Prove that the group of all isometries is abelian

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SUMMARY

The discussion centers on proving that the group of all isometries, denoted as G, is not abelian. The participant initially attempted to demonstrate that isometric functions commute but later clarified that the group of isometries does not exhibit commutativity. The key property discussed is that if f: Z → Z is bijective and preserves distances, then f is isometric. The conclusion reached is that the isometric functions do not satisfy the condition f(g(n)) = g(f(n)) for all n in Z.

PREREQUISITES
  • Understanding of group theory concepts, particularly groups and their properties.
  • Familiarity with isometric functions and their definitions.
  • Knowledge of bijective functions and distance preservation.
  • Basic mathematical proof techniques, including the use of QED in proofs.
NEXT STEPS
  • Study the properties of non-abelian groups in group theory.
  • Explore the implications of distance preservation in isometric functions.
  • Learn about bijective mappings and their role in mathematical proofs.
  • Investigate examples of isometric transformations in different mathematical contexts.
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, group theory, and isometric functions, will benefit from this discussion.

Jamin2112
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Homework Statement



The only thing I need to do now is show that isometric functions commute. I've shown the 3 properties that prove the the set G of isometric functions is a group.

Homework Equations



If f:Z-->Z is bijective and preserves distances, then f is isometric.

The Attempt at a Solution



f, g in G and n in Z

--------> |f(n) - g(n)| = |f(f(n)) - g(f(n))| = |g(f(n) - g(g(n))|
--------> ?
--------> |f(g(n) - g(f(n))| = 0
--------> f(g(n)) = g(f(n)). QED.

Can you give me an oh-so-subtle hint? I just started class this week, so I'm not yet in my proper mindset.
 
Physics news on Phys.org
Hi, guys! Apparently I read the question wrong. The question was "Is the group [Gamma] of all isometries abelian?" It isn't abelian and the proof is quite easy.
 

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