non-residually finite group Let $G$ be the subgroup of $\text{Bij}(\mathbb{Z})$ generated by $\sigma : n \mapsto n+1$ and $\tau$ which switches $0$ and $1$. 
How can we prove that $G$ is not residually finite? Is it hopfian?
 A: *

*$G$ is not residually finite. Specifically, any map $f:G\to H$ with $H$ finite sends $g=(123)$ to the identity. To see this, pick $n$ with $n!>\max(5,|H|)$ and note that $f$ restricts to a map $S_n\to H$, which must have a non-trivial kernel, but the only normal subgroups of $S_n$ are $1$, $A_n$ and $S_n$.

*$G$ is hopfian. Consider an epimorphism $f:G\to G$. If $\ker(f)$ contains an element $g$ of infinite order, write $g=g'\tau^k$ where $g'$ is a finitary permutation. There exists $h$ with $f(h)=\tau$; write $h=h'\tau^\ell$ similarly. Then $h^kg^{-\ell}$ is a finitary permutation, so $f(h^kg^{-\ell})=\tau^k$ has finite order, but this is a contradiction. If $\ker(f)$ contains a non-trivial element $g$ of finite order then the restriction $f:S_X\to G$ has a non-trivial kernel for some finite set $X=\{-n,\dots,n\}$. But we can choose $|X|\geq 5$ and derive a contradiction as before unless the kernel is the group of even permutations. But in the last case, since $n$ can be chosen arbitrarily large, $\ker(f)$ is the group of even permutations, so $G/\ker(f)$ is isomorphic to $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ (I think), contradicting the surjectivity of $f$.
So $f$ must be an isomorphism.
