Proof that $\left | \left \langle g^{m} \right \rangle \right | = n$, where $\left | G \right |=mn$ confusion

Question

I'm trying to prove that if $\left | G \right |>1$ is not prime, there exists a subgroup of $G$ which is non-trivial, where $G$ is finite.

I know that the question has been answered multiple times, for example. But for many of the answers given, they say that "$|G|=mn$ with $m,n>1$, then $\langle g^m\rangle$ and $\langle g^n\rangle$ are proper subgroups", in order to complete the proof.

However, I've come accross a theorem on Proof Wiki stating that Group does not Necessarily have Subgroup of Order of Divisor of its Order, so how are we certain that in the case above, that $\langle g^m\rangle$ even exists, and is it always true that $\left | \left \langle g^{m} \right \rangle \right | = n$ ?

edit: For the Proof Wiki theorem, does it not apply to cyclic groups, as this University of Washington PDF states that "3. Suppose $G$ is a finite cyclic group. Let $m = |G|$. For every positive divisor $d$ of $m$, there exists a unique subgroup $H$ of $G$ of order $d$."

Answer 1

Here is a bit unsatisfactory answer. I'd be curious to hear a more beautiful answer using e.g. Lagrange. Nevertheless, here goes:

Pick a random element $g$, not equal to the identity and look at the group $\langle g \rangle$ generated by $g$.

We know it is not the one-element group because it contains at least two elements: $e$ and $g$ itself.

We are now left with two possibilities. The first one is that $\langle g \rangle$ is not equal to all of $G$. In that case we are done and have a group of the type we were looking for without putting in any effort. This is the unsatisfactory case.

In the second case we have that $\langle g \rangle$ is all of $G$. But we also know that $\langle g \rangle$ is cyclic. It follows that $G$ is cyclic of order $mn$. Now, using cyclicity it is easy to show that the subgroup $\langle g^m \rangle$ has order at most $n$ and hence is a proper subgroup: just write down all its elements!

We have $g^m, g^{2m}, \ldots, g^{nm}$ and we know that $g^{mn} =e$ because $G$ was cyclic of order $mn$ by assumption, so this means that these elements are all the elements of $\langle g^m \rangle$. It looks like that there are $n$ of them, but as you point out in the comments we might have written down the same $k < n$ elements a number of times (actually $n/k$ times, showing that this is a whole number, showing in turn that $k|n$ but we do not really need this). What is not possible is that there are more elements than the $n$ written down here. So the number of elements of $\langle g^m \rangle$ is at most $n$, as claimed.

Finally you might worry that $\langle g^m \rangle$ could still be the one element group in this case, so that $g^m = e$ but then, since $g$ generates all of $G$ by assumption this would mean that $G$ had exactly $m$ elements and hence that $n =1$.

It feels like cheating because the assumption that $g$ generates $G$ does all the heavy lifting, but remember: we could make that assumption because in the alternative, when $g$ does not generate all of $G$, it would be even easier to find a proper subgroup.

Answer 2

I think you are missing several clauses in the argument.

Take a finite group $G$ of order $mn$, with $1\lt m,n\lt |G|$. We want to show that it has a proper nontrivial subgroup.

Let $g\in G$ be any nontrivial element. Now, if $\langle g\rangle\neq G$, then we are done: the subgroup $\langle g\rangle$ works, because it is a proper subgroup (since $\langle g\rangle\neq G$), and it is nontrivial (since $e\neq g\in\langle g\rangle$).

So now assume that this is not the case; that is, that $\langle g\rangle$ is, unfortunately for us, actually equal to $G$. That means that $G$ is cyclic, and the order of $g$ is $mn$. In that case, $\langle g^m\rangle$ is nontrivial, since $m\lt mn$, so $g^m\neq e$ and $e\neq g^m\in\langle g^m\rangle$. Moreover, since $(g^m)^n = g^{mn}= e$, the order of $\langle g^m\rangle$ is a divisor of $n\lt mn$, so $\langle g^m\rangle\neq G$ (it has order $|g^m|$, which is at most $n$, which is less than $mn$). Thus, you have a proper nontrivial subgroup. Similarly if you take $\langle g^n\rangle$ instead of $\langle g^m\rangle$.

But this is all predicated on $\langle g\rangle = G$. Without that assumption, you do not know that $\langle g^m\rangle$ and $\langle g^n\rangle$ are nontrivial. For example, if you take the Klein $4$-group of order $4 = 2\times 2$, with $m=n=2$, and you take $g$ a nontrivial element, then $\langle g^m\rangle = \langle g^n\rangle = \langle g^2\rangle = \langle e\rangle = \{e\}$ is a trivial subgroup. But again, in this case, you get that $\langle g\rangle$ is a subgroup of order $2$ and hence a proper nontrivial subgroup.

So what you are missing is the assumption that $\langle g\rangle = G$, which is the "bad" case (for what you are trying to show). But just one of the two possibilities you are exploring.

Answer 3

Let $g$ be any non-identity element of $G$. Then the order of $g$ is not $1$ and so is $pk$ where $p$ is a prime and $k$ is a positive integer.

Then $g^k$ has order $p$ and so generates a cyclic subgroup of $G$ of prime order. This is therefore the required non-trivial subgroup.