direct product

by barbiemathgurl
Tags: product
barbiemathgurl is offline
Jul29-07, 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.
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matt grime
matt grime is offline
Jul29-07, 07:22 PM
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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)....
Kummer is offline
Jul29-07, 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]
[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.

mathwonk is offline
Jul30-07, 12:50 AM
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P: 9,421

direct product

but the point is to prove that theorem.

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