Is a Random Perfect Group Nontrivial?

20 Mar 2026 - Tags: sage

So it’s finals season, and earlier today some of the younger grad students were asking me for help studying for their topology finals. One of their practice problems was to build a cell complex with one 0-cell, two 1-cells, and two 2-cells which has nontrivial $\pi_1$ but trivial $H_1$. In principle, this isn’t very hard to do – I encourage you to think about it for a while to see what kind of group you’re looking for… Then see if you can look in the literature for a source of such a group.

Last chance to think about it yourself…

Ok, the fact that we have two 1-cells means our fundamental group will have two generators, say $a$ and $b$. Then the two 2-cells will give us two relations, say $R$ and $S$. Since we know that $H_1$ is the abelianization of $\pi_1$, this means we’re looking for a perfect group with a presentation by two generators and two relations.

If you try to build one by hand for a while (and I encourage you to try!), it’s actually somewhat difficult to do! I ended up looking up “small presentations of perfect groups” (or something like this) and quickly found Campbell, Kawamata, Miyamoto, Robertson, and Williams’s Deficiency Zero Presentations for Certain Perfect Groups which is full of examples¹. I also posted about this on mastodon, and Omar Antolín had a characteristically helpful response! He mentioned that the binary icosahedral group gets the job done², and these relations are the kind of thing you have a chance of finding by hand!

This brings up an interesting question, though – In some sense you would expect that two “random” relations should get the job done… Obviously things can’t be completely random, since we need the abelianized relations to be a matrix with determinant $\pm 1$ – otherwise our $H_1$ will have torsion… But what if we condition on this property. Does a “random” pair of relations work?

It’s been a minute since I’ve written a genuinely short blog post, and I was curious enough to write up some sage code anyways³, so I thought it could be fun to do it together!

Ok, so here’s the plan:

We’ll take as input a number $N$. Then we’ll iterate through all pairs of relations $R,S$ which are words in the alphabet \(\{a, a^{-1}, b, b^{-1} \}\) of length $N$ then we’ll see what fraction of those with vanishing $H_1$ also have nonvanishing $\pi_1$. If our conjecture is right then this fraction should get closer to $1$ as $N$ gets bigger.

Let’s do it!

This basically does exactly what you expect, haha. The main things to take note of are the @fork decorator when checking if a group is trivial. Since this is undecidable in general we bail on the computation if it lasts longer than thirty seconds… This creates some serious overhead on each loop, which is basically the same overhead involved in parallelising… So we might as well parallelise! This is what the @parallel decorator does. My laptop only has a measly 8 cores, so that’s how many I tell it to use. The reseed_rng flag is to make sure each process gets its own RNG, otherwise our 10,000 “random” runs will all be the same!

So what does the final scatter plot look like?

The trend is definitely upwards, which makes sense. Interestingly the code only starts seeing examples once the relations have length $7$. The smallest example I’m aware of is the binary icosahedral group that Omar told me about, which happens to have two relations of length $7$!

Even with relations of length $17$, though, only a tiny $0.4\%$ of the samples with trivial $H_1$ had an interesting $\pi_1$… I still think that this ratio should approach $1$ as the size of the relations gets large? But I was really hoping this data would be more suggestive, haha. That said, there’s a lot of problems with the data!

My laptop starts to struggle after $N=8$, which corresponds to relations of length $17$. I’ve uploaded the raw output if you’re interested in looking at it, but the main thing to note is how often we bail on computations. This is almost certainly throwing off our statistics, but I don’t really know how. After all, we’re now conditioning on both “trivial $H_1$” and “can be checked to be (non)trivial in at most 30 seconds on my ten year old laptop”. Here’s a table:

Relation Length	Nontrivial $\pi_1$	Total Runs	Killed Computations
1	0	10000	0
3	0	10000	0
5	0	10000	0
7	8	10000	0
9	10	10000	2
11	18	10000	19
13	27	10000	183
15	34	9998	664
17	47	9995	1726

So at this point the possible error we’re incurring by killing computations that take longer than $30$ seconds is dwarfing the number of groups we’re able to quickly prove are (non)trivial. Plus when we kill computations there seems to be a small chance that GAP throws some kind of exception that I’m not sure how to handle… This is the reason some later runs have slightly fewer than $10,000$ trials. I thought about increasing the timeout to a minute, or even five minutes, and running it again overnight to see if we can push things a bit further? But I decided I don’t really care, haha. I’m supposed to be writing a thesis, after all.

Further optimizing this would make a great project for somebody else to try, though! I think that it should be quite easy to get better data, and a lot of it! If you decide to look into this, definitely reach out and let me know what you find ^_^. In that vein, here’s a few take-home problems that you might want to play around with. They all look fun (at least to me) and if I had the time to spend I would probably think about them for a few weeks.

solution

We know there's $2n$ many letters in each of the relations $R$ and $S$, coming from the alphabet $\{a,A,b,B\}$. We'll write $R_a$ for the number of times $a$ shows up in $R$, and similarly for $S_B$, etc. Then working mod $2$ we have $$R_a + R_A + R_b + R_B = 2n \equiv_2 0$$ so that (remembering that $+$ and $-$ are the same mod $2$) $$R_a - R_A \equiv_2 R_b - R_B$$ Recall that the abelianization of $G = \langle a,b \mid R, S \rangle$ vanishes if and only if the following matrix has determinant $\pm 1$: $$ \begin{pmatrix} R_a - R_A & S_a - S_A \\ R_b - R_B & S_b - S_B \end{pmatrix} $$ but working mod $2$ again and using the above observation say that $R_a - R_A \equiv_2 x \equiv_2 R_b - R_B$ and $S_a - S_A \equiv_2 y \equiv_2 S_b - S_B$. Then the mod-2 determinant of the above matrix is $xy - xy = 0$ so that this matrix can never have determinant $\pm 1$!

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Alright! This is my first really short blog post in… quite a while, haha. I wrote the code pretty quickly, and once I figured out how to get the long computation to time out properly (and let it run all evening yesterday) the rest of the post came together in like two hours. I love problems like this, so it was easy to get nerdsniped by it.

Now it’s back to checking some details for my thesis. Next week is spring break, so I’m hoping to really sit down and get a lot done before teaching starts back up.

OH! And that reminds me! I mentioned this in a draft for a blog post on representation theory in analysis, but I haven’t mentioned it anywhere that’s live yet, so I should say it here! I got a postdoc!! 🎉🎉

This Fall I’ll be going to Montana State University to work with Sam Gunningham, David Ayala, and probably Ryan Grady too. I’ll be thinking about all sorts of fun things like factorization homology, “quantum” geometric langlands, topological field theories, and more! Everyone at the MSU campus has been so nice to me, and even though I’m an island girlie through and through I’m oddly excited to experience real winter for the first time in my life, haha. I’m tearing up a little bit writing this because I’m so happy to get to go there and work with them.

Alright, thanks for reading everyone! It’s back to the dissertation grind now, but this was really fun to think about for a day or two.

Stay safe, and we’ll talk soon 💖

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