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!
 A: 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 $D_5$.
A: For part c) you can take the generators $r=(12345)$ and $s=(25)(34)$ of $D_5$ and map them via the identity to $S_5$. This yields an injective homomorphism $\phi\colon D_5\rightarrow S_5$. (Note that $srs=s^{-1}$).
