I am aware that i have posted this question before, but the comments i received did not really help silly old me. So i would appreciate it if someone can walk me through some parts of the problem. I have revised my attempt from the previous post, accounting for the comments i have received as i understood them.

Let $n>1$ be a natural number. the m-cycle $\sigma=(a_1,a_2,a_3,\dots,a_m)$ denotes a permutation in $S_n$ where $\sigma(a_1)=a_2,\sigma(a_2)=a_3,\dots, \text{ and }\sigma(a_m)=a_1$.

Prove by induction on $k \geq 1$ that if $k+i\equiv j (mod\text{ m})$, then $\sigma^k(a_i)=a_j$ whenever $1\leq i \leq m$ and $1\leq j \leq m$.

Pained Attempt :

Let $n>1$ be a natural number. the m-cycle $\sigma=(a_1,a_2,a_3,\dots,a_m)$ denotes a permutation in $S_n$ where $\sigma(a_1)=a_2,\sigma(a_2)=a_3,\dots, \text{ and }\sigma(a_m)=a_1$.

[Base Step]

For the base case,$k=1$, assume that $1+i\equiv j (mod\text{ m})$ and prove that $\sigma^1(a_i)=a_j$ whenever $1\leq i \leq m$ and $1\leq j \leq m$.

Since $1+i\equiv j (mod\text{ m})$ then $j \equiv 1+i (mod\text{ m})$ and $m | j-(1+i)$ hence $j = mh+i+1$ for some $h \in \Bbb{Z}$.

[case:$1 \leq i \lt m$]

Since $j=mh+i+1$ then $a_j=a_{mh+i+1}$. By assumption $\sigma(a_{mh+i})=a_{mh+i+1}=a_j$.

I am not sure how to proceed further here..

I thought: With $1 \leq i \lt m$ and mh+i+1 being constrained to lie between 1 and m given the definition of the m-cycle, this implies that h must be 0 (?).Therefore $\sigma(a_{m(0)+i})=a_{m(0)+i+1}=a_j$ and $\sigma(a_{i})=a_{i+1}=a_j$ as required (?)

...this feels iffy.

[case: $i=m$]

Since $j=mh+i+1$ then $a_j=a_{mh+m+1}$ and $a_j=a_{m(h+1)+1}$ . By assumption $\sigma(a_{m(h+1)})=a_{m(h+1)+1}=a_j$.

I am not sure how to proceed further either..

Any help would be much appreciated.

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    $\begingroup$ When reposting a question, you should link to the old question. It's also good to be as specific as possible about what you didn't understand about the given responses. $\endgroup$ – Theo Bendit May 2 '19 at 7:33

You are getting hung up on completely irrelevant details. Once you know $k$ and $i$, you know $j$. There's no need to delve into modular arithmetic and introduce new variables. Here's a complete proof of the base step:

Base case:

Let $k=1$ and let $1\leq i,j\leq m$ be such that $k+i\equiv j\pmod{m}$. We distinguish two cases:

  1. If $1\leq i<m$ then $1<k+i\leq m$ and so $j=k+i=1+i$. Then $\sigma^1(a_i)=\sigma(a_i)=a_{i+1}=a_j$.
  2. If $i=m$ then $k+i=m+1$ and so $j=1$. Then $\sigma^1(a_i)=\sigma(a_m)=a_1=a_j$.

Can you now prove the induction step?

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    $\begingroup$ You are given as the definition of $\sigma$ that $\sigma(a_m)(=\sigma^1(a_m))=a_1$. That step is where the modulo comes in. $\endgroup$ – Robert Shore May 2 '19 at 8:15
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    $\begingroup$ @HalfAFoot Since $k=1$ and $i=m$, it is immediate from the definition that $$1+m\equiv j\pmod{m}.$$ Everything you wrote before that is completely irrelevant. Further, the implication $$m\mid(j-1)\quad\implies\quad m\leq(j-1),$$ is false. Instead simply note that $$1+m\equiv 1\pmod{m}\qquad\text{ so }\qquad j\equiv1\pmod{m}.$$ Because $1\leq j\leq m$ it follows that $j=1$. Do not focus on the modular arithmetic, it is just a formality here. $\endgroup$ – Servaes May 2 '19 at 10:20
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    $\begingroup$ What do you mean by 'artificially choosing'? For any value of $x$ the congruence $$j\equiv x\pmod{m},$$ has a unique solution $j$ with $1\leq j\leq m$. For $x=1$ this is clearly $j=1$. $\endgroup$ – Servaes May 2 '19 at 10:44
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    $\begingroup$ I do not understand your question. It should be more than obvious that $1+m\equiv1\pmod{m}$ by definition. $\endgroup$ – Servaes May 2 '19 at 10:51
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    $\begingroup$ @Servaes, you are ofcourse right..! Thank you very much for walking me through this :) ...i did not realise your point about there being a unique solution j with 1<=j<=m..after flipping through my book the closest result i have found is : For each integer a there is exactly one integer r in the list 0,...,m-1 s.t. a is congruent to r (modulo m) or more precisely : 0<=r<m...Thanks again $\endgroup$ – HalfAFoot May 2 '19 at 11:02

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