Explicit example showing non-residual finiteness

This is mostly a continuation on the group I gave in the last post, which is given by the presentation
$$G = \langle a,t ~|~ t^{-1}a^2t = a^3\rangle.$$ At the risk of beating a dead horse, I proved that the homomorphism $f:G\to G$ given on generators by $f(t) = t$ and $f(a) = a^2$ is surjective but not injective. Groups for which surjective homomorphisms are isomorphisms are called Hopfian, and so our group $G$ is not Hopfian.

As I've been talking about frequently in the past little while, a group $G$ is called residually finite if for every nontrivial $x\in G$ there exists a homomorphism $\varphi:G\to F$ such that $F$ is finite and $\varphi(x)$ is not the identity of $F$. In the post on residually finite groups, I explained the classic proof that a finitely-generated, residually finite group is Hopfian.

Now the particular group $G$ here that is not Hopfian is finitely-generated, and so of course it can't be residually finite. I was wondering, can we find an explicit nontrivial element $x\in G$ such that for every homomorphism $\varphi:G\to F$ where $F$ is finite, the element $\varphi(x)\in F$ is the identity of $F$?

Yes, that's actually quite easy. It is because in the last post we already proved that the commutator $[t^{-1}at,a]$ is sent to the identity under the endomorphism $f:G\to G$ (recall, which was given by $f(t) = t$ and $f(a) = a^2$). But if we carefully examine the proof of the statement "every finitely-generated residually finite group is Hopfian", we see that the kernel of $f$ is actually contained in every finite-index normal subgroup of $G$. Therefore, in particular, the commutator $[t^{-1}at,a]$, which is nontrivial by Britton's lemma, is mapped to the identity under every homomorphism $G\to F$ where $F$ is a finite group!

Yet another group that is not Hopfian

A few weeks ago I gave an example of a non-Hopfian finitely-presented group. Recall that a group $G$ is said to be Hopfian if every surjective group homomorphism $G\to G$ is actually an isomorphism. All finitely-generated, residually finite groups are Hopfian. So for example, the group of the integers $\Z$ is Hopfian.

Another example of a group that is not Hopfian was given by Gilbert Baumslag and Donald Solitar. Their group is the one-relator group
$$G = \langle a,t ~|~ t^{-1}a^2t = a^3\rangle.$$ …read the rest of this post!

A zero-dimensional ring that is not von Neumann regular

An associative ring $R$ is called von Neumann regular if for each $x\in R$ there exists a $y\in R$ such that $x = xyx$.

Now let $R$ be a commutative ring. Its dimension is the supremum over lengths of chains of prime ideals in $R$. So for example, fields are zero dimensional because the only prime ideal in a field is the zero ideal.

Theorem. Let $R$ be a commutative ring. If $R$ is von Neumann regular, then it is zero dimensional.

The proof follows directly from the definition: suppose $P\subset R$ is a prime ideal of a von Neumann regular ring. If $x\not\in P$ and $y\in R$ is an element such that $x = xyx$, then $x(1 – yx) = 0$. Since $x\not\in P$, we must have $1 = yx$. Therefore, $P$ is maximal.

What about the converse? That's what this counterexample is all about.
…read the rest of this post!

A finitely generated flat module that is not projective

Let's see an example of a finitely-generated flat module that is not projective!

What does this provide a counterexample to?

If $R$ is a ring that is either right Noetherian or a local ring (that is, has a unique maximal right ideal or equivalently, a unique maximal left ideal), then every finitely-generated flat right $R$-module is projective.

So what happens if we drop the Noetherian and local hypotheses?

The Example

Let $R = \prod_{j=1}^\infty F_j$ be an infinite product of fields and let $I = \oplus_{i=1}^\infty F_j$ be the ideal that is the direct sum of all the fields. Then the module $R/I$ is finitely generated. It is also flat, because $R$ is von Neumann regular and in such rings, every module is flat. Why is it not projective?

To see that it is not projective, consider the exact sequence
$$0\to I\to R\to R/I\to 0.$$ If $R/I$ were projective, that would mean that the map $R\to R/I$ splits, which gives a direct sum decomposition $I\oplus R/I\xrightarrow{\sim} R$ where the composition of the map $I\to I\oplus R/I\to R$ is the inclusion $I\to R$. The image of $R/I$ then corresponds to a nonzero ideal in $R$. But any nonzero ideal intersects $I$, so such a splitting is impossible.

This example is part of my new counterexamples project.

Kourkovka Notebook: Open problems in group theory

Every once in a while I spot a true gem on the arXiv. Unsolved Problems in Group Theory: The Kourkovka Notebook is such a gem: it is a huge collection of open problems in group theory. Started in 1965, this 19th volume contains hundreds of problems posed by mathematicians around the world. Additionally, problems solved from past volumes are also included with references.

For example, F.M. Markel proved that if $G$ is a finite supersolvable group with no two conjugacy classes having the same number of elements, then $G$ is actually isomorphic to the symmetric group $S_3$. Pretty cool right? Jiping Zhang extended this theorem by replacing 'supersolvable' by just 'solvable'. Problem 16.3 in the Kourkovka notebook asks the obvious: if $G$ is any finite group where no two conjugacy classes have the same size, is $G\cong S_3$? There are of course many more problems of varying technicality, but there should be something in here for any group theorist.

I've always thought that you can gauge the health of a discipline by the quality of open problems in it. If that's true, then the Kourkovka notebook shows that group theory is thriving very well.

Britton's lemma and a non-Hopfian fp group

In a recent post on residually finite groups, I talked a bit about Hopfian groups. A group $G$ is Hopfian if every surjective group homomorphism $G\to G$ is an isomorphism. This concept connected back to residually finite groups because if a group $G$ is residually finite and finitely generated, then it is Hopfian. A free group on infinitely many generators is an example of a residually finite group that is not Hopfian.

Are there examples of finitely generated groups that are not Hopfian? Such an example would then of course give us an example of a group that is not residually finite.

In this post, we'll see an example of a group that is finitely presented and not Hopfian. Not only that, but I promise the construction is actually not even scary, unlike those finitely presented groups with unsolvable word problem.
…read the rest of this post!

How to make your own WordPress theme

This is a meta post on blogging, not mathematics. Recently, I got it into my head that I should design my own WordPress theme from scratch. As a consequence, you may have noticed that the theme of this blog has changed a little. I don't know if many other math bloggers will want to try this (given that there aren't too many active math bloggers these days), but if you really want the ultimate in customization, it's the only way to go.

The reason why I did this is that I wasn't totally happy with how customizable my last theme was, even though I like that theme generally. By designing your own WordPress theme, you can pretty much change anything and the final product is only limited by your own technical limitations. If you do design your own theme, you'll realize that designing a WordPress theme is really not that scary after all. Given that, you need a little basic knowledge to get your theme started:
…read the rest of this post!

Book Review: Riot at the Calc Exam by Colin Adams

When it comes to math humour, there's not much out there. There is a good list of jokes on MathOverflow. There's also Mathematical Apocrypha by Krantz, many of whose folklore stories are also amusing. The other day at the library I found another one: Riot at the Calc Exam by Colin Adams.
book cover
Adams' book is a collection of 33 short stories involving teaching, exams, research, and the math anxiety of students. Most, but not all of these are strictly in the short story format. "The Theorem Blaster" for example is an advertisement for a machine to simplify theorems. One of my favourites, "The Mathematical Ethicist", is a series of Dear X style letters with responses from an unethical ethicist.

So, is it funny? I found it to be so, and laughed aloud or smiled frequently. Most of the humour does indeed revolve around math themes, some of them advanced, so you really need an advanced math degree to get all the jokes on the first reading. That being said, there are end of chapter notes that I skimmed that give an explanation of some of the more esoteric concepts.

I liked also that the stories can be enjoyed on multiple levels. The "Deprogrammer's Tale" is a story about young students being drawn into mathematics as though it were a vice to beware. This is one of the few stories that is less humour and more caricature of truth. Being currently looking for jobs myself, I have had recent experience with the odd view that non-academic employers take towards a math PhD that vaguely mirrors some of the reactions of the characters in this story.

Riot at the Calc Exam is certainly a fun read and I recommend it to anyone looking for a collection of good stories imbued with the curious theme of mathematics.

Commutators and the Ore Conjecture

In a talk yesterday by Boris Kunyavski at the University of Ottawa, I learned a little about the Ore conjecture, which in 2010 was proved a theorem in:

Liebeck, Martin W.; O'Brien, E. A.; Shalev, Aner; Tiep, Pham Huu. The Ore conjecture. J. Eur. Math. Soc. (JEMS) 12 (2010), no. 4, 939-1008.

It's quite a fascinating result that arises by considering commutators in groups. If $G$ is a group, its commutator subgroup $[G,G]$ is the subgroup of $G$ generated by all the commutators $[g,h] = ghg^{-1}h^{-1}$ of $G$. It's easy to see that the commutator subgroup is normal. A group $G$ is said to be perfect if $G = [G,G]$.

So let's assume $G$ is perfect. This implies that every element of $G$ can be written as a product of commutators. But can every element of $G$ be written as a single commutator? That's really far from obvious. For example, take your favourite perfect group and an element in it: can you prove that this single element is a commutator? Not so easy, right?

In fact, we can define the commutator length of any $g\in G$ to be the minimum number of commutators in all products of commutators equal to $g$. If $g$ can't be written as the product of commutators, then its commutator length is infinite.

The commutator width of a group is defined to be the supremum over commutator lengths of all the elements of $G$. (Note: I think this should just be called the commutator length of $G$ as well, but that's how the terminology ended up!)

It turns out that finding a perfect group $G$ with commutator width greater than one is quite tricky. In fact, the theorem proved in loc. cit. is:

Former Ore Conjecture/Now Theorem. If $G$ is a finite nonabelian simple group, then every element of $G$ is a commutator.

That's pretty cool, though the proof is very long. That's not surprising since it is a theorem about all finite nonabelian simple groups. What's perhaps even more surprising is that there are examples of finitely-generated infinite simple groups containing elements that are not commutators. In fact, examples exist of such $G$ with infinite commutator length, as given in Alexey Muranov's paper:

Finitely generated infinite simple groups of infinite square width and vanishing stable commutator length. J. Topol. Anal. 2 (2010), no. 3, 341-384.

This result makes use of small cancellation theory, which is a geometric group theory machinery that studies presented groups whose relations don't have too much in common.