Let $G = \langle a,b | a^3=b^3=(ab)^3 =1\rangle$. I'm trying to compute centralizers in $G$; in particular, I'm interested in the centralizers of $ab$, $ba$, $a^2$, and $b^2$. Does anyone know a good way to compute these centralizers? Will GAP do it for me or anything? Thanks.
computing centralizers in the von dyck group $\langle a,b | a^3=b^3=(ab)^3 =1\rangle$
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0@Steve: How can GAP han$d$le the group? I have to hand GAP a polycyclic presentation and somehow tell GAP that the group is polycyclic? – 2011-09-21
1 Answers
EDIT: It's actually not hard to see that $a$ generates its own centralizer, and similarly for $b$ or $ab$; the GAP code below is more useful for computing centralizers of longer words. To see that $C_G(a)$ has order $3$, one could first begin to understand the structure of G': it is generated by $[a,b]$ and $[a^2,b]$ and we have G'\cong \mathbb{Z}^2. [This is not surprising: your group is a Euclidean triangle group, and so should be virtually $\mathbb{Z}^2$.] In this way, $a$ acts as the matrix $\begin{pmatrix} -1 & -1\\ 1 & 0\end{pmatrix}$, and thus fixes no nonzero vector of $\mathbb{Z}^2$. That is, C_G(a)\cap G'=\{1\}. So $C_G(a)$ injects into G/G', an elementary abelian group of order $9$, and since $G$ is not abelian, certainly $b\notin C_G(a)$; thus $C_G(a)$ has order $3$.
Here's a way to do it in GAP (you will need the polycyclic package):
>ftl:=FromTheLeftCollector(4); <> >SetRelativeOrder(ftl,1,3); >SetRelativeOrder(ftl,2,3); >SetConjugate(ftl,2,1,[2,1,4,-1]); >SetConjugate(ftl,3,1,[4,-1]); >SetConjugate(ftl,4,1,[3,1,4,-1]); >SetConjugate(ftl,4,3,[4,1]); >SetConjugate(ftl,3,2,[4,-1]); >SetConjugate(ftl,4,2,[3,1,4,-1]); >UpdatePolycyclicCollector(ftl); >IsConfluent(ftl); true >g:=PcpGroupByCollector(ftl); Pcp-group with orders [ 3, 3, 0, 0 ] >c:=GeneratorsOfGroup(g); [ g1, g2, g3, g4 ] >a:=c[1];b:=c[2];u:=c[3];v:=c[4]; g1 g2 g3 g4
$a$ and $b$ are your generators; we also have $u=[a^2,b]$ and $v=[a,b]$:
>Comm(a^2,b); g3 >Comm(a,b); g4
Now we can compute centralizers; for example:
>w:=a*b*a; g1^2*g2*g4^-1 >h:=Centralizer(g,w); Pcp-group with orders [ 3, 0, 0 ] >IsAbelian(h); true >AbelianInvariants(h); [ 0, 0 ]
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0In particular, all of $a$, $b$, $ab$, and $ba$ generate their own centralizer (of order 3). – 2011-09-21