Is m the smallest order for an m-cycle in the symmetric group?

[Mr Davis 97]

Homework Statement

Let $n$ be a natural number and let $\sigma$ be an element of the symmetric group $S_n$. Show that if $\sigma$ is an m-cycle $(a_1a_2 \dots a_m)$, then $|\sigma|=m$

Homework Equations

The Attempt at a Solution

First, we want to show that $\sigma ^m = id$. To this end, we claim that for all $i \in \mathbb{N}$, $\sigma ^i (a_k) = a_{(k+i) \bmod m}$. We proceed by induction. The base case holds by the fact that $\sigma$ is an m-cycle. Next, suppose that for some $j$ we have $\sigma ^j (a_k) = a_{(k+j) \bmod m}$. Then $\sigma ^{j+1} (a_k) = \sigma (\sigma ^{j} (a_k)) = \sigma (a_{(k+j) \bmod m}) = a_{(k+(j+1)) \bmod m}$. So we have shown that for all $i \in \mathbb{N}$, $\sigma ^i (a_k) = a_{(k+i) \bmod m}$, and if we let $i=m$, we see that $\sigma ^m = id$.

Second, we want to show that $m$ is the smallest positive integer such that $\sigma ^m = id$. To the contrary, suppose that there is a $p \in [1, m)$ such that $\sigma ^p = id$. Then $\sigma ^p (a_p) = a_p$ and also $\sigma ^p (a_p)= a_{(p+p) \bmod m} = a_{(2p) \bmod m}$. So $p \equiv 2p \bmod m \implies p \equiv 0 \bmod m$, which contradicts the assumption that $p \in [1, m)$.

 

[fresh_42]

This is correct.

 

[Mr Davis 97]

This is correct.

One quick question though. Have I justified that $\sigma (a_{(k+j) \bmod m}) = a_{(k+(j+1)) \bmod m}$? Do I need to show that $[(k+j) \bmod m] + 1 \bmod m = [k+(j+1)] \bmod m$

 

[fresh_42]

One quick question though. Have I justified that $\sigma (a_{(k+j) \bmod m}) = a_{(k+(j+1)) \bmod m}$? Do I need to show that $[(k+j) \bmod m] + 1 \bmod m = [k+(j+1)] \bmod m$

Depends on what you take as a basis for your proof. $\mathbb{Z} \longrightarrow \mathbb{Z}/m\mathbb{Z} = \mathbb{Z}_m\; , \;i\longmapsto i \operatorname{mod} m$ is a ring homomorphism, or here a group homomorphism, so with that it is clear. Without it, prove this first. For me this is still better than to distinguish the cases.