Rotman's Group Theory Problem 1.14 (ii) The problem is: Let $\alpha$ and $\beta$ be cycles in $S_n$. If there is $i_1$ moved by both $\alpha$ and $\beta$ and if $\alpha^k(i_1) = \beta^k(i_1)$ for all positive integers $k$, then $\alpha = \beta$.
Here is what I have at the moment. I can't see anything wrong, but I didn't use part (i) at all so I'm not sure. Please let me know if anything is wrong.
Let $\alpha = (i_1\ i_2\ ...\ i_n)$ and $\beta = (i_1\ j_2\ ...\ j_m)$ (the assumptions give us that $i_1$ is moved by both). Then
$$i_k = i_{k \mod n} = i_{1 + (k - 1)(\mod n)} = \alpha^{k - 1}(i_1).$$
So
$$\alpha(i_k) = \alpha^k(i_1) = \beta^k(i_1).$$
Next, for $k \geq 2$,
$$j_k = i_{1 + (k - 1)(\mod n)} = \beta^{k - 1}(i_1)$$
so 
$$\beta(j_k) = \beta^k(i_1)$$
and so $\beta(j_k) = \alpha(i_k)$ for each $k \geq 2$. But then we must have that $m = n$ and $j_k = i_k$ for all $k \geq 1$.
 A: This is correct but somewhat sloppy in my opinion. Here's how I would have gone about it.

From $\alpha = (i_1, i_2, \dots, i_n)$ and $\beta = (i_1, j_2, \dots, j_m)$ we prove $m = n$.
Note that $n$ is the smallest number $\ge 2$ such that $\alpha^{n}(i_1) = i_1$ and $m$ is the smallest number $\ge 2$ such that $\beta^{m}(i_1) = i_1$. Since $\alpha^k(i_1) = i_1$ iff $\beta^k(i_1) = i_1$, we conclude that $m = n$.
Next for $k \in \{2,\dots,n\}$ one has $$i_k = \alpha^{k - 1}(i_1) = \beta^{k - 1}(i_1) = j_k.$$
Hence $\alpha = \beta$.

In particular, I'm not a fan of your use of $\bmod n$ and $\bmod m$. The typesetting is funny for one. For two, we already know that $i_k = \alpha^{k - 1}(i_1)$ for $1 \le k \le n$ (and similar for $\beta$ and $j_k$) by definition so I'm not sure why you spend 4 lines doing this.
I will also comment that the statement Let "$\alpha = (i_1, i_2, \dots, i_n)$ and $\beta =(i_1, j_2, \dots, j_m)$ (the assumptions give us that $i_1$ is moved by both)." Could use some more justification. Here's how to justify it:

Let $(i_1, i_2, \dots, i_n)$ be the unique cycle in $\alpha$ containing $i_1$. We know $n \ge 2$ since $\alpha(i_1) \ne i_1$. Since $\alpha$ only has one non-trivial cycle it follows that $\alpha = (i_1, i_2, \dots, i_n)$. Similarly $\beta =(i_1, j_2, \dots, j_m)$.

