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Let $m\in\Bbb N$ be the exponent of a finite group $G\ $ ($|G|=n$). It' the smallest integer such that $g^m=e_G\ \forall g\in G$

Proving that $m\mid |G|$ is the same as proving that the least common multiple of the orders of the elements of $G$ divides the order of the group once you realize the exponent is the LCM of $o(g)$ for $g\in G$.

But how to conclude the proof?

I tried to reason with $q_i\cdot o(g_i)=|G|\ \forall i\in[n]$ but it doesn't seem to lead me enywhere

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  • $\begingroup$ Have you thought of showing that the prime-power divisors of the exponent divide the order of the group? $\endgroup$ – ancientmathematician Nov 15 '18 at 14:36
  • $\begingroup$ @ancientmathematician no, I'll try that $\endgroup$ – John Cataldo Nov 15 '18 at 14:39
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Some hints: By Lagrange's Theorem, the order of every element of a finite group must divide the order of the group. In addition it implies that $a^{|G|}=e$ for all $a\in G$.

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