Prove that if $d \leq n$, then $S_n$ contains elements of order $d$.

Question

I've came up with a proof to the above problem, but I'm stuck on whether or not this should be done with induction instead. I get stuck in this pit a lot (the "is or isn't this enough for a proof") so I wanted to see what others think.

Here's my proof:

Let $d \leq n$. Consider the the element $x\in S_n$ where

$$x= \begin{bmatrix} 1 & 2 & 3 & \dots & (n-d) & (n-d+1) & (n-d+2) & \dots & (n-1) & n \\ 1 & 2 & 3 & \dots & (n-d) & n & (n-d+1) & \dots & (n-2) & (n-1) \\ \end{bmatrix}$$ which cycles the last $d$-many elements, that is, shifts the last $d$-many elements 'one spot' to the right, with the very last element being mapped to the start of the cycle again (i.e. $(n-d+1)$). It clearly follows that $x^{d}$ corresponds to shifting each element $d$-many times, returning it to its starting position, that is, $x^d=e$. Since any choice of $0 \leq d \leq n$ can be made, this concludes the proof. $\square$

As for the matrix notation (which is used in Chapter 0 by Aluffi), the element in the top row is mapped to the element directly below it -- so the entire "matrix" is merely notation for the element its representing (i.e. the bijective function in $\text{Aut}_{\textbf{Set}}(A)$ for some set $A$.)

Just in case it's not clear what I'm asking: I just want someone to verify this proof is valid, and if not, what I could do to improve it.

Answer 1

"Obviously $x$ is of order $d$" is clearly enough for a homework assignment or a paper or an exercise. However, to prove it formally, I see no other way to do so than describing $x^k$ by induction, and showing, with this description, that $x^k$ isn't the identity unless $k=0$ mod $d$.

Answer 2

If it is of your interest, here is another aproach:

By Cayley's Theorem, there exists a monomorfism $f:\mathbb{Z}/d\mathbb{Z}\to S_d$ and there is a natural monomorfism $g:S_d\to S_n$. $g(f(1))$ have order $d$ in $S_n$

Answer 3

There are several copies of $S_d$ inside $S_n$.

The most natural one is the set of permutations $\sigma$ such that $\sigma(i)=i$ for $i>d$.

Your choice is just the set of permutations $\sigma$ such that $\sigma(i)=i$ for $i\le n-d$.

In both cases, the permutation in $S_n$ corresponding to the $d$-cycle $(12\cdots d)$ in $S_d$ has order $d$.