Adv Alegbra HW Solutions 66

Adv Alegbra HW Solutions 66 - 66 (vii) If G and H are...

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66 (vii) If G and H are abelian groups, then G × H is an abelian group. Solution. True. (viii) If G and H are cyclic groups, then G × H is a cyclic group. Solution. False. (ix) If every subgroup of a group G is a normal subgroup, then G is abelian. Solution. False. (x) If G is a group, then { 1 } C G and G / { 1 } = G . Solution. True. 2.96 Prove that U ( I 9 ) = I 6 and U ( I 15 ) = I 4 × I 2 . Solution. The 6 elements in U ( I 9 ) are 1, 2, 4, 5,7, 8, and they form a cyclic group with 2 as a generator. The 8 elements of U ( I 15 ) are 1, 2, 4, 7, 8, 11, 13, 14. Now 2 has order 4: h 2 i= 1 , 2 , 4 , 8 , and 11 has order 2. The intersection of these two subgroups is { 1 } and they generate all of U ( I 1 5 ) (if S =h 2 , 11 i , then | S |≥ 5; by Lagrange s theorem, | S | is a divisor of 8; hence, | S |= 8 and S = U ( I 1 5 ) . 2.97 (i) If H and K are groups, prove, without using the f rst isomorphism theorem, that H ={ ( h , 1 ) : h H } and K ={ ( 1 , k ) : k K } are normal subgroups of H × K with H = H and K = K . Solution. First we show that H is a subgroup of H × K .I t contains ( 1 , 1 ) , which is the identity of H × K
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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