Basis Theorem for Finite Abelian Groups

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SUMMARY

The discussion focuses on proving a specific aspect of the Basis Theorem for Finite Abelian Groups, particularly the assertion that if the product \( a^{l_0}b_1^{l_1}...b_n^{l_n}=e \), then each component must equal the identity element \( e \). The user proposes using the orders of the elements, specifically that if \( |a| > |b_1| > |b_2| > ... > |b_n| \), raising both sides of the equation to the power of \( |b_1| \) leads to the conclusion \( a^{l_0} = e \). However, the user expresses uncertainty about proving the strict inequality of the orders of \( a \) and \( b_i \).

PREREQUISITES
  • Understanding of group theory, specifically finite abelian groups
  • Familiarity with the concept of element order in group theory
  • Knowledge of induction proofs in mathematics
  • Basic algebraic manipulation of group elements
NEXT STEPS
  • Study the properties of finite abelian groups and their structure
  • Learn about the implications of element orders in group theory
  • Review induction techniques in mathematical proofs
  • Explore related exercises in group theory to reinforce understanding
USEFUL FOR

This discussion is beneficial for students and researchers in abstract algebra, particularly those focusing on group theory and the properties of finite abelian groups.

Kiwi1
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I am attempting to answer the attached question. I have completed parts 1-4 and am struggling with part 5.

5. Prove that if a^{l_0}b_1^{l_1}...b_n^{l_n}=e then a^{l_0}=b_1^{l_1}=...=b_n^{l_n}=e

If |a|>|b1|>|b2|>...>|bn| then I could raise both sides of a^{l_0}b_1^{l_1}...b_n^{l_n}=e to the power of |b1| and would get (a^{l_l})^{|b_1|}=e from which I conclude: (a^{l_0})=e carrying on by induction gives me the required result.

But I don't think I can prove that the order of a is STRICTLY greater than the order of all b's?
 

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Hi,
What is needed is exercise O, which I have included in the following:

51ujk9.png
 

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