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let $G$ a group, prove that $o(ab)\mid [o(a),o(b)]$.It is trivially easy if $G$ is abelian; I can't prove it in general.

Thanks.

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    $o(x)$ meaning order of $x$? If so, do you mean $o(ab) \mid o([a, b])$? – 2017-02-09
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    It's not true in general. The product of two transpositions can be a $3$-cycle, for example. – 2017-02-09
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    This was the reason of my failure. – 2017-02-09
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    Could someone tell me what the question means, please? – 2017-02-09
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    Yes excuse me: $o(x)$ was the order of $x$. So I ask a way to demonstrate the (false) property of the order of product: $o(a)\mid lcm(o(a),o(b))$. It is clear now? – 2017-02-09
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    It can be event more extreme than in Daniel Fischer's example: in the free product of two copies of a 2-element group, the order of $ab$ (where $a$ and $b$ are the generators of the factors) is infinite, while the lcm of their orders is 2. – 2017-02-09
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    @G.Cantisani: thank you: the notation $[m, n]$ for an lcm is not well-known (at least to me $\ddot{\smile}$!). I genuinely thought you were interested in the order of the commutator $[a, b]$. – 2017-02-09
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    Every $A_n$ is generated by elements of order $3$. Iteration of your claim would yield that every element of $A_n$ satisfies $x^3=e$, which is not true for $n\ge 5$, because $5\mid\#A_n=\frac{n!}{2}$. – 2017-02-09

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