- #1

- 207

- 0

## Homework Statement

In a finite group, show that the number of nonidentity elements that satisfy the equation x^5=e is a multiple of 4. If the stipulation that the group is finite is omitted, what can you say about the number of nonidentity elements that satisfy the equation x^5=e?

## Homework Equations

Not sure anything should be here

"e" is the identity in Gallian's notation

x^5=e?

## The Attempt at a Solution

I began typing that I wasn't sure how to show this, and now I'm doubting I understand why it's true. Involution (for groups) was covered, but I missed it and I'm hoping that I didn't miss anything relevant.

I tried some finite groups just to see how it would go:

So say we're looking at a simple group like, {1, 3, 5, 7} under multiplication modulo 8. The operation associative, it has an identity element (1), and has inverses for all of the components (each element is it's own inverse).

So besides 1 (or e here) I'll look at each element:

3^5=3 mod 8

5^5=5 mod 8

7^5=7 mod 8

(this make sense since they are all their own inverses, all even powers will be e and all odd power will be the element)

So I have 0 nonidentity elements...which is a multiple of 4, but gets me nowhere. I could try experimenting with larger groups (;O Monster group, anyone?) but I'm not sure that this would help as much as it would be tedious.

I don't know where I should go, conceptually. This is a good book, so I'm pretty sure all the information I need is present...I'm just not sure what I should be thinking about. I keep getting drawn to Cyclic groups and subgroups...I really wish I could show that like, break it into two cases. One where there are no nonidentity x such that x^5=e, and one where where there are. I have no idea if this would work though, and I can't see any easy way to test this.

Direction would be wonderful, this is the end of the problem set and I'm just burnt out.

Thanks for any and all advice,

Elwin