h6 - | H K | = 1, and so H K = { 1 } . 2 . 58 . Prove that...

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1 Homework VI: July 14, 2009 2 . 46 . If G is a group with an even number of elements, prove that the number of elements in G of order 2 is odd. In particular, G must contain an element of order 2. G is the disjoint union of X and Y , where X = { g G : g - 1 6 = g } and Y = { g G : g - 1 = g } . Now | X | is even, for it is the disjoint union of 2-points subsets, each of the form { g,g - 1 } consisting of an element and its inverse. Since | G | is even, we conclude that | Y | is even. But 1 Y , and so there is an odd number of elements g with g 6 = 1 and g = g - 1 . The last condition says that g 2 = 1; since g 6 = 1, we have g of order 2. 2 . 57 . If H and K are subgroups of a group G and | H | and | K | are relative prime, prove that H K = { 1 } . Observe that H K is a subgroup of H and, also, a subgroup of K . By Lagrange’s Theorem, | H K | | | H | and | H K | | | K | . Since | H | and | K | are relative prime, it follows that
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Unformatted text preview: | H K | = 1, and so H K = { 1 } . 2 . 58 . Prove that an innite group G contains innitely many subgroups. Every group is the union of its cyclic subgroups: in particular, G = [ a G h a i . Suppose that G has only nitely many subgroups: G = h a 1 i h a n i . If every h a i i is nite, then G is nite, a contradiction. Hence, at least one a i is innite. But the subgroups h a n i i , for n = 1 , 2 , 3 ,... are all distinct: if n < k , then a n / h a k i . Therefore, h a i i , and hence G , has innitely many subgroups, a contradiction. We conclude that G has innitely many subgroups....
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This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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