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While this is a straightforward result which follows almost immediately from Lagrange's theorem (by considering the subgroup generated by $\langle g\rangle $), is there any way this can be proven without using Lagrange or any other theorem which derives from it?

One approach I thought of is that if it can be proven that for $ g \in G$ we have $ g^{|G|} = e$ then the result follows as $ g^n = e \iff o(g) \mid n $ , but I've had no success going down this line of attack.

While I have no idea if it is possible to do so, I suspect it is as it's from a past paper question which says we can't use Lagrange's theorem or the Orbit-Stabiliser theorem to prove the result without proving these beforehand. If the result could only be deduced from Lagrange, then I would assume the question would first ask for a proof of it and then have this as a corollary.

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up vote 4 down vote accepted

Considering the free action of $G$ on itself by left multiplication, the permutation induced by $g$ must be a product of disjoint cycles each having length $o(g)$. Therefore $|G|$ is a multiple of $o(g)$. However, this is very similar to how one proves Lagrange's theorem for this special case.

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I believe this doesn't answer the question... OP is not asking for a proof of Lagrange's theorem, but a significantly different approach to prove $o(g) | |G|$ without using Lagrange. –  Patrick Da Silva Jan 7 '13 at 16:17
    
I also don't see why g has to be a product of disjoint cycles all of length o(g). –  Andrew D Jan 7 '13 at 16:46
    
Ah, I see now. While this is similar to a proof of Lagrange, it technically does answer the question without using Lagrange explicity - its a bit of a shame it seems only possible using similar methods though. –  Andrew D Jan 7 '13 at 17:18

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