While taking a practice qual I realized I didn't know how to show that a simple group of order $168$ has no elements of order $14$. Thankfully mse does know, and I don't feel bad about not coming up with this idea.

It did get me thinking, though. I know a lot of techniques for showing subgroups of certain orders do exist (Sylow/Hall Theorems, etc.) but almost no techniques for showing subgroups of certain orders don't exist.

Obviously Lagrange's Theorem is the main obstruction, and even the linked question we take a purported copy of $C_{14}$ in $G$ and build a counterexample to Lagrange. But at least at present the technique in that answer, as well as techniques to similar nonexistence theorems I've found, feel somewhat ad hoc. So my question is

Are there general techniques for showing $G$ has no subgroup of order $k$, even if $k \mid |G|$?

I'm interested both in ways to extend the reach of Lagrange's Theorem, as well as in other invariants (even if they might be complicated). It looks like this has been asked before, but the answer isn't very informative. Maybe that's because there's no great answer? I hope that isn't the case!

Thanks in advance ^_^

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    $\begingroup$ You can also sometimes get a group-action contradiction in the same spirit of Lagrange. $\endgroup$
    – Randall
    Aug 25, 2021 at 18:17
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    $\begingroup$ @Randall -- would you mind elaborating? Are you thinking of letting $G$ act on the cosets of a purported subgroup of size $k$? $\endgroup$ Aug 25, 2021 at 18:21
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    $\begingroup$ Yeah, that's it. Sometimes you can get fancy and act on a quotient, too, or even a set of subgroups. $\endgroup$
    – Randall
    Aug 25, 2021 at 18:25
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    $\begingroup$ Note that $k$ cannot be prime (by Cauchy's Theorem for finite groups). $\endgroup$
    – Shaun
    Aug 25, 2021 at 18:45
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    $\begingroup$ And there is a subgroup of order $k$ for every $k$ such that $\gcd(k,|G|/k)=1$ if and only if $G$ is solvable (Hall's Theorem). $\endgroup$ Aug 25, 2021 at 19:47


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