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Prove or disprove : If every element of a group G are of same order then G is abelian . I am not getting how to show it. I can prove easily that group having elements of order 2 only is abelian. But how to prove it generally ?

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  • $\begingroup$ It is already false for $n=3$, see my answer. $\endgroup$ – Dietrich Burde Jan 18 '18 at 13:34
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This question is closely related to the restricted Burnside problem: given numbers $m$ and $p$, is the restricted Burnside group $B_0(m,p)$ finite?

Proposition: A finite group $G$ has the property that all non-trivial elements have the same order $p$ if and only if $p$ is prime and $G\ne 1$ is a quotient of $B_0(m,p)$ for some $m.$

Another counterexample is the following group:

$$G= \langle x, y, z | x^3=1, y^3=1, z^3=1, [x,z]=1, [y,z]=1, [x,y]=z^{-1} \rangle$$ is non-abelian of order $27$, and all its non-trivial elements have order $3$. The group is exactly the Heisenberg group over $\mathbb{F}_3$ from Robert's answer.

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I assume you mean every non-trivial element.

This is not true in general. A counter example can be found here.

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