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Prove that the group of all isometries is abelian

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

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

    3. 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.
  2. jcsd
  3. Sep 28, 2012 #2
    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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