Does the First Isomorphism Theorem apply in this case?

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SUMMARY

The discussion centers on the application of the First Isomorphism Theorem to the homomorphism between G = (integers modulo 8 direct product integers modulo 2) and H = (integers modulo 4 direct product integers modulo 4). It is established that a nontrivial homomorphism exists, specifically phi(a, b) = (0, 2b), which contradicts the initial assumption of no homomorphism. The participants also highlight the necessity of considering these structures as rings to properly apply the theorem, emphasizing the importance of identity preservation in ring homomorphisms.

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  • Understanding of group theory, specifically the First Isomorphism Theorem.
  • Knowledge of ring theory and the properties of ring homomorphisms.
  • Familiarity with direct products of groups and rings.
  • Basic concepts of modular arithmetic, particularly integers modulo n.
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  • Explore the properties and definitions of ring homomorphisms.
  • Investigate the structure of direct products in both group and ring contexts.
  • Learn about the implications of identity preservation in ring homomorphisms.
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Mathematicians, particularly those specializing in abstract algebra, students studying group and ring theory, and anyone interested in the applications of the First Isomorphism Theorem.

JasonJo
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prove that there does not exist a homomorphism from G:= (integers modulo 8 direct product integers modulo 2) to H:= (intergers modulo 4 direct product integers modulo 4).

Pf:
i tried this route, assume that there is such a homomorphism. then by first isomorphism theorem, G/ker phi is isomorphic to phi(G) but what would the kernel of phi have to be in this case?
 
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Z8 x Z2 => Z4 x Z4??
There is a homomorphism. For example phi(a, b) = (0, 2b) describes a nontrivial homomorphism, unless I've gone blind.
 
Last edited:
Hrm. Maybe he means to consider them as rings, rather than as additive groups?

Ring homomorphisms must preserve the identity. (As well as all integer multiples of the identity...)
 

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