Given $G=\mathbb{Z}_{16}^{\times}$ then how does one go about to determine the isomorphism type of $G/ \langle15 \rangle$ and $G/\langle 9 \rangle$?
Question: When you say $G/⟨15⟩$ has 4 elements, have you removed the elements $1,7,9,$ and $15$ from $G$? That is to say have you removed the elements who form subgroups of the same size as $⟨15⟩$ If so, is this what we want to do when $G/⟨a⟩$ in general?
With regards to comment below
Does this mean that we should count the order of say $3 (mod(16))$ when Dietrich Burde is speaking of "finding an element of order 4". If so, then would I be right to claim $G/⟨15⟩$ is C4 and $G/⟨9⟩$ is this as well?
$G$ may be represented by $\{ 1,3,5,7,9,11,13,15\}$, and has $\phi(16)=8$ elements. The subgroup $\langle 15\rangle$ equals $\langle -1\rangle$, since $15\equiv -1(16)$. Hence $G/\langle 15 \rangle$ has $4$ elements. There are only two different groups with $4$ elements: either $C_4$ or $C_2\times C_2$. If you find an element of order $4$, then your group is $C_4$, otherwise it is $C_2\times C_2$. The same applies for the group $G/\langle 9\rangle$. The Wikipedia page has the details.