Proof of "Quotienting Out" by M: Is it True?

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  • Thread starter Thread starter Mr Davis 97
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SUMMARY

The discussion confirms that if \( N \trianglelefteq H \), \( N \trianglelefteq G \), and \( H \le G \), then it is indeed true that \( H/N \le G/N \). The participants clarify that the condition \( (h_1N)(h_2^{-1}N) = h_2h_2^{-1}N \in H/N \) is sufficient for the proof. Additionally, the importance of the index being \( 1 \) is highlighted, indicating a specific condition for the quotient groups. The reference to the Third Isomorphism Theorem on Wikipedia provides further validation of the conclusion.

PREREQUISITES
  • Understanding of normal subgroups, denoted as \( N \trianglelefteq H \) and \( N \trianglelefteq G \)
  • Familiarity with group theory concepts, particularly quotient groups \( H/N \) and \( G/N \)
  • Knowledge of the Third Isomorphism Theorem in group theory
  • Basic algebraic manipulation involving group elements and their cosets
NEXT STEPS
  • Study the Third Isomorphism Theorem in detail to understand its implications on quotient groups
  • Explore examples of normal subgroups in various groups to solidify understanding of \( N \trianglelefteq H \) and \( N \trianglelefteq G \)
  • Learn about the properties of quotient groups and their indices in group theory
  • Investigate additional proofs involving quotient groups to enhance problem-solving skills in abstract algebra
USEFUL FOR

Mathematicians, particularly those specializing in abstract algebra, students studying group theory, and anyone interested in the properties and applications of quotient groups.

Mr Davis 97
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If ##N \trianglelefteq H ##, ##N \trianglelefteq G ##, and ##H \le G##, then is it true that ##H/N \le G/N##?

I want to use the result for a proof I am currently doing, but I am not sure it is true.
Is it enough just to note that if ##h_1,h_2\in H##, then ##(h_1N)(h_2^{-1}N) = h_2h_2^{-1}N \in H/N##?
 
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Mr Davis 97 said:
If ##N \trianglelefteq H ##, ##N \trianglelefteq G ##, and ##H \le G##, then is it true that ##H/N \le G/N##?

I want to use the result for a proof I am currently doing, but I am not sure it is true.
Is it enough just to note that if ##h_1,h_2\in H##, then ##(h_1N)(h_2^{-1}N) = h_2h_2^{-1}N \in H/N##?
No, it's wrong. The index before last has to be ##1## :biggrin:

The rest is a yes.
E.g.: https://en.wikipedia.org/wiki/Isomorphism_theorems#Third_isomorphism_theorem
 
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