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Let $G$ be the finitely presented group with two generators $a,b$ and one relation $b^2=1$. First question:

Does that group have a name ?

Perhaps an answer to this question can lead me to interesting literature concerning this group and containing informations on the question below. The questions that interest me about this group are related to the Burnside problem:

For $m>0$ an integer, let $G_m$ be the quotient of $G$ by the normal subgroup generated by all $m$-th power in $G$.

Is $G_m$ finite ? if yes, are there known upper bound for its order, and if not, for the order of its finite quotient ?

(I am looking for bounds that are much better that what you get when $G$ is replaced by the free group in two generators) ?

Other kind of question:

Is the word problem solvable for every quotient of $G$ ?

I know that there are groups with two generators with unsolvable word problem but the condition that one generator is an involution seems to simplify the problem quite a bit...

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3  
1. Yes, it is called $Z\star Z_2$. 2. $G_m$ is infinite for some large even $m$'s, this follows, from say, paper by Olshanskii, Minasyan and Sonkin "Periodic quotients of hyperbolic and large groups", since $G$ is nonelementary hyperbolic. –  Misha May 31 '12 at 19:06
    
Since the OP is also interested in finding literature on these subjects, let me mention the phrases "free product" in connection with Question 1 and "Burnside group" and "Burnside problem" in connection with Question 2. –  Andreas Blass May 31 '12 at 20:39
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1 Answer 1

  1. Yes, it is called $Z\star Z_2$.

  2. $G_m$ is infinite for some large even $m$'s, this follows, from say, paper by Olshanskii, Minasyan and Sonkin "Periodic quotients of hyperbolic and large groups" (Groups Geom. Dyn. 3 (2009), no. 3, 423-452), since $G$ is nonelementary hyperbolic, or, alternatively, "large."

  3. I am not sure about this one, but I do not see any reasons why having an order 2 generator would force solvable word problem in a 2-generated (recursively presented) group. A better question, I think, would be:

Let $G$ be an arbitrary nonelementary hyperbolic group. Does $G$ have a recursively presented
quotient with unsolvable word problem?

One way to construct such a quotient would be to take a f.g. free subgroup $F\subset G$, add finitely many relators $R$ to $F$ to get $F/\langle\langle R\rangle\rangle$ which has unsolvable word problem. Now, add the same relators to $G$. I think $G/\langle\langle R\rangle\rangle$ also has unsolvable word problem, but I did not think about the details.

Update: Indeed, as Yves observed, every nonelementary hyperbolic group $G$ is SQ-universal. Namely, find a finitely-generated free subgroup $F\subset G$ with free generators $x_1,...,x_n$ such that the elements $x_i$ and their distinct $G$-conjugates freely generate a free subgroup in $G$. Then for every set $R$ of elements in $F$, the group $F/\langle\langle R\rangle\rangle_F$ embeds in $G/\langle\langle R\rangle\rangle_G$. Here subscripts $F$ and $G$ refer to the normal closures in $F$ and $G$ respectively. Now, SQ-universality for $G$ follows from Higman's embedding theorem. It seems that SQ-universality of hyperbolic groups was first proven by A.Olshanskii in "The SQ-universality of hyperbolic groups", Math. Sbornik, 1995. I did not see his paper, but I assume that his proof was along the above lines.

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1  
@Misha the answer to your question is certainly yes: given a recursive sequence of independent elements $r_n$ in $G$ (in the sense that for every $n$, $r_n$ is not in the normal closure of the set of all others), pick the quotient of $G$ by the $r_n$ for $n$ ranging over a recursively enumerable, not recursive set of indices. The point it the n to find the $r_n$ but this is very basic stuff for specialists of hyperbolic groups. –  YCor May 31 '12 at 21:35
    
@Yves: Yes, indeed, I was thinking about finitely-presented quotient while typing the answer. –  Misha May 31 '12 at 22:53
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OK; the answer is still yes: every nonelementary hyperbolic H group is SQ-universal, i.e. every countable group G embeds into some quotient of H. In particular, if G is f.p. and H is SQ-universal, then G embeds in a quotient of H by finitely many elements (enough to kill all the relators of G). Picking G with unsolvable word problem, you get what you want. –  YCor Jun 1 '12 at 0:44
    
@Yves: Of course, you are right. –  Misha Jun 1 '12 at 2:41
    
@Misha: SQ-universality of hyperbolic groups was independently proved by Delzant (Duke, published 1996 but the preprint circulated in the early 90s, I saw it once in the bibliography of a 1993 paper I think) –  YCor Jun 1 '12 at 6:23
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