Showing $\mathbf Z_2/8\mathbf Z_2\simeq\mathbf Z/8\mathbf Z$
$2$-adic integers are of the form $a_0+a_12+a_22^2+a_32^3+a_42^4\dots$
does modulo $8\mathbf Z$ means that, the only remaining part is $a_0+a_12+a_22^2$ with $a_i\in\{0,1\}$, so we have $2^3$ possibilites ?
and $\mathbf Z/8\mathbf Z$ has also $8$ elements, is this true or much easier is there an isomorphic mapping ?
Consider the ring homomorphism $\mathbb{Z}\to\mathbb{Z}_2/8\mathbb{Z}_2$ defined by $x\mapsto x+8\mathbb{Z}_2$ (using the embedding of $\mathbb{Z}$ into $\mathbb{Z}_2$).
This map is surjective (why?) and its kernel is $8\mathbb{Z}$ (why?).