代數學問題 show that any finite

Let G be a group of even order. Consider the set S = {g ∈ G, g ≠ g^(-1)}

Part 1: S is a group of even order

if a ∈ S, [a^(-1)]^(-1) = a ≠ a^(-1) and so a^(-1) ∈ S

Thus the elements of S may be exhausted by repeatedly selecting an element and pairing it with its inverse. So, |S| is even.

Part 2: there is an element a ∈ G such that a^2 = e

Since S ∩ (G\S) = Ø and S U (G\S) = G, it implies that

|S| + |G\S| = G => |G\S| is even

Now, e ∈ G\S and because G\S is not empty, there must be at least two distinct elements in G\S. If b ∈ G\S, then b ∉ S and so b = b^(-1). That is b^2 = e and the proof is complete.




2011-11-05 10:50:02 補充：
S的元素可以透過a<->a^(-1)的方式而被窮舉

這句不大懂Thus the elements of S may be exhausted by repeatedly selecting an element and pairing it with its inverse. So, |S| is even.