what's left to prove?
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
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
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
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.
Thanks everybody.
and hatsoff, thanks for the additional input about Cauchy's theorem.....
I just wanted to know more rigorous mathematical way of presenting what I thought. I had taken a class on introductory real analysis few years ago. And the teacher was very strict about presenting logical arguments. Since it was an analysis class , our arguments had to be logically watertight. Before taking the
analysis class, I used think that some things are obvious. During this class, in the beginning I lost many points on assignments because my arguments
were not as watertight. I think a class in real analysis really forces you to think logically. May be other branches of mathematics dont force such
mental habits. I am just trying to be loyal to the rigor as demanded in real analysis.
thanks