Chris11 said:
Well. Perhaps I didn't have enough information to draw my conclusions. But, I'm currently an undergraduate, and it seems like you can get to high level research in graph theory in a shorter space of time than you can in algebra. This is based on the fact that we actually went through a large number of theorems in graph theory that were proved in recent years (≤10). When I go to look up papers online that are on group rep theory, I can't even understand the title of most of them, whereas in graph theory, I can.Its just that I got the impression that you can't start doing what people would consider meaningful research until the phd level. Perhaps the gap isn't as wide as I thought it was. The question above sounds very interesting to me by the way. But, I don't know locally indicable means. I guess there's only one solution. Do as much math as possible.
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
no in general.
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 realize that if you intersect all subgroups of a given finite index then you end up with a
characteristic subgroup - a subgroup $H\leq G$ such that $H\alpha=H$ for all $\alpha\in \operatorname{Aut}(G)$.
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.