Given permutation $\alpha \in S_n$; cycles formed by sequence $1, \alpha, \cdots$, are either identical or disjoint.

Question

In the second chapter of the book by R.P.Burn, titled: Groups: A path to geometry, it is given in q. 16 ( scanned image of question: https://i.stack.imgur.com/vtg24.jpg , scanned image of answer: https://i.stack.imgur.com/7iIxj.jpg ) to show that given a group $S_n, n \in \mathbb{N}$, and an arbitrary permutation $\alpha \in S_n$; the cycles formed by the sequence $1, \alpha, \alpha^2, \alpha^3, \alpha^4, \cdots$, are either identical or disjoint.

The approach taken by the question is to assume two non-disjoint sequences:
(i) $A : \ a, a\alpha, a\alpha^2, a\alpha^3, a\alpha^4, \cdots$,
(ii) $B : \ b, b\alpha, b\alpha^2, b\alpha^3, b\alpha^4, \cdots$,

such that $a\alpha^i = b\alpha^j$.
The book states to show:
(i) If $i\ge j$, identify second sequence with in the first.

(ii) Also, as the two sequences are non-disjoint, so identify first sequence with in the second.

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Issue:

1. The assumption of two non-disjoint sequences $A, B$ with different indexes $i, \ j$ is confusing. I feel at most a contradiction can be used to show that $i=j$.

2. Unable to grasp the two liner answer stated:
   If $a\alpha^i = b\alpha^j$, then $a\alpha^{i -j} = b$ and $a\alpha^{i -j+k} = b\alpha^k$. If $a\alpha^m = a$, then $b\alpha^{j-i+m} = a$.

Answer 1

Since they are not disjoint, we know that we can find $i$ and $j$ such taht

$$a\alpha^i = b \alpha^j$$ We can assume $i \ge j$, suppose not, swap the role of $A$ and $B$.

First thing that we want to check, does $b$ appear in $A$?

Answer is yes, since we can write $a\alpha^{i-j} = b$ and we can see that $i-j \ge 0$, hence $b$ is an element in $B$ and hence all subsequent terms in $B$ appears in $A$ as well since $b \alpha^k$ can be writtean as $a\alpha^{i-j+k}$.

We assume that the cycles in $A$ is of length $m$. We can write $a\alpha^m = a$.

Does $a$ appear in $B$?

Answer is yes again since $a = b\alpha^{j-i}=b\alpha^{j-i+m}=b\alpha^{j-i+rm}$ where we can pick $r$ to be large enough such that $j-i+rm \ge 0$.