= Solution
\b[$\boxed{BS(3,4)\text{ is not residually finite}.}$] We exhibit a nonidentity element killed by every finite quotient.
In any finite image, let $r$ be the order of the image of $a$. The relation conjugating $a^3$ to $a^4$ gives
$$
\frac r{\gcd(r,3)}=\frac r{\gcd(r,4)}.
$$
Thus $\gcd(r,3)=\gcd(r,4)=1$. Since $3$ is invertible modulo $r$, the relation implies that the image of $bab^{-1}$ is a power of the image of $a$. Therefore every finite image kills the <group commutator>
$$
w=[a,bab^{-1}]=a^{-1}b a^{-1}b^{-1}a b a b^{-1},
$$
using $[s,t]=s^{-1}t^{-1}st$.
View $BS(3,4)$ as an <HNN extension> of $\langle a\rangle\cong\mathbb Z$, with associated subgroups $\langle a^3\rangle$ and $\langle a^4\rangle$. A <pinch in an HNN extension> would be $ba^{3j}b^{-1}$ or $b^{-1}a^{4j}b$. The word $w$ has none: its intervening exponents are $-1,1,1$, incompatible with the required divisibilities $3,4,3$. By <Britton's lemma>, $w\ne1$. Hence finite quotients fail to separate this nonidentity element, proving the conclusion.
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