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Possible Duplicate:
Is Lagrange's theorem the most basic result in finite group theory?

I can't seem to find a simple proof of this in my textbook, not can I figure out a good way to search for it online.

Basically, I'm trying to show that $\forall g \in G$, $g^{\#(G)}=1$.

Obviously I can show this from Lagrange's theorem, but this requires introducing cosets to prove.

I'm wondering if there's some way to prove this from basic manipulations, perhaps relying on the additional properties of cyclic subgroups. Or is there no shortcut through this proof besides using Lagrange's theorem?

Note: This is not homework, but I'm actually looking to explain some basic group theory to someone else.

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    @JeremySalwen, the proof I outline in [my question](http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory) does not introduce several abstractions in row, though it does use the concept of coset in disguise. I argue there that this may be of some pedagogical value.2012-01-30

1 Answers 1

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If all you want to do is

demonstrate something nontrivial that one can prove with just the abstraction of a group, without introducing further abstractions

then restrict to the abelian case and consider the following proof: if $g_1, ... g_n$ is a list of all the elements of the group, then so is $g g_1, g g_2, ... g g_n$. So $g_1 ... g_n = (g g_1) ... (g g_n)$

hence $g_1 ... g_n = g^n (g_1 ... g_n)$

hence $g^n = 1.$

Notice that the proof breaks down badly in the non-abelian case.