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

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

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

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

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