Is G_nm Isomorphic to G_n x G_m?

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    Group Isomorphism
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Discussion Overview

The discussion centers on the mathematical properties of the groups defined by the integers coprime to a given integer, specifically exploring whether the group G_nm is isomorphic to the direct product G_n x G_m. The scope includes theoretical aspects of group theory and the application of the Chinese Remainder Theorem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant defines G_k as the set of integers coprime to k and asserts that G_k is a group under multiplication modulo k.
  • Another participant questions the approach taken for proving the isomorphism between G_nm and G_n x G_m, suggesting that there is a specific way to define a mapping.
  • A third participant proposes using the Chinese Remainder Theorem to define a mapping from G_nm to G_n x G_m, stating that the homomorphism aspect is straightforward and that the bijection is guaranteed by the theorem.
  • A later reply emphasizes that the goal is to prove the Chinese Remainder Theorem itself, indicating a focus on foundational aspects of the argument.

Areas of Agreement / Disagreement

Participants express differing views on the approach to proving the isomorphism, with some advocating for the use of the Chinese Remainder Theorem while others emphasize the need to establish the theorem itself. No consensus is reached on the method of proof.

Contextual Notes

The discussion involves assumptions about the properties of the groups and the applicability of the Chinese Remainder Theorem, which may not be universally accepted without further proof. The steps required to establish the isomorphism remain unresolved.

barbiemathgurl
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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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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)...
 
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.
 
but the point is to prove that theorem.
 

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