a group of even order contains an element of order 2
Every group of even order contains an element of order .
Let be a group of even order, and consider the set . We claim that is even; to see this, let , so that ; since , we see that as well. Thus the elements of may be exhausted by repeatedly selecting an element and it with its inverse, from which it follows that is a multiple (http://planetmath.org/Divisibility) of (i.e., is even). Now, because and , it must be that , which, because is even, implies that is also even. The identity element of is in , being its own inverse, so the set is nonempty, and consequently must contain at least two distinct elements; that is, there must exist some , and because , we have , hence . Thus is an element of order in . ∎
Notice that the above is logically equivalent to the assertion that a group of even order has a non-identity element that is its own inverse.
|Title||a group of even order contains an element of order 2|
|Date of creation||2013-03-22 17:11:37|
|Last modified on||2013-03-22 17:11:37|
|Last modified by||azdbacks4234 (14155)|