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Problem: Show that a finite group of even order has elements of order 2

Attempt:

The book gives a suggested approach that lead me to write the most round about, ugly proof I've ever written.

Can't I just say:

1.) If G has even order, G/{1} has odd cardinality.

2.) Assume that no elements of G/{1} has order 2.

3.) Then for each x in G/{1}, x^-1 is a distinct element of G/{1}.

4.) Then G/{1} has even cardinality.

Contradiction

5.) Therefore, there exists a y in G/{1} such that y = y^-1, and thus has order 2.