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Direct productby barbiemathgurl
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#1
Jul2907, 06:47 PM

P: 12

Let k be a positive integer.
define G_k = {x 1<= x <= k with gcd(x,k)=1} prove that: a)G_k is a group under multiplication modulos k (i can do that). b)G_nm = G_n x G_m be defining an isomorphism. 


#2
Jul2907, 07:22 PM

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P: 9,396

What have you done for b)? There is only one possible way you can think of to write out a map from G_nm to G_n x G_m, so prove it is an isomorphism. Remember, G_n x G_m looks like pars (x,y)....



#3
Jul2907, 08:22 PM

P: 291

We can use the Chinese Remainder Theorem on this one.
Define the mapping, [tex]\phi: G_{nm}\mapsto G_n\times G_m[/tex] As, [tex]\phi(x) = (x\bmod{n} , x\bmod{m})[/tex] 1)The homomorphism part is trivial. 2)The bijection part is covered by Chinese Remainder Theorem. 


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