How can I find the order of $\overline {15}+ <\overline 4>$ in $\Bbb Z_{45}/<\overline 4>$? find the order of $\overline {15}+ <\overline 4>$ in $\Bbb Z_{45}/<\overline 4>$
how can I start? is it $\Bbb Z_{45}/<\overline 4>\cong \Bbb Z$

As $\;\bar 4$ is a unit in $\mathbf Z/45\mathbf Z$, the subgroup generated by $\bar 4$ is $\mathbf Z/45\mathbf Z$, hence $(\mathbf Z/45\mathbf Z)/\overline4(\mathbf Z/45\mathbf Z)=0$. So the order is $1$.
