Understanding Proof in Algebra: How to Prove H is Contained in N | Group Theory

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SUMMARY

The discussion centers on proving that a finite subgroup H of a group G is contained in a normal subgroup N of finite index. The proof utilizes the natural projection f: G -> G/N, demonstrating that the order of the image f(H) must divide both |G/N| and |H|. Given that the greatest common divisor (gcd) of |H| and the index [G:N] is 1, it follows that |f(H)| equals 1, leading to the conclusion that H is contained in N. This conclusion is supported by the isomorphism theorem, which relates the sizes of groups and their kernels.

PREREQUISITES
  • Understanding of group theory concepts, particularly normal subgroups and finite index.
  • Familiarity with the isomorphism theorem in group theory.
  • Knowledge of natural projections and their implications in group structures.
  • Basic understanding of greatest common divisor (gcd) in the context of group orders.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Learn about the isomorphism theorem and its applications in algebra.
  • Explore the concept of finite index and its significance in group theory.
  • Investigate the implications of gcd in the context of group orders and subgroup relationships.
USEFUL FOR

This discussion is beneficial for mathematicians, particularly those specializing in abstract algebra and group theory, as well as students seeking to deepen their understanding of subgroup containment and normal subgroups.

moont14263
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Can someone help to understand the under lined part of the proof:
Let G be a group and let N be a normal subgroup of G of finite index. Suppose that H is a finite subgroup of G and that the order of H is relatively prime to the index of N in G. Prove that H is contained in N

Proof:
Let f : G -> G/N be the natural projection. Then f(H) is a subgroup of G/N, so its order must be a divisor of |G/N|. On the other hand, |f(H)| must be a divisor of |H|. Since gcd (|H|,[G:N]) = 1, we must have |f(H)| = 1, which implies that H c ker (f ) = N.
Thanks in advance
 
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This follows from the isomorphism theorem: for every group G and every homomorphism f, we have an isomorphism

f(G)\cong G/Ker(f)

thus

|G|=|f(G)||Ker(f)|
 
Thank you micromass, you are helpful.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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