If H is a subgroup of Sn (Symmetric group of n elements), show that either a) every member is even, or b) exactly half are even.
a) We know every permutation can be broken down into a composition of 2-cycles. If every element in H has an even number of 2-cycles, then all of H is even. (I know that sounds simple, but it works because of part b)
b) Let H contain k even elements. If H contains just one odd element, that odd element can be composed with all of the even elements to form k additional odd elements. Thus there are k + 1 odd elements. However, the original odd element can be composed with itself to form an even element, thus making k + 1 even elements. Since k + 1 = k + 1, exactly half are even.
Thus, either H is all even, or exactly half of H is even. --Bcaulkin 04:28, 29 January 2009 (UTC)
