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Let $G = \langle x , y \ | \ x^{12}y=yx^{18} \rangle$. I want to continue a discussion here on the normal closure of $y$ in $G$.
How can we determine a group presentation for $N$?

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$N$ is free, so you are really asking for generators. –  Mariano Suárez-Alvarez Oct 13 '10 at 17:30
    
From Schreier's formula it follows that $N$ can be generated by 7 elements ([G:N]=6), maybe $\{y,\ldots,x^{-6}yx^6\}$ will do. I'm sorry but why does $N$ have no relation? –  Florian Pei Oct 13 '10 at 20:23
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$N$ is not free. –  user641 Oct 14 '10 at 3:43
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2 Answers 2

up vote 5 down vote accepted

This is a much harder question than your last one, and involves two separate parts: finding generators for $N$ and finding relations for $N$. I'll try to sketch the answers.

For context, recall from the previous thread that every element of $G$ can be written as a word in $x$ and $y$, and $N$ is the subgroup consisting of all words for which the total power of $x$ is a multiple of $6$. In particular, $N$ is a subgroup of $G$ with index $6$.

Generators

Finding generators for $N$ involves something called a Schreier graph. This is a directed graph defined as follows:

  • It has one vertex for each coset of $N$ in $G$ (so six vertices total)

  • There is a directed edge labeled $x$ from each coset $aN$ to the coset $xaN$, and a directed edge labeled $y$ from each coset $aN$ to the coset $yaN$. (Thus there will be $12$ edges total, $6$ labeled $x$ and $6$ labeled $y$.)

For the case you are looking at, the six vertices are $N,xN,x^2N,\ldots,x^5N$. These six vertices are arranged in a directed hexagon of $x$ edges), and each vertex has a loop labeled $y$: alt text

Note that each path in this graph corresponds to a word in $x$ and $y$, with backwards travel along directed edges corresponding to inverse generators. In particular, the elements of $N$ are the paths that start and end at the vertex $N$.

Now, the statement that is true is that generators for $N$ are the same as generators for the fundamental group of this graph, using $N$ as the basepoint. In this case, what that means is that $N$ is generated by the following elements: $$ A=x^6,\quad B_0 = y,\quad B_1 = xyx^{-1},\quad B_2 = x^2yx^{-2},\quad\ldots,\quad B_5 = x^5yx^{-5} $$ (In general, one can obtain generators for the fundamental group of a graph by first choosing a spanning tree for the graph, and then considering all closed paths starting at the basepoint that pass through exactly one edge not in the spanning tree. In this case, the spanning tree consists of the first five edges of the hexagon.)

Relations

Next, the relations for $N$ come from the relations for $G$. There is one relation for $G$, namely: $$ x^{12}yx^{-18}y^{-1} = 1. $$ This will lead to six different relations for $N$, obtained by conjugating this relation by representatives for the six cosets of $N$. Thus, the relations in $N$ can be written as $$ x^{12}yx^{-18}y^{-1} = 1, \quad x(x^{12}yx^{-18}y^{-1})x^{-1} = 1, \quad\ldots,\quad x^5(x^{12}yx^{-18}y^{-1})x^{-5}=1. $$ Of course, these are expressed in terms of $x$ and $y$, and we must rewrite them in terms of the generators $A,B_0,\ldots,B_5$: $$ A^2B_0 A^{-3} B_0^{-1} = 1,\quad A^2B_1 A^{-3} B_1^{-1} = 1,\quad\ldots,\quad A^2B_5A^{-3}B_5^{-1} = 1 $$ (The justification for this comes from algebraic topology. Each group $G$ has an associated 2-complex, which has a single vertex, one loop at the vertex for each generator, and one 2-cell for each relation. The fundamental group of this 2-complex is equal to $G$. The finite-index subgroup $N$ corresponds to a finite-sheeted cover of this 2-complex. This cover has a Schreier graph as its 1-skeleton, and the six 2-cells of the cover are attached along the six words listed above.)

Final Result

In any case, we have now obtained a presentation of the group $N$: $$ N = \langle A,B_0,\ldots,B_5 \mid A^2B_k = B_kA^3\text{ for each $k$}\rangle $$

By the way, the group $G$ that you seem to be interested in is called a Baumslag-Solitar group, and is usually denoted $B(12,18)$.

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Wow, many many thanks! Could you please suggest some reference to these results you mentioned between parentheses. Do you recommend that Johnson's book Max said? –  Florian Pei Oct 14 '10 at 1:08
    
I'm sorry but I cannot vote you up. –  Florian Pei Oct 14 '10 at 1:14
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I'm not familiar with the Johnson book. I think most of this stuff is in Stillwell's "Classical Topology and Combinatorial Group Theory", or any modern book on Algebraic Topology. –  Jim Belk Oct 14 '10 at 3:26
    
Yes, this is all in Johnson's book, as well as Stillwell's, and other books on combinatorial group theory. The general keyword is Reidemeister-Schreier rewriting. –  user641 Oct 14 '10 at 3:32
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I don't know the answer myself but you can take a look on Chapter 9 (Presentation of Subgroups) in Johnson's "Presentations of Groups" - I just discovered this very nice book. I'll do that when I get to university.
Unfortunately not many people participated in the first question, so thanks for continuing the discussion... I'll be voting you up when I can.

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Thanks. I'll look for the book. –  Florian Pei Oct 14 '10 at 1:00
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