# Homework Help: Group set proof

1. Dec 6, 2012

### rideabike

1. The problem statement, all variables and given/known data
Let (G,*) be a finite group of even order. Prove that there exists some g in G such that g≠e and g*g=e. [where e is the identity for (G,*)]

2. Relevant equations
Group properties

3. The attempt at a solution
Let S = G - {e}. Then S is of odd order, and let T={g,g^-1: g$\in$S}.
Then $\exists$h$\in$S such that h$\notin$T. Since G is a group, h*h must equal e.

Does that work?

2. Dec 6, 2012

### Michael Redei

You have a good idea here, nearly perfect, but the last step seems not obvious at first. Who says that h*h must be equal to e? All we know (since G is a group) is that h*h must lie in G. To show that h*h = e, you also need h∉S. (Then h*h ∈ G - S = {e}.)

3. Dec 6, 2012

### jbunniii

How is that possible? You have defined T in such a way that $S \subset T$. You have the right idea, though. If you partition G such that each element of g is partnered with its inverse, then e will be partnered with itself. If you assume that no other element has this property, then you should be able to obtain a contradiction.

4. Dec 6, 2012

### rideabike

If h$\notin$S, then h=e, which isn't what we want. I was trying to show that if you take all the elements and their inverses out of S, then you're left with a single element whose inverse must be itself. There's probably a better way to show this.

5. Dec 6, 2012

### Michael Redei

How about collecting all elements that aren't their own inverses in one set, i.e. S = { g∈G : g-1≠g }. Then for any element x∈G, either x ∈ G - S, or x and x-1 are both in S. "Both" is important here, since it means that S has an even number of elements.

So what's left in G - S?

6. Dec 6, 2012

### rideabike

That's good, thanks!

7. Dec 6, 2012

### micromass

It might be interesting for the OP that this result has a significant generalization: If G is a group and if p is a prime number such that p divides the order of G, then there exists an element g in G with order p.

This is called Cauchy's theorem.

8. Dec 6, 2012

### rideabike

So |x|$\in$G = p? Would this relate to a group being cyclic?

9. Dec 6, 2012

### micromass

I don't really get your notation of $|x|\in G=p$

I mean to say that there exists a $x\in G$ such that $|x|=p$ (assuming |x| means order for you).

If G has prime order, then this result shows indeed that there exists an element in G with prime order. So G must be cyclic. Is this a relationship that you are talking about?

10. Dec 6, 2012

### rideabike

Yeah that's exactly it, I was being sloppy. My understanding of a cyclic group is that there exists some element of G, say a, such that, for every x in G, x=a^n for some integer n.

How does this follow from |x|=p?

11. Dec 6, 2012

### micromass

I'm saying the following: If G has order p, then there exists an element a in G that has order p. So $\{1,a,a^2,...,a^{p-1}\}$ has p elements and is a subset of G. Since G also has p elements, we must have $G=\{1,a,a^2,...,a^{p-1}\}$. So G is cyclic.