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

• Mr Davis 97
In summary, the conversation discusses proving that if an element in the symmetric group is an m-cycle, then its order is equal to m. The proof involves showing that the element raised to the mth power is the identity, and that m is the smallest positive integer for which this is true. The justification for this proof is also discussed, with the need to show the relationship between modular arithmetic and the group homomorphism.
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##

## 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
fresh_42 said:
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##

Mr Davis 97 said:
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.

## 1. What does it mean for an m-cycle to have order m?

For an m-cycle to have order m means that the cycle consists of m elements and that applying the cycle m times will result in the identity element (e). In other words, the cycle will return to its original state after m repetitions.

## 2. How is the order of an m-cycle determined?

The order of an m-cycle is determined by the number of elements within the cycle. For example, a 3-cycle will have an order of 3, while a 5-cycle will have an order of 5.

## 3. Can an m-cycle have an order greater than m?

No, an m-cycle cannot have an order greater than m. This is because the cycle will repeat itself after m repetitions, so any additional repetitions will simply result in the same elements being cycled through again.

## 4. Is the order of an m-cycle always an integer?

Yes, the order of an m-cycle is always an integer. This is because the order is determined by the number of elements in the cycle, and the number of elements must be a whole number.

## 5. How does the order of an m-cycle relate to its group?

The order of an m-cycle is a factor of the order of the group in which it belongs. This means that the order of the group can be divided evenly by the order of the m-cycle. For example, in a group of order 12, an m-cycle with an order of 3 would be a subgroup of order 4.

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