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Let be $G$ a group. Is the following statement true?

If every $x\in G, x\neq e=1$ has order at most 3 (i.e. $x^3=1$), then $G$ is abelian.

I wanted to prove that $xy=yx\ \forall x,y\in G$.

$$xy=x1y = x(xy)^3y=xxyxyxyy \neq xxxyxyyy=x^3yxy^3=1yx1=yx.$$

So, I've proved that the group with the following properity is not abelian.

But I'm not sure whether my proof is correct.

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    $\begingroup$ What makes you assume that $xxyxyxyy\neq xxxyxyyy$? This has not been proven. Maybe they still are equal, despite having different representations. $\endgroup$
    – JMoravitz
    Apr 10, 2016 at 15:50
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    $\begingroup$ We say that the group has *exponent $3$ in this case. There are nonabelian groups of exponent $3$: mathoverflow.net/questions/32116/exponent-of-a-group. $\endgroup$
    – lhf
    Apr 10, 2016 at 15:54

1 Answer 1

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This statement is false. For a counter example, consider the group of unipotent upper triangular matrices with coefficients in $\mathbf{F}_3$ (under matrix multiplication): $$U(3, \mathbf{F}_3) = \left\{\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1\end{pmatrix} : a, b, c \in \mathbf{F}_3\right\}.$$

Some computational details:

Write $[a, b, c]$ for the element $$\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1\end{pmatrix} $$ of the group $U(3, \mathbf{F}_3)$. Then, the group law can be written as: $$[a_1, b_1, c_1] \cdot [a_2, b_2, c_2] = [a_1 + a_2, b_1 + a_1 c_2 + b_2, c_1 + c_2].$$ From here, you may quickly verify that every element has order $3$ in $U(3, \mathbf{F}_3)$ and that it is not abelian. Thus, this group has exponent $3$ but is not abelian.

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