- Thread starter
- #1

- Thread starter Chris11
- Start date

- Thread starter
- #1

- Feb 1, 2012

- 57

How do you know its much much more active than algebra?

Are you talking about Abstract Algebra?

Are you talking about Abstract Algebra?

I know very little about research in graph theory, but I do know a bit about research in Algebra. And I wouldn't say it is `inactive'. For instance, in the past two-to-three years three long-standing conjectures have been proven in geometric group theory: the Hanna Neumann conjecture (Metaftsis), that one-relator groups with torsion are residually finite (Wise), and something called the virtually Haken conjecture which I don't understand but the proof was announced just the other week by Ian Agol (I just looked it up and found this post about it).

With finite group theory/rep. theory, the second approach at the classification of finite simple groups is continuing, and I believe a third attempt is started. Getting a foot in here will provide a job for life!

Semigroup theory is also a very active, if small, field.

Now, if by "Algebra" you mean "non-commutative algebra" then this is also a very large field of research. For example,

Q. What properties $\mathcal{P}$ of a group $G$ imply that the group algebra $\mathbb{C}G$ has zero-divisors?

It is relatively easy to find zero divisors if $G$ is finite, while it is a classical result of Higman that $G$ being locally indicable works...but what else?

Also, what do you mean by "start your research life until your late 20s"? I mean, I am 25 and study group theory, and although I have no papers published yet I

Of course, are you talking about doing research without doing PhD (although you do mention grad school, so I doubt this)? Well, there is a mathematician in Edinburgh called Agata Smoktunowicz who does research into non-commutative algebra stuff. She had a student who did amateur research in that area before he studied for his PhD with her.

So, in short, I do not think your reasons for studying graph theory over algebra are very good. This is pure maths we are talking about - the only reason you need is that you enjoy it!

Last edited:

- Thread starter
- #4

I should have said - locally indicable is basically the property required to make Higman's proof work. If I remember correctly, it means every subgroup of your group maps onto the integers.

I understand what you mean about there being a gap. However, this gap isn't necessarily a bad thing - it just means you have to do a lot of learning when you start your postgraduate studies, and so you spend all your time reading books and papers and stuff. However, this is good as it teaches you to think like a researcher!

Now, two of my three favourite group theory proofs are not overly high-level.

The first is about the subgroup membership problem. Basically, if I give you a group via a (recursively defined - it doesn't matter what this means though!) presentation $G=\langle X; mathbf{r}\rangle$ then for every subgroup $H\leq G$ is it possible to determine if a given element is in $H$? The answer is

The proof I like relies on the fact that there are $2$-generated groups with undecidable word problem. The word problem is similar - if I give you a group via a (recursively defined) presentation $G=\langle X; \mathbf{r}\rangle$ then if I give you a word $W$ over the letters of $X^{\pm 1}$ is it possible to determine if $W=_G 1$. Again, in general this is insoluble as there exists groups where this problem is insoluble. Indeed, there exist two-generated groups with this insolubility property (essentially because every group can be embedded in a two-generated group. Look up HNN-extensions for more details).

You can find an obscenely neat proof of the insolubility of the subgroup membership problem for $F_2\times F_2$ here.

The second is to do with a finiteness condition called "residually finite". A group is residually finite if for all $1\neq g\in G$ there exists a homomorphism $\phi$ from $G$ to some finite group $H_g$ ($H_g$ is dependent on $g$), so $\phi: G\rightarrow H_g$, such that $g\phi\neq 1_{H_g}$. This is a very nice property, and as I said in my earlier post, it is a recent result that every one-relator group with torsion ($G\cong \langle X; R^n\rangle$, $n>1$) satisfies this property. Note that $H_g\cong G/N$ for some finite-index subgroup $N$ of $G$, and $g\not\in N$.

Theorem: The automorphism group of a finitely-generated residually finite group is itself residually finite.

To understand the proof (due to G. Baumslag), you only need to know two things. Firstly, that if you intersect to finite index subgroups in a finitely generated group then you get another finite index subgroup. Secondly, there are only finitely many subgroups of a given finite index in a finitely generated group. Combining these, you should realise that if you intersect all subgroups of a given finite index then you end up with a

Proof: Let $G$ be a finitely generated residually finite group, and let $id\neq \alpha\in\operatorname{Aut}(G)$. We want to prove that there exists a homomorphism from $\operatorname{Aut}(G)$ to some finite group, $K$ say, such that the image of $\alpha$ under this homomorphism is non-trivial.

As $\alpha\neq id$ there exists $g\in G$ such that $g\alpha\neq g$. So, take $h=g(g^{-1}\alpha)\neq 1$. As $G$ is residually finite, there exists a finite index subgroup of $G$ not containing $h$. Intersecting all subgroups of this index, we see that there exists a characteristic subgroup $N\leq G$ such that $h\not\in N$. Then, because $N$ is characteristic in $G$, $\operatorname{Aut}(G)$ induces a finite group $A$ of automorphisms of $G/N$. However, $h\not\in N$ so $\alpha$ induces a non-trivial automorphism of $G/N$. Thus, $\psi: \operatorname{Aut}(G)\rightarrow A$ and $\alpha\psi\neq id_A$, as required.

This proof is from the 60s. It leaves us with a natural question: If $G$ is conjugacy separable, is $\operatorname{Out}(G)$ residually finite? (Conjugacy separable is basically residually finite but for conjugacy - $G$ is conjugacy separable if for every non-conjugate pair $u$ and $v$ there exists a homomorphism from $G$ to a finite group $H$ such that the images of $u$ and $v$ are non-conjugate in $H$, while $\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G)$.) E. Grossman proved that if $G$ satisfies something she called "Property A" and $G$ is conjugacy separable then one can edit Baumslag's proof and get that $\operatorname{Out}(G)$ is residually finite. So people do this - they take a group $G$ and prove that it is conjugacy separable and satisfies property A.

Now, my point isn't that you can understand these proofs and ideas, but that you should be able to understand them after only a little bit of reading.

I'm doing it at the moment.hey, swalbr are you a phd student at the moment or have you already finished it? Agata Smoktunowicz is my lecturer for my numbers and rings course this year (or was, my course is just about to finish)