2
$\begingroup$

Prove that the groups $\langle a,b \mid ababa=babab \rangle$ and $\langle x,y \mid x^2=y^5 \rangle$ are isomorphic.

I managed to solve this problem using by using Tietze transformations. However, I'm not familiar with that Tietze theory and it doesn't seem plausible to me. So I was wondering how to solve this by more standard methods? That is, the first group is the quotient of the free group on the elements $a,b$ by the normal subgroup generated by $ababab^{-1}a^{-1}b^{-1}a^{-1}b^{-1}$, and similarly for the second group. How to establish the isomorphism explicitly? (Or how to show it exists directly, without making use of Tietze transformations?)

$\endgroup$
  • $\begingroup$ I'm guessing the generator $x$ is identified with $ababa$ and $y$ is identified with $ab$? $\endgroup$ – Theo Bendit Jul 3 '18 at 2:42
  • $\begingroup$ @TheoBendit I used $x=babab, y=ab$. $\endgroup$ – user419669 Jul 3 '18 at 2:45
  • $\begingroup$ They are the same, as $ababa=babab$. $\endgroup$ – Kenny Lau Jul 3 '18 at 2:47
3
$\begingroup$

We implicitly apply the universal property of group presentations.

$x=ababa$ and $y=ab$ satisfies $x^2=y^5$, so we get a surjection from the second group to the first group.

To have the inverse map, we need to solve for $a$ and $b$ in terms of $x$ and $y$:

  • $a = ababab(babab)^{-1} = y^3x^{-1}$
  • $b = babab(ab)^{-2} = xy^{-2}$

This gives a surjection from the first group to the second group.

By an abstract nonsense argument, the maps are mutually inverse.

$\endgroup$
0
$\begingroup$

Your groups both have a single defining relator. In this setting there is a relatively simple check for isomorphism (here, $F(\mathbf{x})$ denotes the free group on the set $\mathbf{x}$, etc):

Lemma. Suppose that there exists an isomorphism $\phi:F(\mathbf{x})\rightarrow F(\mathbf{y})$ such that $\phi(R)=S$ or $\phi(R)=S^{-1}$. The the one-relator groups $\langle \mathbf{x}\mid R\rangle$ and $\langle \mathbf{y}\mid S\rangle$ are isomorphic.

The proof is via (ahem) Tietze transformations. It is easy use this lemma to show that your two groups are isomorphic.

It is an extremely interesting question to ask when the above lemma is an "if and only if" condition. That is:

Question. For which classes $\mathcal{C}$ of one-relator groups is it true that $\langle \mathbf{x}\mid R\rangle\cong\langle \mathbf{y}\mid S\rangle$ if and only if there exists an isomorphism $\phi:F(\mathbf{x})\rightarrow F(\mathbf{y})$ such that $\phi(R)=S$ or $\phi(R)=S^{-1}$?

Showing that this question has a positive answer for a class of one-relator groups solves the isomorphism problem for these groups. Which is big news! The question is known to have a positive answer for the following classes of groups:

  • Two generator, one-relator groups with torsion (that is, groups of the form $\langle a, b\mid R^n\rangle$ with $n>1$). Pride, 1977 (link).
  • "Generic" one-relator groups. Kapovich-Schupp, 2005 (link).
  • No other classes of groups which I know of (and this is one of my favourite questions!).

This question is related to an old conjecture of Magnus (subsequently shown to be false) that freely-indecomposable $n$-generator one-relator groups have a single Nielsen equivalence class of generating $n$-tuples. Both Pride and Kapovich-Schupp prove this "single Nielsen equivalence class" property for the classes of groups they are studying.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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