This post is part of a three-post set (1 2 3) about groups of order $pqr$, where $p<q<r$ are prime.
In this post, I'm reflecting on Ben Blum-Smith's classic the things in proofs are weird. I encountered such a proof while thinking about $pqr$-groups, and an even weirder one earlier in the semester.

A theorem about $pqr$-groups

Here's a weird chain of implication I ran into. Suppose $\vert G\vert = pqr$, where $p<q<r$ are primes. Let's see what we can learn about its $r$-Sylow subgroup(s).

Let $R$ be an $r$-Sylow subgroup of $G$. Possible values of $n_r$ (the number of $r$-Sylow subgroups in $G$) are $1$, $p$, $q$, or $pq$. $p$ and $q$ are both out because they're too small to be $1 \mod r$. Case 1: $R$ is normal, okay, neat. Case 2: Let's see what happens if $n_r = pq$.

Well, that takes up a hell of a lot of the space in $G$; specifically, there are $pq(r-1)$ elements of order $r$. This only leaves room for exactly one $p$-Sylow subgroup $P$, and exactly one $q$-Sylow subgroup $Q$; therefore, both $P$ and $Q$ must be normal in $G$.

Since $Q\triangleleft G$, the set $QR := {qr \mid q\in Q, r\in R}$ is a subgroup of $G$, and since $Q\cap R = 1$, $\vert QR\vert = qr$. Therefore, $[G:QR] = p$, and since $p$ is the smallest prime dividing $\vert G\vert $, $QR$ must be normal in $G$ by the small-index lemma.

Let's now zoom in on the subgroup $QR$. Since $r$ divides $\vert QR\vert $, $QR$ has an $r$-Sylow subgroup; indeed, our original guy $R$ must be such a thing. $n_r$ within $QR$ can only be $1$ or $q$, and $q$ is too small, so $R$ is normal in $QR$. Further, $R$ is characteristic in $QR$, since it's the unique subgroup of order $r$.

Therefore, since $R \operatorname{char} QR \triangleleft G$, $R$ is normal in $G$ (indeed, characteristic in $G$). \qed

That's a fine enough proof, and we used a lot of interesting facts along the way, but let me summarize the chain of implication to point out that something weird happened:

$R$ not normal → $P$, $Q$ normal → $QR$ is a subgroup (and since $[G:QR] = p$, $QR$ is normal) → $R$ is an $r$-Sylow subgroup of $QR$ → $R$ is normal in $QR$ → $R$ is characteristic in $QR$ → $R$ is normal in $G$

In fewer words: “Suppose that $R$ is not normal in $G$. Surprise, $R$ is normal after all!”

“Just” proof by contradiction?

So, a reasonable complaint about me calling this thing about $pqr$-groups the same kind of weirdness as the proof of the vanishing quotient theorem is that this one is ultimately “just” a proof by contradiction. If one was being very punctilious, one might write the statement of the theorem as “$R$ is normal in $G$”, begin the proof with “suppose not; then $n_r = pq$,” and end the proof with “therefore $n_r = 1$, which is a contradiction.” Okay, sure; this is now a pretty textbook proof-by-contradiction setup.

But the reason I didn't originally frame the theorem this way is because I didn't originally encounter this line of reasoning in such a pure form. In real life, I really did stumble into this from saying “hey, what do $pqr$-groups look like?”.

And further, even a textbook proof by contradiction is weird. Ben Blum-Smith writes in the aforementioned blog post that mathematical objects have a peculiar ontology:

The place where this is easiest to see is in proofs by contradiction. When you read a proof by contradiction, you are spending time with objects that you expect will eventually be revealed never to have existed, and you expect this revelation to furthermore tell you that it was impossible that they had ever existed. That’s bizarro science fiction on its face.

But just in case you've been doing math long enough that you are inured to the weirdness inherent in the technique of proof by contradiction (and whomst among us is not so jaded), here's an even weirder proof, in which no contradiction occurs.

The vanishing quotient theorem

Here is a theorem that is often given as an exercise in a group theory class: If $G/Z(G)$ is cyclic, then $G$ is abelian. This is a good exercise to give in a group theory class precisely because the proof is one of these ones that just falls out as soon as you write down all the relevant definitions.

Write $Z$ for $Z(G)$ and suppose that $G/Z$ is cyclic. Then it is generated by one element, and elements of $G/Z$ are cosets of $Z$, so say $G/Z = \langle gZ \rangle$.

Now our goal is to show that $G$ is abelian, so let's take some arbitrary $x, y\in G$ and argue that they commute. Since the cosets of $Z$ partition $G$, $x$ lives in some coset of $Z$, and since $G/Z$ is cyclic, every coset of $Z$ looks like $g^n Z$ for some integer $n$. Therefore, $x=g^n z$ for some element $z\in Z$; similarly, for some arbitrary $y\in G$, $y = g^m z'$ for some element $z' \in Z$. Both $z$ and $z'$ commute with everything since they were elements of $Z(G)$, and $g^n$ and $g^m$ certainly commute with each other, so:

\[\begin{aligned} xy &= (g^n z) \cdot y \\ &= g^n (y\cdot z) = g^n (g^m z') z \\ &= (g^m g^n) z'z = g^m (z'\cdot g^n) z \\ &= yx. \end{aligned}\]

Therefore, $G$ is abelian, hooray, and this is a direct proof in which no contradiction has occurred.

The things in this proof are secretly quite weird

But now take a look at the consequences of what we've just proved. Since $G$ is abelian, $Z(G) = G$ – that's right, $Z(G)$ is all of $G$! Remember how we said $x = g^n z$ for some integer $n$ and some element $z\in Z(G)$? Well, by golly, $n = 0$ and $z=x$! What a bullshit argument this has turned out to be!!

To think our way through this proof, we had to believe implicitly that $x$ was some non-central element of $G$, so that we could think about which implicitly non-identity coset of $Z$ it lived in, so that we could use the structure of the implicitly non-trivial quotient group, so that we could say $x = g^n z$. But all of that was kinda wrong!!!

This is why, thanks to Andrew Stacey, I now refer to this as the “vanishing quotient theorem:” if $G/Z(G)$ is cyclic, then it vanishes, from the sole pressure of your thinking upon it.

The objects in this proof are so weird. We had to assume that they were “generic particulars,” with nothing specific to distinguish them from any ol' element, but then they turned out to be very specific indeed. The “generic” integer $n$ in $x=g^n z$? It was secretly impossible for it to be anything but 0. That $g$ whose coset $gZ$ was a generator for $G/Z$? Actually just the identity. Those implicitly possibly-non-central elements of $G$? Actually central after all.

All this is very weird! It should actually be amazing that all of us mathematicians have developed the capacity to work with these implicitly-generic-but-actually-special objects as if they were sensible, well-behaved things like the number 2 or the area of a circle1. It should be even more amazing to us that this weirdness emerges from a pretty elementary proof of a pretty elementary theorem. And when students find these things difficult, we should remember that they very much are.

  1. This is a joke. ↩