Is it possible to prove $g^{|G|}=e$ in all finite groups without talking about cosets?

Question

Let $G$ be a finite group, and $g$ be a an element of $G$. How could we go about proving $g^{|G|}=e$ without using cosets? I would admit Lagrange's theorem if a proof without talking about cosets can be found.

I have a proof for abelian groups which basically consists in taking the usual proof of Euler's theorem and using it in a group, I do not know if it can be modified to work in arbitrary finite groups.

The proof is as follows: the function from $G$ to $G$ that consists of mapping $h$ to $gh$ is a bijection. Therefore

$\prod\limits_{h\in G}h=\prod\limits_{h\in G}gh$ but because of commutativity $\prod\limits_{h\in G}gh=\prod\limits_{h\in G}g\prod\limits_{h\in G}h=g^{|G|}\prod\limits_{h\in G}h$.

So we have $\prod\limits_{h\in G}h=g^{|G|}\prod\limits_{h\in G}h$.

The cancellation property yields $e=g^{|G|}$.

I am looking for some support as to why it may be hard to prove this result without talking about cosets, or if possible an actual proof without cosets.

Thank you very much in advance, regards.

Answer 1

Here is a non-rigorous justification of why it should be very difficult to make this argument without using cosets.

First of all, the statement is precisely Lagrange's theorem for a subgroup $H\leq G$, under the additional assumption that $H$ is cyclic.

A proof that the cardinality of a finite set $H$ divides the cardinality of a finite set $G$, consists of a partition of $G$ along with bijections between $H$ and each element of the partition. We could also construct a surjection $G\to H$ and a bijection between its fibers, which amounts to the same thing.

We have to use somewhere the assumption that $H$ is a subgroup of $G$, i.e. that the group structure on $G$ is compatible with the group structure on $H$. So at some point, we should construct a bijection between $H$ and another subset of $G$ using the group operation on $G$.

There are two ways to do this: multiplication and conjugation. But conjugation is trivial in abelian groups! We are left with considering the set $gH$ (or $Hg$), and we get the coset argument.

We could potentially try to capitalize on the fact that $H$ is cyclic, by taking a generator $h\in H$ and looking at the permutation $g\mapsto hg$. But the orbits of this permutation are exactly the right cosets of $H$, so this turns into the same argument.

Answer 2

Hints.

First observe that the set $$ H=\{g,g^2,\ldots,g^n,\ldots\} $$ is finite.

If $\lvert H\rvert=n$, then $g^n=e$.

Finally, $\lvert H\rvert$ divides $\lvert G\rvert$.

Answer 3

Let me take a shot-

Let $o(g)=n$ for some arbitrary $g \in G$, then $g^n=e$ (and $n$ is least such positive integer), now if suppose $g^{|G|}\neq e$, then there exist there exist $t \in \mathbb{Z}$ which is also greater than $1$ such that $g^{|G|t} = e$, but then by division algorithm $\exists \ $ $q,r \in \mathbb{Z}$ such that $|G|t=nq+r$ and $0\leq r <n \implies g^{nq+r}=g^r=e$ $\implies$ $r=0$ $\implies$ $|G| = \frac{nq}{t} \implies g^{|G|}=g^{n(q/t)} \neq e $ (by hypothesis) but $g^n=e$.

Now the question is why does $t$ has to divide $q$, but I argue as (avoiding order of element divides order of $G$, which is the question itself) that it must divide, as once $g^n=e$, then raising $e$ to the power $\frac{q}{t}$ doesn't make sense if $t$ does not divide $q$.