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direct product |
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| Jul29-07, 06:47 PM | #1 |
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direct product
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. |
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| Jul29-07, 07:22 PM | #2 |
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Recognitions:
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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)....
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| Jul29-07, 08:22 PM | #3 |
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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. |
| Jul30-07, 12:50 AM | #4 |
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Recognitions:
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direct product
but the point is to prove that theorem.
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