# Abstract Algebra: Dummit and Foote Exercise

1. Jun 25, 2015

### nateHI

This isn't homework, I'm just trying to refresh my memory on cyclic groups.

My question is, in this problem solution, how does ${\sigma_i}^m=1$ follow from $\sigma_i$ being disjoint?

2. Jun 25, 2015

### geoffrey159

It rather follows from $\sigma_i$ being a cycle.

It says that a permutation can be written as a product of disjoint cycle $\sigma = \sigma_1 \circ ... \circ \sigma_p$.
The question asks you to find the smallest $m \ge 1$ such that $\sigma ^ m = \text{id}$.
Since the cycles are disjoint, you can commute all the cycles, and rewriting things nicely, $m$ has to be such that $\sigma_i ^ m = \text{id}$, for all $i = 1 ... p$.
Since the order of a cycle is its length, $m$ must be a multiple of the length of each cycle $\sigma_i$. The smallest such $m$ is the least common multiple of all the lengths.

3. Jun 26, 2015

### nateHI

Got it thanks!