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$G$ is a group, if ∀a,b∈G, $a^2b=ba^2$, how to make a counterexample show that $G$ is NOT Abelian?

What's the counterexample when $a^nb=ba^n$?

user290115
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    Note that for some $n$ this is easier than others - if there is a nonabelian group of order $n$, then that's an example for that $n$. More generally, it's enough to get a nonabelian group in which every element has order $n$. For 2, however, this won't work - if every element of a group has order 2, then the group is abelian, see http://math.stackexchange.com/questions/275544/order-of-nontrivial-elements-is-2-implies-abelian-group. – Noah Schweber Nov 14 '15 at 22:56
  • @NoahSchweber I assume you mean "...every element has order dividing $n$." – Alex G. Nov 14 '15 at 22:59
  • @AlexG. To be fair, I didn't say you needed order exactly $n$ - what I wrote is still correct. :P – Noah Schweber Nov 14 '15 at 23:00
  • @NoahSchweber For $n>1$ there is no group in which every element has order $n$, since in particular the identity element has order $1$. This is really semantics, since I would guess that by "every element has order n" you mean "every $g \in G$ satisfies $g^n = 1$," which is equivalent to saying that every element has order dividing $n$. – Alex G. Nov 14 '15 at 23:02

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A simple counterexample is the group $G = \{\pm 1, \pm i, \pm j, \pm k\}$ whose multiplication is that of the quaternions. $G$ is not abelian since $ij = k$ and $ji = -k$, but the square of any element is $1$ or $-1$ and therefore commutes with all elements of the group.

Alex G.
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    Note that what's being used here is that every element of the quaternion group almost has order 2: for each $g$, $g^2$ is in the center of the group (= commutes with everything). – Noah Schweber Nov 14 '15 at 23:01
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To get a counterexample, you need an element which is not the square of another element. A good example would be the quaternion group, where $i^2=j^2=k^2=-1$, and $(-1)^2=1$. $-1$ commutes with all the elements in the quaternion group. Read the Wikipedia article here.

Since every element in the quaternion group raised to the $4$th power is the identity, it follows that for all $n|4$, your group does not need to be Abelian.

For $n\equiv2\mod4$, you can also use the quaternion group, as every element raised to the $n$th power would be either $-1$ or $1$, which commutes.

(This still leaves out the case when $n$ is odd.)

Element118
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