the number of elements x in a group that do not satisfy x=x^{-1}

• Oct 8th 2011, 11:50 AM
VonNemo19
the number of elements x in a group that do not satisfy x=x^{-1}
Hi. I'm trying to prove the following:

Let $G$ be a finite group, and let $S=\{x\in{G}|x\neq{x}^{-1}\}$. Show that the set $S$ has an even number of elements.

So, Here's my thoughts.

Let the number of all elements $x$ such that $x\neq{x}^{-1}$ be $m$ so that there are $m$ elements in $G$ not equal to their own inverses. But, $G$ is a group and so there exists for every element $x\in{G}$ a $x^{-1}\in{G}$ such that $x*x^{-1}=e$ the identity element of $G$. Thus, there are also $m$ inverses corresponding to the $m$ elements belonging to $S$, but these elements also belong to $S$ since $(x^{-1})^{-1}=x$, Therefore, there are $2m$ elements in $S$. Therefore, there are an even number of elements in $S$

Is this Right. The thing that concerns me is the fact that from this argument $m=2m$. Can someone help me phrase this better? I'm new to this subject and I've only recently became mature enough to construct proofs like this.
• Oct 8th 2011, 12:24 PM
Deveno
Re: the number of elements x in a group that do not satisfy x=x^{-1}
you have the right idea. since the elements of S do not equal there own inverses, elements of S occur in G "in pairs" {x,x^-1}.

by pairing every element of S with it's inverse (which must also be in S, right?), if we had an odd number of things in S, we would wind up with "one left over".

but this one "left over" would then have to be its own inverse, a contradiction.
• Oct 9th 2011, 07:00 AM
VonNemo19
Re: the number of elements x in a group that do not satisfy x=x^{-1}
Yes, I'm pretty sure that my argument is sound in its essence, but ...
Quote:

Originally Posted by VonNemo19
The thing that concerns me is the fact that from this argument $m=2m$.

Do you see what I mean?
• Oct 9th 2011, 07:28 AM
Deveno
Re: the number of elements x in a group that do not satisfy x=x^{-1}
well it doesn't work out that way. let's say you start pairing elements from S. when you get half-way through (at m/2), you already got the other m/2 paired up with the half you took to be the x's.

look, i'll show you with an example:

say G = <x> = {e,x,x^2,x^3,x^4,x^5,x^6}.

well e = e^-1, so it's outta there!

x^-1 = x^6 ≠ x, so we have the pair (x,x^6).

(x^2)^-1 = x^5, so we have the pair (x^2,x^5), giving us S = {x,x^6,x^2,x^5,....} so far

(x^3)^-1 = x^4, so we have the pair (x^3,x^4), so now S = {x,x^6,x^2,x^5,x^3,x^4.....}

continuing, we have (x^5)^-1 = x^2, so...wait a minute! we've already counted x^5 and x^2, when we paired x^2 with x^5.

so we don't wind up with 2m elements, we wind up with 2(m/2) = m.

but what we DO establish, is that m/2 is an integer, which means m must have been even.

(that is, if we had one left over (if m/2 wasn't an integer), it would have to be its own inverse, so how could it have been in S?).