Let G be a group of exponent n and mgenerated. Suppose further that every finite quotient of the Burnside group B(m,n) can be occurred as a finite quotient of G. Can we say that G≅B(m,n) ?

$\begingroup$ Did you want to assume also that $G$ is infinite order? $\endgroup$– Ian AgolCommented Feb 22, 2012 at 15:37

1$\begingroup$ @Agol I would guess that if you take my answer and take a direct product of it with some infinite simple group which is generated by less than $m$ generators you might be able to manufacture an infinite example. $\endgroup$– Yiftach BarneaCommented Feb 22, 2012 at 21:34

$\begingroup$ yes, G has is infinite order. $\endgroup$– ali tavakoliCommented Feb 24, 2012 at 6:29
1 Answer
The answer is no. Take $\widehat B$ to be the profinite completion of $B(m,n)$. Then every finite quotient of $B(m,n)$ is still a quotient of $\widehat B$. However, from Zelmanov's solution of the Restricted Burnside Problem $\widehat B$ is finite.
In other words, take the intersection of all the normal subgroups of finite index in $B(m,n)$. From Zelmanov's solution of the RBP there are only finite number of them. Thus, the intersection is of finite index and the quotient satisfies your request, but it is finite (and of course generally, $B(m,n)$ is infinite).
Edit: Indeed as I commented above, let $q$ be a prime not dividing $n$ and let $S$ be a monster in which every element has order $q$. Then $G=\widehat B \oplus S$ satsifies the requirements. Clearly, $G$ has the same finite quotients as $B(m,n)$ since $S$ is simple. Now, $S$ is generated by any two elements not in the same cyclic subgroup. So let $y_1,y_2$ be two such generators, each has order $q$, and let $x_1,\ldots,x_m$ be generators of $B(m,n)$. Then $(x_1,y_1),(x_2,y_2),(x_3,1), \ldots, (x_m,1)$ generate $G$ since $(x_1,y_1)^n,(x_2,y_2)^n$ generate $S$.