If $a^3=1$, is $G$ abelian? If $G$ is a group that satisfies $a^3=1$ for every $a\in G$, then is $G$ abelian?
This is an exercise I found in Jacobson's Basic Algebra. It is analogous to the question: If $G$ is a group that satisfies $a^2=1$ for every $a\in G$, then $G$ is abelian. I tried to multiply on both sides of $ab$ ($a,b\in G$) by some appropriate ($b***$) and ($***a$) three times to yield $ba$ but failed. Nor could I give a counterexample. Can someone give me an answer to the question and some hints to solve it?
 A: The multiplicative group of matrices
$$G = \left\{ \begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix} \middle\vert\,  a,b,c \in \mathbb{F}_3\right\} \subset \operatorname{Mat}(3 \times 3, \mathbb{F}_3)$$
is a counterexample (it is isomorphic to the group in Joanpemo's answer).

For all those who are not willing to check this example on there own, here are the arguments:


*

*Every element $g \in G$ has characteristic polynomial $\chi_g(\lambda) = \lambda^3 - 1$, and thus has order $3$ by the Cayley-Hamilton theorem.

*The elements $\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}$ and $\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$ do not commute:
$$ \begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix} \cdot \begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix} = \begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix} $$
$$ \begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix} \cdot \begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$$
A: Nope. The (non-trivial) semidirect product $\;C_3\ltimes(C_3\times C_3)\;$ has exponent $\;3\;$ and it is certainly non-abelian.
