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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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Jeremy Salwen
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    I am reasonably certain someone has asked this exact question before (including the specification that Lagrange's theorem be avoided). As I recall, the discussion was inconclusive. I really, really do not see the point of avoiding Lagrange's theorem here. – Qiaochu Yuan Jan 29 '12 at 07:31
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    (There is a cute proof under the additional assumption that $G$ is abelian that you should try to think of if you aren't familiar with it, but I don't see any way to salvage it in general.) – Qiaochu Yuan Jan 29 '12 at 07:35
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    @Qiaochu: Do you have this post in mind? – Srivatsan Jan 29 '12 at 07:35
  • @Srivatsan: ah, yes, there it is. – Qiaochu Yuan Jan 29 '12 at 07:38
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    @QiaochuYuan: I tried to make it clear why I want to avoid Langrange's theorem: I'm trying to demonstrate something nontrivial that one can prove with just the abstraction of a group, without introducing further abstractions (like equivalency classes and cosets). In general I find that if you have to introduce several abstractions in a row without showing something interesting about them in their own right, the listener tends to be bored. – Jeremy Salwen Jan 29 '12 at 10:04
  • One could always avoid abstraction until one has sufficiently many interesting examples that one might actually have a use for a theorem. Rosette groups are pretty, and the cosets involved have names, so there is no need for abstraction. – Jack Schmidt Jan 29 '12 at 19:00
  • @JeremySalwen, the proof I outline in my question 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. – lhf Jan 30 '12 at 12:47

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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.

Qiaochu Yuan
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