# Thread: counting inverses of finite group

1. ## counting inverses of finite group

Hi

I am using Charles Pinter's book on abstract algebra . I have to prove that
for any finite group G, if we define the set

$S=\{x\in G : x\neq x^{-1} \}$

i.e. set of elements in G which are not equal to its own inverse. I have to prove that
set S has even no of elements. Now I can see that if some x is in S then its inverse
has to be there since

$(x^{-1})^{-1}\neq x^{-1}$

so members of S appear in pairs. So I can see that , its cardinality has to be even.
But how can I prove this ? any hints ?

thanks

2. ## Re: counting inverses of finite group

what's left to prove?

3. ## Re: counting inverses of finite group

Hi Deveno, I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...

4. ## Re: counting inverses of finite group

Originally Posted by issacnewton
I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...
Suppose that $\{x,y\}\subseteq\mathcal{S}~\&~x\ne y,~x\ne y^{-1}$ then is it true that $\left\{ {x,x^{ - 1} } \right\} \cap \left\{ {y,y^{ - 1} } \right\} = \emptyset~?$

Is it true that $\mathcal{S}=\bigcup\limits_{x \in S} {\left\{ {x,x^{ - 1} } \right\}}~?$

What does that tell about the cardinality of $\mathcal{S} ~?$

5. ## Re: counting inverses of finite group

Originally Posted by issacnewton
Hi Deveno, I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...
since $x^{-1} \in S$ whenever $x \in S$ and for all $x \in S,\ x \neq x^{-1}$, the elements of S occur in distinct pairs.

so if S has k such pairs, |S| = 2k, which is even.

unless you are required to put this in the form of a well-formed statement in some first-order formal language, that is all you need say.

6. ## Re: counting inverses of finite group

Thanks Deveno.......thats better...

Plato, its true that

$\left\{x,x^{-1}\right\}\cap \left\{y,y^{-1}\right \}=\emptyset$

so we can say that cardinality of S is some power of 2 since each subset
$\{x,x^{-1}\}$ of S has two distinct elements.... hence
the proof........right ?

7. ## Re: counting inverses of finite group

Originally Posted by issacnewton
Hi Deveno, I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...
As others have suggested, there's really nothing more to do once you notice that S is composed of pairs of inverses. I can understand the desire to express the point in a more computational manner, but that will only make the proof harder to write---and more importantly, harder to read.

By the way, an interesting result from considering S is that we can prove Cauchy's theorem for the case p=2. For if G is a group of even order, then S and the identity element make up an odd number of elements, which means there must be at least one non-identity element in G which is its own inverse, i.e. has order 2.

8. ## Re: counting inverses of finite group

Originally Posted by issacnewton
Thanks Deveno.......thats better...

Plato, its true that

$\left\{x,x^{-1}\right\}\cap \left\{y,y^{-1}\right \}=\emptyset$

so we can say that cardinality of S is some power of 2 since each subset
$\{x,x^{-1}\}$ of S has two distinct elements.... hence
the proof........right ?
Not a power of 2... a multiple of 2.

9. ## Re: counting inverses of finite group

Thanks everybody.