmetric group of degree $n$) by:
$rho(k)=k+1$, for $k in I_<:=lbrace 1,...,r-1 rbrace$
- $rho(r)=1$
$rho(k)=k$, for $kin I_>:=lbrace r+1,...,n rbrace$
I've thought to prove that $rho^r=iota_{S_n}$ -the identical permutation- in the following manner (reductio ad absurdum). Could you double check if the proof is correct, please?
Proof. Let's suppose that $rho^r ne iota_{S_n}$; then, $exists tilde k in I:=lbrace 1,...,n rbrace$, such that $rho^r(tilde k)=rho(rho^{r-1}(tilde k)) ne tilde k$. By the very same definition of $rho$, it follows that $rho^{r-1}(tilde k) in I_<^*:=lbrace 1,...,r-1 rbrace cup lbrace r rbrace$. Now, let's assume that $rho^j(tilde k) in I_<^*$ for some $0 le j le r-2$ (inductive hypothesis); then $rho^{j-1}(tilde k)=rho^{-1}(rho^j(tilde k)) in I_<^*$, because the restriction $rho_{|I_<^*}$ is a bijection of $I_<^*$ into itself; so, finally, $rho^j(tilde k) in I_<^*$ $forall j=0,dots,r-1$. This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$, which in its turn implies $tilde k=1$: contradiction, because by definition of $rho$, it is $rho^r(1)=1$.
abstract-algebra group-theory symmetric-groups
asked 2 days ago
Luca Carlotto
204
New contributor
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given $n ge 3$ and $2 le r le n$, let's define the permutation $rho in S_n$ (the symmetric group of degree $n$) by:
$rho(k)=k+1$, for $k in I_<:=lbrace 1,...,r-1 rbrace$
- $rho(r)=1$
$rho(k)=k$, for $kin I_>:=lbrace r+1,...,n rbrace$
I've thought to prove that $rho^r=iota_{S_n}$ -the identical permutation- in the following manner (reductio ad absurdum). Could you double check if the proof is correct, please?
Proof. Let's suppose that $rho^r ne iota_{S_n}$; then, $exists tilde k in I:=lbrace 1,...,n rbrace$, such that $rho^r(tilde k)=rho(rho^{r-1}(tilde k)) ne tilde k$. By the very same definition of $rho$, it follows that $rho^{r-1}(tilde k) in I_<^*:=lbrace 1,...,r-1 rbrace cup lbrace r rbrace$. Now, let's assume that $rho^j(tilde k) in I_<^*$ for some $0 le j le r-2$ (inductive hypothesis); then $rho^{j-1}(tilde k)=rho^{-1}(rho^j(tilde k)) in I_<^*$, because the restriction $rho_{|I_<^*}$ is a bijection of $I_<^*$ into itself; so, finally, $rho^j(tilde k) in I_<^*$ $forall j=0,dots,r-1$. This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$, which in its turn implies $tilde k=1$: contradiction, because by definition of $rho$, it is $rho^r(1)=1$.
abstract-algebra group-theory symmetric-groups
abstract-algebra group-theory symmetric-groups
asked 2 days ago
Luca Carlotto
204
New contributor
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Luca Carlotto
204
New contributor
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Luca Carlotto
204
New contributor
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Luca Carlotto
204
asked 2 days ago
Luca Carlotto
204
204
New contributor
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Luca Carlotto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
"This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$," Why? If you have intermediate claims, it makes sense to delimit them (and their proofs) clearly, particularly when you are trying to formalize the proof.
– darij grinberg
2 days ago
@darij grinberg - You're right, that conclusion is undue, and frankly I'm stuck there. Any chance to conclude on that way?
– Luca Carlotto
2 days ago
A quick way is by proving (by induction) that $rho^i left(kright) in left{1,2,ldots,rright}$ and $rho^i left(kright) equiv k+i mod r$ for each $k in left{1,2,ldots,rright}$
– darij grinberg
2 days ago
add a comment |
"This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$," Why? If you have intermediate claims, it makes sense to delimit them (and their proofs) clearly, particularly when you are trying to formalize the proof.
– darij grinberg
2 days ago
@darij grinberg - You're right, that conclusion is undue, and frankly I'm stuck there. Any chance to conclude on that way?
– Luca Carlotto
2 days ago
A quick way is by proving (by induction) that $rho^i left(kright) in left{1,2,ldots,rright}$ and $rho^i left(kright) equiv k+i mod r$ for each $k in left{1,2,ldots,rright}$
– darij grinberg
2 days ago
"This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$," Why? If you have intermediate claims, it makes sense to delimit them (and their proofs) clearly, particularly when you are trying to formalize the proof.
– darij grinberg
2 days ago
"This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$," Why? If you have intermediate claims, it makes sense to delimit them (and their proofs) clearly, particularly when you are trying to formalize the proof.
– darij grinberg
2 days ago
@darij grinberg - You're right, that conclusion is undue, and frankly I'm stuck there. Any chance to conclude on that way?
– Luca Carlotto
2 days ago
@darij grinberg - You're right, that conclusion is undue, and frankly I'm stuck there. Any chance to conclude on that way?
– Luca Carlotto
2 days ago
A quick way is by proving (by induction) that $rho^i left(kright) in left{1,2,ldots,rright}$ and $rho^i left(kright) equiv k+i mod r$ for each $k in left{1,2,ldots,rright}$
– darij grinberg
2 days ago
A quick way is by proving (by induction) that $rho^i left(kright) in left{1,2,ldots,rright}$ and $rho^i left(kright) equiv k+i mod r$ for each $k in left{1,2,ldots,rright}$
– darij grinberg
2 days ago
add a comment |
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Luca Carlotto is a new contributor. Be nice, and check out our Code of Conduct.
Luca Carlotto is a new contributor. Be nice, and check out our Code of Conduct.
Luca Carlotto is a new contributor. Be nice, and check out our Code of Conduct.
Luca Carlotto is a new contributor. Be nice, and check out our Code of Conduct.
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"This means that $tilde k+j in I_<^*$, $forall j=0,dots,r-1$," Why? If you have intermediate claims, it makes sense to delimit them (and their proofs) clearly, particularly when you are trying to formalize the proof.
– darij grinberg
2 days ago
@darij grinberg - You're right, that conclusion is undue, and frankly I'm stuck there. Any chance to conclude on that way?
– Luca Carlotto
2 days ago
A quick way is by proving (by induction) that $rho^i left(kright) in left{1,2,ldots,rright}$ and $rho^i left(kright) equiv k+i mod r$ for each $k in left{1,2,ldots,rright}$
– darij grinberg
2 days ago