Counterexample so that (ab)^i=a^ib^i for two consecutive integers

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    Counterexample Integers
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Discussion Overview

The discussion revolves around finding a counterexample to the assertion that if \((ab)^i = a^i b^i\) holds for two consecutive integers \(i\) in a group \(G\), then \(G\) must be abelian. Participants are exploring group theory concepts and seeking specific examples or counterexamples related to this property.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that finding a counterexample for two consecutive integers \(i\) does not imply that \(G\) is abelian, referencing a problem from Herstein.
  • Another participant hints that the choice of consecutive integers can simplify the problem.
  • A question is raised about the meaning of \(i\), with clarification that \(i\) refers to an integer in the context of group theory.
  • One participant expresses a desire to find a nontrivial counterexample, indicating that they have ruled out symmetric and dihedral groups as candidates, but are considering the quaternion group for specific values of \(i\).
  • A link to an external resource is shared, suggesting that the quaternion group \(Q\) may work for \(i=4\) and \(i=5\).

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on a specific counterexample, and multiple competing views regarding suitable groups and values for \(i\) remain present in the discussion.

Contextual Notes

There are unresolved assumptions regarding the properties of groups being discussed, particularly concerning the implications of the condition \((ab)^i = a^i b^i\) for different integers \(i\) and the nature of the groups being considered.

SiddharthM
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counterexample so that (ab)^i=a^ib^i for two consecutive integers for any a and b in a group G does not imply that G is abelian.

this is a problem in herstein and I'm struggling to find an example. The previous problem to show that if (ab)^i=a^ib^i for 3 consecutive integers then G is abelian is a starred problem but seems to be easier.
 
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Hint: It's trivial if you pick the right two consecutive integers.
 
im sorry but is i the square root of 1 here?
 
no, i is an integer. It's a group theory problem
 
I wonder if we can find an easy nontrivial counterexample (i.e. where we choose a pair of values for i other than 0 and 1). I scribbled down some stuff and managed to convince myself that the symmetric and dihedral groups aren't going to be good candidates, but I haven't had the energy to pursue this any further.

Edit:
I did some searching and found http://www.groupsrv.com/science/about12002.html . Apparently the quaternion group Q works with i=4 and i=5.
 
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