# G is a group and $(ab)^3=a^3b^3$ for all $a,b \in G$. Prove (or disprove with a counterexample) that if $(ab)^3=(ba)^3$, then $ab=ba$.

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 $$a^3$$.

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.
• @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!) – M. Vinay Mar 18 at 12:19
• 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 $3$. – M. Vinay Mar 18 at 12:20
• @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 $3$. – M. Vinay Mar 18 at 12:40
• @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! – M. Vinay Mar 18 at 13:38