determine the number of homomorphisms from $D_5$ to $\mathbb{R^*} $ and from $D_5$ to $S_4$ I'm trying to do these three exercises for my math study:
a) Determine the number of homomorphisms from $D_5$ to $\mathbb{R^*}$
b) Determine the number of homomorphisms from $D_5$ to $S_4$
c) give an injective homomorphism form $D_5$ to $S_5$
I think I solved a):
The amount of homomorphisms is 1. The group $D_5$ is generated by a rotations and reflections. every reflection has order 2, and the rotations have order 1(the rotation about nothing) and order 5(the 4 rotations left). so we have to sent the reflection to -1 or 1, because the order of $f(x)$ has to divide the order of $x$ and the rotation about 0 degrees has only 1 as option to send it to. Combining that gives us only 1 homomorphism.
For b), I think the answer is 6! = 720, because every reflection in $D_5$ has to go to a 2-cycle in $S_4$, because the 2-cycles generate $S_4$. There are six of them, so the number of different homomorphisms are 6!.
Can you tell me if this is correct, and explain c) to me?
Thanks in advance!

For a) the answer is two because :
$$Hom(D_5,\mathbb{R}^*)=Hom(D_5/[D_5,D_5],\mathbb{R}^*)=Hom(Z/2Z,\mathbb{R}^*) $$
The cardinal of the last one is $2$. 
Edit : another way without abelianization.
Take $r$ a rotation and $s$ a reflection generating $D_5$. Take $f\in Hom(D_5,\mathbb{R}^*)$. You have $f(r)^{5}=1$, the only real number verifying this is $1$. 
Now every element of $D_5$ is written as $r^k$ or $r^ks$. The image of $r^k$ by $f$ must be $1$. The image of $r^ks$ by $f$ is $f(r^ks)=f(r)^kf(s)=f(s)$. 
We then see that the morphism $f$ is only determined by its value in $s$. This value must verify $f(s)^2=1$ so it is either $-1$ or $1$. This shows that there are at most $2$ such morphisms. To justify that there exist a non-trivial morphism from $D_5$ to $\mathbb{r}^*$ one can find a non-trivial morphism from $D_5$ to $Z/2Z=\{\pm 1\}$ by :
$$D_5\rightarrow D_5/<r> $$
For b) you see that if $f:D_5\rightarrow S_4$ is a morphism then $f(r)=Id$ (because there are no element of order $5$). Now such morphisms have then image in a subgroup of order $2$ of $S_4$. It follows easily that the number of such morphisms is the number of elements of order dividing $2$ in $S_4$, you have :
$$\frac{4\times 3}{2}=6\text{ transpositions and } 3\text{ double transpositions and the identity.} $$
you then get $10$ morphisms from $D_5$ to $S_4$.
Now for c), the idea is to find a subgroup  $H$ of $S_5$ isomorphic to $D_5$ (then the isomorphism from $D_5$ to $H$ gives you the injective homomorphism from $D_5$ to $H$).
It suffices (because $D_5=Z/5Z\rtimes_{-1} Z/2Z$) to exhibit an element $\tau\in S_5$ such that :
$$\tau^2=Id\text{ and }  \tau(1,2,3,4,5)\tau^{-1}=(1,2,3,4,5)^{-1}=(1,5,4,3,2) $$
Then you can verify that :
$$<(1,2,3,4,5)>\rtimes_{conj} <(2,5)(3,4)> $$
Is a subgroup of $S_5$ isomorphic to $D_5$.
