Consider pairing up elements with their inverses.
Let (G,*) be a finite group with an even number of elements. Show that there must exist at least one element such that .
I have no clue how to start and what ticks me off about that is that I think I've seen it before. (And I don't recall it was that difficult, either.) However my class notes (from years ago) doesn't contain the proof.
-Dan
Got it. Let me formalize it then.Originally Posted by rgep
Let (G,*) be a finite group with an even number of elements. Consider the subset of G: where e is the identity in G. It may be that A is empty. If not, the inverse of any element is also in A and is distinct from b. So b and its inverse form a pair of elements in A and this is true of any element in A. Thus we know that A has no elements or has an even number of them.
Now form the set G - A. (For others, I believe the notation is G\A?) We know that G - A is not empty because . We also know that the set G - A has an even number of elements since both G and A do (calling 0 an even number). Because of this, the set G - A - {e} cannot be empty, since this would imply that G has an odd number of elements. Since any element is not in A we know that .
Thus there is at least one element a in G that is not the identity that has the property . (End of proof.)
Wow. That took a lot more writing than I thought it would.
Thanks for the tip!
-Dan
Don't you need to add to the definition of ?Originally Posted by topsquark
Take the group , where . Then but is not distinct form .where e is the identity in G. It may be that A is empty. If not, the inverse of any element is also in A and is distinct from b.
So b and its inverse form a pair of elements in A and this is true of any element in A. Thus we know that A has no elements or has an even number of them.
Now form the set G - A. (For others, I believe the notation is G\A?) We know that G - A is not empty because . We also know that the set G - A has an even number of elements since both G and A do (calling 0 an even number). Because of this, the set G - A - {e} cannot be empty, since this would imply that G has an odd number of elements. Since any element is not in A we know that .
Thus there is at least one element a in G that is not the identity that has the property . (End of proof.)
Wow. That took a lot more writing than I thought it would.
Thanks for the tip!
-Dan