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

In summary, the problem at hand is to find a counterexample where (ab)^i = a^ib^i for two consecutive integers does not imply that the group G is abelian. While the previous problem for three consecutive integers was easier, the current one seems to be more challenging. It is suggested that choosing the right two consecutive integers can make the problem trivial. The value of i is not the square root of 1, but an integer. After some exploration, it is discovered that the quaternion group Q with i=4 and i=5 serves as a nontrivial counterexample.
  • #1
SiddharthM
176
0
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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  • #2
Hint: It's trivial if you pick the right two consecutive integers.
 
  • #3
im sorry but is i the square root of 1 here?
 
  • #4
no, i is an integer. It's a group theory problem
 
  • #5
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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1. What is a counterexample?

A counterexample is an example that proves a statement or hypothesis to be false. In this case, we are looking for two consecutive integers that do not follow the pattern of (ab)^i = a^i b^i.

2. Can you give an example of a consecutive integer counterexample?

Yes, for example, if we take a=2 and b=3, the statement (ab)^i = a^i b^i would be false for the consecutive integers 2 and 3, since (2x3)^1 = 6, but 2^1 x 3^1 = 2 x 3 = 6.

3. Are there any special cases where the statement holds true?

Yes, if either a or b is equal to 1, then the statement will be true for any consecutive integers. For example, if a=1 and b=2, then (1x2)^i = 1^i x 2^i = 2^i, which holds true for any consecutive integers.

4. How do you prove that a counterexample exists for a given statement?

To prove that a counterexample exists, we can simply show one example where the statement is not true. This will prove that the statement is not universally true and that a counterexample exists.

5. Can you explain why it is important to find a counterexample for this statement?

Finding a counterexample for a statement can help us identify where the statement may not hold true and can also lead to a better understanding of the concept. In this case, finding a counterexample for (ab)^i = a^i b^i can help us understand the limitations of this equation and when it may not be applicable.

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