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I'm trying to find an isomorphism of a group with the following presentation:$$\langle a,b \mid (ab)^2=(abaa)^2=(abbb)^2=e\rangle$$

Basically, I'm not that experienced with groups so I'm wondering if I'm missing something obvious. I figured out some identities, but I can't seem to simplify the generators any further.

Shaun
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theelk801
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1 Answers1

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By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by

$$\langle a, y\mid y^2, y^{-1}a^2=a^{-2}y\rangle,$$

where $y=ab$. This presentation is equivalent (via Tietze transformations) to

$$\langle a, y\mid y^2, y^{-1}a^2y=a^{-2}\rangle,\tag{P}$$

which defines the group

$$\boxed{\operatorname{BS}(2, -2)/Y,}\tag{1}$$

where $\operatorname{BS}(2, -2)$ is a Baumslag-Solitar group and $Y$ is given by the normal subgroup $$\langle y^2\rangle^{\operatorname{BS}(2, -2)}$$ of $\operatorname{BS}(2, -2)$ generated by $y^2$.

Since $G$ and $(1)$ are defined by equivalent presentations, they are isomorphic.

NB: I could be wrong.

NB 2: I think $(\text{P})$ could be rewritten as an amalgamated free product. That might define a better (or even more accurate) group that $G$ is isomorphic to.

Shaun
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  • Here's the start of (what I hope would be) a chat on the accuracy of this answer. Feel free to comment below though. – Shaun Dec 11 '18 at 00:26
  • $y^2$ can not generate a subgroup isomorphic to $\Bbb Z$ – janmarqz Dec 12 '18 at 04:06
  • Thank you, @janmarqz; I've changed it to $Y$. – Shaun Dec 12 '18 at 11:02
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    This group splits as $1\rightarrow\mathbb{Z}\rightarrow G\rightarrow D_{\infty}\rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=\langle a^2\rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{\infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group). – user1729 Dec 12 '18 at 12:28
  • I'm trying to find $Y_p=\langle y^2\rangle^{\operatorname{BS}(2,-2)}$ using GAP. So far, I've done F:=FreeGroup(2); rels:=[ (F.2)^(-1)*(F.1)^2*(F.2)*(F.1)^2 ]; G:=F/rels; S:=Subgroup(G, [ (G.2)^2, (G.1)*(G.2)^2*(G.1)^(-1)]); then checked things like the abelian invariants of S but, so far, things have been inconclusive. I'd really like to pin down what $Y_p$ is. I might ask a question about it here in the near future. – Shaun Dec 13 '18 at 14:37
  • What d'you think, @user1729? – Shaun Dec 13 '18 at 14:38
  • @janmarqz, same question . . . – Shaun Dec 13 '18 at 14:38
  • are you taking $H^G$ as the set of maps $G\to H$ or the set ${ g^{-1}hg}$? and you only have $\langle y^2\rangle=e$, right? – janmarqz Dec 13 '18 at 15:02
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    @janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup..."). – user1729 Dec 13 '18 at 15:05
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    @Shaun I believe the normal closure of $y^2$, $Y_p:=\langle\langle y^2\rangle\rangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $\langle a\rangle$, and if $Y_p\cap\langle a\rangle^g\neq1$ for some $g\in BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_p\cap\langle a\rangle\neq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment. – user1729 Dec 13 '18 at 15:07
  • but if $y^2=e$ then $\langle y^2\rangle^{G}={e}$...? – janmarqz Dec 13 '18 at 15:12
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    @janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $\langle a, y; y^{-1}a^2y=a^2\rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $\langle y^2\rangle^{BS(2, -2)}$ (I much prefer the notation $\langle\langle y^2\rangle\rangle$, but anyway). The question in Shaun's comment is: describe this kernel. – user1729 Dec 13 '18 at 15:21
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    (If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).) – user1729 Dec 13 '18 at 15:24
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    Also, $\langle a\rangle$ is infinite cyclic (hence, $Y_p$ is free). You can see this by proving that the subgroup $\langle a^2\rangle$ is infinite cyclic using Reidemeister-Schreier to compute the kernel of the map in my above comment about splittings (for high-rep users, fellow the method of my deleted answer whilst not making the mistake I made...!). – user1729 Dec 13 '18 at 17:51
  • Thank you, @user1729; that gives me some insight into what's going on. By the way, to correct this comment of yours, $y^{-1}a^2y=a^{\color{red}{-}2}.$ – Shaun Dec 13 '18 at 20:27