How many homomorphisms are there from $D_6$ to $D_5$?

Question

I know that:

$D_6$={$e,a,a^{2},a^{3},a^{4},a^{5},b,ab,a^{2}b,a^{3}b,a^{4}b,a^{5}b$}, with $a^{6}=e$ and $ba^{k}b=a^{-k}$.

$D_5$={$e,r,r^{2},r^{3},r^{4},s,rs,r^{2}s,r^{3}s,r^{4}s$}, with $r^{5}=e$ and $sr^{k}s=r^{-k}$.

Let $\varphi:D_6\rightarrow D_5$, a homomorphism is an injective function such that: $\forall x,y\in D_6:\varphi(xy)=\varphi(x)\varphi(y)$

I am now asked to determine how many homomorphisms there are from $D_6$ to $D_5$ but the only one i can find is the trivial one thus: $\varphi(e)=\varphi(a)=\varphi(a^{2})=\varphi(a^{3})=\varphi(a^{4})=\varphi(a^{5})=\varphi(b)=\varphi(ab)=\varphi(a^{2}b)=\varphi(a^{3}b)=\varphi(a^{4}b)=\varphi(a^{5}b)=r^{5}$

Can anyone help me with this i'm sure there have to be more homomorphism but i can't find them.

Answer 1

Such a homomorphism $\phi$ is completely determined by $\phi(a)$ and $\phi(b)$. One easily checks that $\phi(a)$ and $\phi(b)$ must be in $\{e,s,rs,r^2s,r^3s,r^4s\}$ and that we must have $a=e$, $b=e$ or $a=b$. In total, this yields $6+6+6$ homomorphisms. However, the trivial homomorphism is now counted three times (no other homomorphism is counted more than once), so in total there are $6+6+6-2=16$ different homomorphisms from $D_6$ to $D_5$.

Answer 2

Hint: In light of the first isomorphism theorem, you want to find normal subgroups $N\triangleleft D_6$ and then find subgroups of $D_5$ isomorphic to $D_6/N$.