I was reading a text about permutations and cycles. At a certain point, the author says that
Let $n,k \in \mathbb{N},$ $X = \{a_1,a_2,\dots,a_n\}$ and $\gamma \in S_{n}$ such that $\gamma = \begin{pmatrix}a_1 & a_2 & \dots & a_k \end{pmatrix}.$ Observe that $ \gamma = \begin{pmatrix}a_1 & a_2 & \dots & a_k \end{pmatrix} = \begin{pmatrix}a_2 & a_3 & \dots & a_k & a_1 \end{pmatrix} = \dots = \begin{pmatrix}a_k & a_1 & a_2 & \dots & a_{k-1} \end{pmatrix}.$
This seemed pretty intuitive but the proof of this equality is omitted.
I tried to prove this fact, but I’m really uncertain of the validity of my proof, so I’m gonna share with you what I have worked and I am really looking for a feedback of what needs to be improved.
Let $n,k \in \mathbb{N}$, such that $k \leq n$, $X = \{a_1,a_2,\dots,a_n\}$ and $\gamma \in S_n$ such that $\gamma = \begin{pmatrix}a_1 & a_2 & \dots & a_k \end{pmatrix}.$
Let $i \in \mathbb{N}$ such that $1 < i \leq k$ and consider $\pi \in S_n$ such that $\pi = \begin{pmatrix}a_i & \dots & a_k & a_1 & \dots & a_{i-1} \end{pmatrix}.$ We want to prove that $\pi = \gamma.$
Since $\gamma$ is a cycle (of length $k$), we know that $\gamma : X \to X$ is such that $\gamma(a_j) = a_{j+1}$ for any $j \in \mathbb{N}$ such that $j < k-1$ and $\gamma(a_k) = a_1.$
Now, we note that, as well, $\pi:X \to X$ is such that $\pi(a_{i}) = a_{i+1}, \dots \pi(a_{k}) = a_1, \dots, \pi(a_{i-2}) = a_{i-1}$ and $\pi(a_{i-1}) = a_{i}.$ Then, $\pi$ is such that $\pi(a_{j}) = a_{j+1}$ for every $j \in \mathbb{N}$ such that $1 < j \leq k-1$ and $\pi(a_k) = a_1.$
Then, we have that $\pi(x) = \gamma(x)$ for every $x \in \{a_1,a_2\dots,a_k\}.$
Since, $\gamma$ and $\pi$ are both cycles, we also know that $\gamma(y)=y$ and $\pi(y)=y$ for any $y \in X \setminus \{a_1,a_2,\dots,a_k\}.$
Hence, $\gamma(x) = \pi(x),$ for every $x \in X.$ Therefore, $\gamma = \pi. \quad \square$
My doubt is that this proof seems so simple and comes really straightforward from the definition that it made me feel uncomfortable.
What do you think? Any suggestions? Thank you for your attention.
Your proof is fine.
The key is, as you have explained, to keep track of where the cycles in question send the elements of the underlying set, then demonstrate that the cycles do the same thing to each of them. It really is that simple.
Well done.