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

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.

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 $\lvert G\rvert$.

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 $q$.