# An m-cycle has order m

## 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##

## 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

fresh_42
Mentor
2021 Award
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
Mentor
2021 Award
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.