In this problem we work out step-by-step the number of group homomorphisms from a dihedral group to a finite cyclic group. Recall that the dihedral group $D_{15}$ is generated by an element $a$ of order $15$, and an element $b$ of order 2, satisfying the relation

$(\ast)$ $ba = a^{15-1}b$

Let $f: D_{15} \rightarrow \mathbb{Z}_{6}$ be a group homomorphism.

Since $a^{15} = e$, properties of homomorphisms then imply that

$15 f(a) \equiv f(a^{15})$ (remember that $\mathbb{Z}_{6}$ is an additive group)
$\phantom{15 f(a)} \equiv f(e)$
$\phantom{15 f(a)} \equiv 0 \pmod{6}$

That means the possible choices for $f(a)$ are (please answer here as a comma-separated list of integers $\ge 0$ and $<6$).

Similarly, from $b^2 = e$ we see that the possible choices for $f(b)$ are (please answer here as a comma-separated list).

But there is an additional restriction: the group elements $a, b$ in $D_{15}$ satisfy the relation $ba = a^{15-1}b$. Apply the group homomorphism $f$ and use the fact that $\mathbb{Z}_{6}$ is Abelian, we see that this implies $(15 - 2) f(a)=0$. All together, we see that the number of group homomorphisms $f: D_{15} \rightarrow \mathbb{Z}_{6}$ is .