Adv Alegbra HW Solutions 70

# Adv Alegbra HW Solutions 70 - G m G n = { } . Solution. If...

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70 xy 1 = h 0 k 0 ( hk ) 1 = h 0 k 0 k 1 h 1 .B u t k 0 k 1 = k 0 K and k 0 k 1 h 1 = k 0 h 1 = h 1 k 1 for h 1 H and k 1 K because KH = HK . Therefore, xy 1 = ( h 0 h 1 ) k 1 HK , as desired. (ii) If HK = KH and H K ={ 1 } , prove that HK = H × K . Solution. Absent. 2.107 Prove the converse of Lemma 2.112: if K is a subgroup of a group G , and if every left coset aK is equal to a right coset Kb , then K C G . Solution. Absent. 2.108 Let G be a group and regard G × G as the direct product of G with itself. If the multiplication µ : G × G G is a group homomorphism, prove that G must be abelian. Solution. If a , b G , then ( a , 1 ) and ( 1 , b ) commute in G × G ,for ( a , 1 )( 1 , b ) = ( a , b ) = ( 1 , b )( a , 1 ). Now µ : ( a , b ) 7→ ab . Hence, ab = µ(( a , b ) = µ(( b , 1 ))(( 1 , a )) = µ(( b , 1 ))µ(( 1 , a )) = ba . Therefore, G is abelian. 2.109 Generalize Theorem 2.128 as follows. Let G be a f nite (additive) abelian group of order mn , where ( m , n ) = 1. De f ne G m ={ g G : order ( g ) | m } and G n ={ h G : order ( h ) | n } . (i) Prove that G m and G
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Unformatted text preview: G m G n = { } . Solution. If x G m G n , then the order of x is a common divisor of m and n . As the gcd ( m , n ) = 1, the element x has order 1 and so x = 1. (ii) Prove that G = G m + G n = { g + h : g G m and h G n } . Solution. Since G is abelian, every subgroup is normal, and so the second isomorphism theorem applies. As G m G n ) = { } , we have G m = ( G m + G n )/ G n , so that | G m + G n | = | G M || G N | = mn = | G | . Therefore, G m + G n = G . (iii) Prove that G = G m G n . Solution. The result now follows from Proposition 2.127....
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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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