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Proposition. Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a,b \in G$. If $(ab)^3=(ba)^3$, then $ab=ba$.

Is it true or false? So far I've only been able to prove that powers of $a$ commute with $b^3$ and powers of $b$ with $a^3$.

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Hint:

Let $U(3, \mathbb F_3)$ be the group of all $3 \times 3$ upper triangular matrices with all diagonal entries $1$, over the field $\mathbb F_3$. Thus, the elements are all the matrices of the form $$\begin{bmatrix}1 & a & b\\0 & 1 & c\\0 & 0 & 1\end{bmatrix}$$ where $a, b, c \in \mathbb F_3 = \{0,1,2\}$, the field of order $3$.

  1. What is the exponent of the group [the least positive $n$ such that $g^n = 1$ for all group elements $g$]? Or: Determine the order of each element.
  2. Is the group Abelian?
  3. What do Points 1 and 2 imply about the status of the Proposition in this group?
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    $\begingroup$ @TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, \mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H \times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!) $\endgroup$ – M. Vinay Mar 18 at 12:19
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    $\begingroup$ So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$. $\endgroup$ – M. Vinay Mar 18 at 12:20
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    $\begingroup$ @TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1\ \forall g \in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$. $\endgroup$ – M. Vinay Mar 18 at 12:40
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    $\begingroup$ @TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent! $\endgroup$ – M. Vinay Mar 18 at 13:38
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    $\begingroup$ @TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory). $\endgroup$ – M. Vinay Mar 19 at 1:38

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