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

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SUMMARY

The discussion centers on finding a counterexample to the equation (ab)i = aibi for two consecutive integers in the context of group theory, specifically within non-abelian groups. Participants noted that while the problem is more straightforward for three consecutive integers, identifying suitable pairs for two consecutive integers remains challenging. The quaternion group Q was identified as a potential candidate, particularly for i=4 and i=5, indicating its relevance in this context.

PREREQUISITES
  • Understanding of group theory concepts, particularly non-abelian groups.
  • Familiarity with the properties of the quaternion group Q.
  • Knowledge of integer sequences and their implications in group operations.
  • Experience with mathematical problem-solving techniques in abstract algebra.
NEXT STEPS
  • Research the properties and applications of the quaternion group Q in group theory.
  • Explore examples of non-abelian groups and their characteristics.
  • Study the implications of the equation (ab)i = aibi for various integer values.
  • Investigate the relationship between consecutive integers and group operations in depth.
USEFUL FOR

This discussion is beneficial for students and researchers in mathematics, particularly those focused on abstract algebra, group theory, and the exploration of non-abelian structures.

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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