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Unformatted text preview: | K | . Assume that H is a subgroup of G having order | K | . Since K C G , the quotient group G/K is dened; let : G G/K be the natural map (of course, ( a ) = aK for all a G ). If h H , then ( h ) = hK has nite order, say, n . By Exercise 2.76(i), we have n | | H | = | K | . But n | | G/K | = [ G : K ], by Lagranges Theorem, and so n is a common divisor of | K | and | G/K | . But ( | K | , [ G : K ]) = 1, so that n = 1. Thus, h ker = K , and so H K . As H and K have the same (nite) order, weh ave H = K ....
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- Spring '08