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

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.

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.

• Thank you very much, I am going to have to take some time to chew on this. – Jorge Fernández Hidalgo May 2 '15 at 22:21

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 $\lvert G\rvert$.

• Hmm, okay but for the last part I need to use a result similar to Lagrange's theorem , perhaps lagrange debilitated to cyclic subgroups? How would I go about doing the last part? Perhaps there is a proof of Lagrange or debilitated lagrange that does not use cosets? – Jorge Fernández Hidalgo May 2 '15 at 21:12

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 $q$.

• How do you know $\frac{q}{t} <1$? It could be that $t=n$ and $q=|G|$, – Mastrel May 2 '15 at 21:43
• yeah, sorry, lemme think and edit – Bhaskar Vashishth May 2 '15 at 21:49
• Well I am not sure that this proof will work or not as it boils down to forcing us to use that order of element divides order of $G$, which is the question itself – Bhaskar Vashishth May 2 '15 at 22:00