# exponent of a finite group divides order of the group

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

• Have you thought of showing that the prime-power divisors of the exponent divide the order of the group? – ancientmathematician Nov 15 '18 at 14:36
• @ancientmathematician no, I'll try that – John Cataldo Nov 15 '18 at 14:39

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 $$a\in G$$.