I ran across a group with presentation $$. Does this group have a name?
Can someone tell me what group this is?
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0I don't know, but $a^2=b^4$ implies $a^2b=ba^2$, so you can remove the first relation from the definition. – 2011-01-13
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0$b$ can be seen as a 90 degree rotation of a square, and $a$ as a reflection across a line perpendicular to a side and cutting the square in half. So basically, the group of plane symmetries of a square. – 2011-01-13
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4@Raskolnikov: The group has the plane symmetries as a quotient, but is not the group of plane symmetries; to see this, consider the map to the infinite cyclic group by sending $a$ to the square of the generator and $b$ to the generator; then all the relations are satisfied, so the infinite cyclic group is a quotient of this group; however, the group of plane symmetries of a square is finite, so it cannot have the infinite cyclic group as a quotient. – 2011-01-13
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1Yeah, you're right, I don't know why but I implicitly set $a^2=b^4=e$ with $e$ being the identity. But that is not included in the definition so that was unwarranted. – 2011-01-13
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0Thanks everybody for your comments. This group arose in a fashion similar to the Klein Bottle group when I was looking at groups of tree automorphisms, so I was wondering if there were some clever Tietze transformations I was missing to show this is the Klein Bottle group. Otherwise I didn't know what to do with it. – 2011-01-13
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0The subgroup is central, and the quotient is the infinite dihedral group $D_\infty$. Don't know if that helps or not. Anyway, this is most definitely not the Klein bottle group (this group has torsion, the Klein bottle group does not). – 2011-01-13
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0you're right about the torsion, thanks for pointing it out. – 2011-01-13
2 Answers
Let $G = \langle a, b\ |\ [a,b^2]=1, a^2=b^4\rangle$ be your group. Consider the map of $G$ to $\mathbb{Z}$, sending $b$ to $1$ and $a$ to $2$. This has a kernel which is $D_\infty$ (found via Reidemeister-Schreier rewriting). That is, the kernel is $\langle x,y\ |\ x^2=y^2=1\rangle$, where $x=ab^{-2}$ and $y=bab^{-3}$. Thus your group splits as a semidirect product $D_\infty \rtimes\mathbb{Z}$, where if $D_\infty=\langle x,y\ |\ x^2=y^2=1\rangle$ and $\mathbb{Z}=\langle t\rangle$, then the action of $t$ is: $x^t=y$, $y^t=x$.
EDIT: So we've gotten the following presentation of your group: $$ G= \langle x,y,t\ |\ x^2=y^2=1, x^t=y, y^t=x\rangle.$$
Let $K=\langle xy,t\rangle$ be a subgroup of $G$; it is not hard to see $K$ has index 2, and is isomorphic to the Klein bottle group. Thus your group $G$ is also a semidirect product $K\rtimes C_2$ (here $C_2$ is the cyclic group of order two), where if $C_2=\langle u\rangle$, and $K=\langle r,s\ |\ s^{-1}rs=r^{-1}\rangle$, then $r^u=r^{-1}$ and $s^u=sr^{-1}$.
Letting $W$ be an arbitrary word, using the third relation $a^2 = b^4$, we can assume that $W$ contains alternating $a$ and $b^k$ factors. Using the second relations $ab^2 = b^2a$ you can transform $W$ to the form
$$ W = b^{k} (ab)^l \mbox{ or } b^k (ba)^l $$
But now note $(ba)(ab) = b^6 \implies ba = b^6(ab)^{-1}$, so every word $W$ is equivalent to
$$ W = b^k (ab)^l $$
for some pair of integer $k,l$. Let us write out the multiplication table for this group in the presentation $(k,l)$. First, the identity is $(0,0)$. Observe that $(2k,0)$ commutes with all other elements. And $(k,l) = (k,0)(0,l)$, but $(0,l)(1,0) = (1+6l,-l)$. In other words, you can reduce to the group $\langle u,b | u^kb = b^{1+6k}u^{-k}\rangle$.
Or, you can think of the group as the free abelian group $\mathbb{Z}^2 = \langle u,v | uv = vu\rangle$ plus the additional generator $b$ where $b^2 = v$ and $b^{-1}ub = b^6 u^{-1}$. (Nope, I also don't know what this group is, but maybe someone else can take it from here.)
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2partial answers are of course fine (and BTW, I don't know what the group is either), but I think it's helpful to indicate at the beginning that you're giving a partial answer only. – 2011-01-14
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0@Pete: good point. Will keep that in mind in the future. – 2011-01-14