2
$\begingroup$

I was recently trying taking a look at the paper "Some two-generator one-relator non-hopfian groups" by Baumslag and Solitar where they introduce the groups now known as the Baumslag-Solitar groups given by the presentation $$ G = \left< a,b \mid a^{-1}b^la = b^m \right> $$ The paper is only 3-pages long and a bit skimpy on details (for me as someone without much group-theory know-how). I have a few questions about some of the claims in the paper.

In considering the case where there is a prime $p$ that divides $l$ but not $m$, the authors claim that the homomorphism given by $a \mapsto a, b \mapsto b^p$ has a nontrivial kernel, and namely, that the element $$ [b^{l/p}, a]^p b^{l-m} $$ which is in the kernel of the above map is actually nontrivial in $G$ (here the commutator convention is $[g_1,g_2] = g_1^{-1}g_2^{-1}g_1g_2$). With a little fussing, we can see that this map is surjective, and thus, if we see that that element is nontrivial, we have a non-hopfian group (which was a punchline/motivation in the above paper). My first question is, why is this element nontrivial?

My second question is with regards to a claim made on the first page. Taking $l = 2, m = 3$ we can see that $a$ and $b^4$ generate $G$. The claim is that for the surjection \begin{align*} F\langle x,y\rangle &\to G \\ x &\mapsto a \\ y &\mapsto b^4 \end{align*} the kernel is finitely normally generated, but not by a single element. Why is this kernel finitely normally generated but not by a single element?

I bit of playing around yields some elements of the kernel, and I imagine that there is some folding-style way of seeing if a given element of a free group is in the normal closure of some finite list of other elements, but I don't know how to do that. If I did, I could probably show that the kernel is not normally generated by a single element.

$\endgroup$
8
  • 1
    $\begingroup$ The modern answer to your first question uses Bass—Serre theory. Baumslag—Solitar groups are HNN extensions and, with respect to that structure, the given element is in normal form. This implies it’s non-trivial. $\endgroup$
    – HJRW
    May 17, 2021 at 14:23
  • $\begingroup$ The usual way (back then) to show that an element in any one-relator group is non-trivial is to just use the Magnus breakdown procedure, i.e. to solve the word problem. This is not very hard. This is spelled out in e.g. Magnus-Karrass-Solitar (and uses no geometric group theory). This is even easier if one uses HNN-extensions as HJRW suggests, but the original solution does not. $\endgroup$ May 17, 2021 at 14:53
  • $\begingroup$ @Carl-FredrikNybergBrodda: The "Magnus breakdown procedure" just involves developing the relevant bit of Bass--Serre theory in a special case. I see no reason to avoid the full theory, which is not really any harder than the special case. It's a very unfortunate feature of combinatorial group theory that some standard textbooks (eg Rotman) develop ideas according to historial, rather than conceptual, order. I haven't checked M--K--S, but I fear it's the same. $\endgroup$
    – HJRW
    May 17, 2021 at 15:32
  • 1
    $\begingroup$ Thank you both for the comments - I have enjoyed looking at some of those classic references in the past and I really appreciate pointers on how they fit together with our current understanding. I'll start looking at Serre tomorrow :-). $\endgroup$
    – user101010
    May 17, 2021 at 20:30
  • 1
    $\begingroup$ @HJRW I don't have much to add to what you write, other than saying that I certainly second learning from Serre! (Bogopolski is also very nice) :-) $\endgroup$ May 17, 2021 at 20:52

2 Answers 2

1
$\begingroup$

I'll write out the answer to your first question (solving the word problem in one-relator groups) for how one might do this in practice. I'll focus on the case $\ell = 2, m=3, p=2$, but you'll hopefully be satisfied that the methods are sufficiently general!

Let $G = \langle a, b \mid a^{-1} b^2 a b^{-3} =1 \rangle$. Then we wish to check that the word $$ W = [b, a]^2 b^{-1} = b^{-1} a^{-1} b a b^{-1} a^{-1} b a b^{-1} $$ is non-trivial in $G$. To do this, we first notice that the letter $a$ has exponent sum zero in the defining relation $a^{-1} b^2 a b^{-3}$, which is usually denotet by $\sigma_a(a^{-1} b^2 a b^{-3}) = 0$. Now any word $w$ with $\sigma_a(w) =0$ can be rewritten, by an easy trick, as a product of conjugates $b_i = a^i b a^{-i}$. For example, $$ a^{-1} b^2 a b^{-3} = (a^{-1} b a)(a^{-1}b a)(b^{-3}) = b_{-1}^2 b_0^{-3}. $$ The way this factorisation is found in practice is to look at the word $a^{-1} b^2 a b^{-3}$ and replace every occurrence of a letter $b$ by $b_i$ when the exponent sum of the $a$'s preceding that occurrence equals $i$, and then removing all $a$'s.

Now it is not hard to check that $G$ is an HNN-extension of the (torus knot) group $$ H = \langle b_{-1}, b_0 \mid b^2_{-1} b_0^{-3} = 1\rangle, $$ where the stable letter is $a$, identifying the two (free) subgroups $\langle b_{-1} \rangle$ and $\langle b_0 \rangle$. Note that this new group is (always!) a one-relator group with a shorter defining relation than $G$.

Now the key insight is that a word equals $1$ in $G$ if and only if it can be written as a product of $b_i$'s and this product equals $1$ in $H$. So to check if $W = 1$ in $G$, we note (again) that a word $v$ can be written as a product of $b_i$'s if and only if $\sigma_a(v) = 0$. As indeed $\sigma_a(W) = 0$, we can rewrite $W$, using the same trick as above if necessary, as $$ W = b^{-1} a^{-1} b a b^{-1} a^{-1} b a b^{-1} = (b^{-1})(a^{-1} b a)(b^{-1})(a^{-1}ba)(b^{-1}) = b_0^{-1} b_{-1} b_0^{-1} b_{-1} b_0^{-1}. $$

Now all we have to do is to solve the word problem in $H$ -- but this is easy (see e.g. this answer for the basics), as $H$ is just an amalgamated free product of two copies of $\mathbb{Z}$. In particular, we check quickly that the word $b_0^{-1} b_{-1} b_0^{-1} b_{-1} b_0^{-1}$ is reduced, and so is non-trivial in $H$. Hence it is non-trivial in $G$, and we are done.


This idea is very reminiscent of Magnus' 1932 solution to the word problem for all one-relator groups. This did not use HNN-extensions, of course (predating the HNN paper) by nearly two decades!), but rather considered chains of normal subgroups -- the same idea is used however, i.e. to keep passing to a shorter defining relation and solving the word problem (and some more problems) there, then proceeding by induction. The same reduction, using $b_i = a^i b a^{-i}$, is also there. An exposition of this is given in Magnus--Karrass-Solitar. Finally, sometimes one must deal with the case that there is no letter with exponent sum zero -- but this is not hard to deal with; one constructs a rewriting and then cranks the induction handle again, see e.g. [1]. For all Baumslag-Solitar groups, however, this is certainly not needed.

[1] McCool, James; Schupp, Paul E., On one relator groups and HNN extensions, J. Aust. Math. Soc. 16, 249-256 (1973). ZBL0288.20046.

$\endgroup$
2
  • 1
    $\begingroup$ The proof using normal forms for HNN extenions is two lines. The given presentation exhibits $BS(2,3)$ as an HNN extension of $\mathbb{Z}\cong \langle b\rangle$, with stable letter $a$ conjugating $\langle b^2\rangle$ to $\langle b^3\rangle$. The word $W=b^{-1}a^{-1}bab^{-1}a^{-1}bab^{-1}$ doesn't contain any pinches $a^{-1}b^{2n}a$ or $ab^{3n}a^{-1}$, so is in normal form and hence non-trivial. $\endgroup$
    – HJRW
    May 20, 2021 at 15:00
  • $\begingroup$ @HJRW Yes, that’s a nice compact way of writing the argument in this case! $\endgroup$ May 20, 2021 at 16:03
1
$\begingroup$

The assertion about the number of relations seems to be a theorem of Brunner, proved in the paper

Brunner, A. M. Transitivity-systems of certain one-relator groups. Proceedings of the Second International Conference on Theory of Groups (Australian Nat. Univ., Canberra, 1973), pp. 131–140. Lecture Notes in Math., Vol. 372, Springer, Berlin, 1974.

An electronic copy doesn't seem to be easy to come by, but I'll quote from Burns' MathSciNet review:

In the paper under review it is shown ... that the presentation $\langle x,y\mid x=[x,y]^2,[x,x^{y^n}]=1\rangle$ is associated with the generating pair $(a^{2^n},b)$ of $G$ for each postive $n$.

Here, $G=BS(2,3)$, of course.

$\endgroup$
3
  • $\begingroup$ Thanks for the reference - I'll check it out! I thought a bit more about proving this by brute force. The elements $y^{-3}x^{-1}y^2x$ and $y^{-6}x^{-1}y^4x$ are in the kernel. I'd guess that given a cyclically reduced word in a free group, that word is the shortest element in the normal group generated by that element. Assuming that, it would remain to look at words $w$ shorter than both of these and see that they are not in the kernel or if they are that both these elements are not in the normal closure of $w$. But I'm not sure if that works. $\endgroup$
    – user101010
    May 20, 2021 at 15:27
  • 1
    $\begingroup$ I wonder how Baumslag and Solitar proved the result - perhaps there is an easier argument. An electronic copy of the proceedings from that conference can be found on whichever pseudo-legal book source you want :-) (there is an excellent paper in there by Baumslag on the history of one-relator groups). $\endgroup$ May 20, 2021 at 16:01
  • $\begingroup$ @user101010: I think the fact you want is true; it's very similar to Magnus' famous result that $\langle\langle u\rangle\rangle=\langle\langle v\rangle\rangle$ in a free group implies that $u$ is conjugate to $v^{\pm1}$. But you still then have to apply the solution to the word problem in multiple different one-relator groups, so I don't think it's going to be easy! $\endgroup$
    – HJRW
    May 20, 2021 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.