Questions on the group with two generators $a,b$ and one relation $b^2=1$ 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...
 A: *

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

*$G_m$ 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."

*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. 
