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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.
 Fall '11
 KeithCornell

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