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I came across the following problem: Let $p$ be a prime, prove that $$(p-1)^np^{(n^2-n)/2}\mid \prod_{k=1}^n(p^n-p^{k-1})$$ for any $n\in \mathbb{N}$. It is quite hard to prove it using elementary number theory. However, the problem becomes trivial if we notice that $|\operatorname{GL}_n(\mathbb{F}_p)|=\prod_{k=1}^n(p^n-p^{k-1})$ and $|\operatorname{B}_n(\mathbb{F}_p)|=(p-1)^np^{(n^2-n)/2}$, where $\operatorname{B}_n(\mathbb{F}_p)$ is the set of $A\in \operatorname{GL}_n(\mathbb{F}_p)$ that is upper triangular. And the proof follows immediately from Lagrange's theorem.

So that got me thinking, can every divisibility problem $m\mid n$ be solved using group theory and Lagrange? More precisely, if $m\mid n$, can we always find a group $G$ of order $n$ and a subgroup $H$ of $G$ of order $m$?

I know that the above is true for $m=p$ a prime by Cauchy's theorem or more generally for $m=p^k, k=\operatorname{ord}_p(|G|)$ by Sylow. But I also know that the converse of Lagrange's theorem is false in general.

I have tried a few different pairs of small $n,m$ and it seems to work out but I can't prove of disprove the above statement.

Bill Dubuque
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Nayr borcherds
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    It is true, but it is not necessarily a universal proof strategy for these types of problems, because you might have a hard time constructing the groups in such a natural way as here, without knowing in advance that the divisibility holds. For example, try to prove $m!n!(m+n)!~|~(2m)!(2n)!$ this way. – Michal Adamaszek Mar 20 '24 at 08:19
  • The proof of the first line using elementary number theory is immediate: there is a factor of $p-1$ in each factor of the product, and a total of $1+2+3+\cdots+(n-1)=\frac{n^2-n}2$ factors of $p$. However I have certainly come across examples where it is easier to prove divisibility using finite fields, rather than elementary number theory. – tkf Mar 20 '24 at 21:31
  • @Eric There are tens if not hundreds of good dupe targets. I did not choose that one, I merely reinstated it. Choose a better one if you like. – Bill Dubuque Mar 21 '24 at 02:14
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    @BillDubuque: So you're saying if a question is a duplicate of some question out there, it should be randomly closed as a duplicate of any other question?? – Eric Wofsey Mar 21 '24 at 02:15
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    @BillDubuque It is entirely unclear to me how your position is distinct from that. – Eric Wofsey Mar 21 '24 at 02:17
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    Your edit is misleading, @BillDubuque. The motivation behind the question is succinctly expressed in its original title (plus "be"). – Shaun Mar 21 '24 at 12:01
  • @Shaun No, I edited the title to match the exact question asked in the text. Having a precise title is important for dupe assessment, searching etc. – Bill Dubuque Mar 21 '24 at 12:04
  • I'm happy with this iteration, @BillDubuque – Shaun Mar 21 '24 at 12:10
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    @Shaun Luckily that compromise actually (barely) fits. – Bill Dubuque Mar 21 '24 at 12:12

1 Answers1

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More precisely, if $m\mid n$, can we always find a group $G$ of order $n$ and a subgroup $H$ of $G$ of order $m$?

Yes.

For each $n$, there exists a cyclic group of order $n$. In the cyclic group of order $n$, for each $d\mid n$, there exists a subgroup of order $d$.

Note that this assumes $m\mid n$, so it doesn't establish that $m\mid n$.

To establish that $m\mid n$, it suffices to find a subgroup of order $m$ of a group of order $n$; then Lagrange's Theorem does indeed kick in.

Shaun
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    Any reason for the downvote? – Shaun Mar 20 '24 at 12:30
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    Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Mar 21 '24 at 00:18
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    Neither of those suggested is a duplicate, @BillDubuque. This question is one of divisibility, not a converse of Lagrange's Theorem per se - though that is close. – Shaun Mar 21 '24 at 11:18
  • It is most certainly a dupe of this (and many others). Please stop destroying site organization. – Bill Dubuque Mar 21 '24 at 11:42
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    Not on its own, it's not, @BillDubuque. The reader needs to (1) know that for each $n$, there is a cyclic group of order $n$ and (2) be able to put them together. Just because an answer involves previous questions, it doesn't mean the question it answers is a duplicate of them. – Shaun Mar 21 '24 at 11:58