Detailed explanation wanted about group homomorphisms I am studying for an algebra exam and I am stuck on the following question:
Let
$D$
be the dihedral group of order 10 and let
$C$
be the cyclic group of order 10
.
Describe explicitly the full set of homomorphisms
$D
→
C$.
(You may assume standard facts about cyclic and dihedral groups but you must state these.)
I think I understand what a homomorphism is and the criteria that a group has to have in order to be a group homomorphism but I get completely lost when looking at there being homomorphisms between groups.
If someone could give me a detailed but simple explanation about how I go about this, it would be much appreciated!
Thanks
 A: The dihedral group of order $10$ is generated by two elements $\sigma$ and $\tau$ where $\tau^5=e$, $\sigma^2=e$, and $\sigma\tau=\tau^4\sigma$.  In other words, every element of $D_{5}$ (some people write this group $D_{10}$) can be written as products of $\sigma$ and $\tau$ and every time you see $\sigma\tau$, you can replace it with $\tau^4\sigma$.
On the other hand, by the structural properties of cyclic groups, $\mathbb{Z}/10$ (sometimes written $\mathbb{Z}/10$) has four subgroups, $\mathbb{Z}/10$, $\mathbb{Z}/5$, $\mathbb{Z}/2$, and the trivial group.
Suppose that we have a homomorphism $\phi:D_5\rightarrow \mathbb{Z}/10$, we need to determine the image of $\sigma$ and $\tau$.  Since $\sigma$ is order $2$, $\sigma$ could go to either $0$ or $5$ in $\mathbb{Z}/10$.  Similarly, since $\tau$ is order $5$, $\tau$ can go to one of $0$, $2$, $4$, $6$, and $8$ in $\mathbb{Z}/10$.  Now, let's look at the possible images of $\sigma\tau$.


*

*By using the properties a homomorphism,
$$
\phi(\sigma\tau)=\phi(\sigma)+\phi(\tau).
$$

*On the other hand, using the identity for $\sigma\tau$, 
$$
\phi(\sigma\tau)=\phi(\tau^4\sigma)=4\phi(\tau)+\phi(\sigma).
$$
Since these are supposed to be equal and $\mathbb{Z}/10$ is abelian, it follows that in $\mathbb{Z}/10$, $3\phi(\tau)=0$.  However, none of the nontrivial possibilities for $\phi(\tau)$ satisfy this equation.  Therefore, $\phi(\tau)=0$.
Hence, there are (at most) two homomorphisms from $D_5$ to $\mathbb{Z}/10$, the trivial map and the map that takes $\sigma$ to $5$ and $\tau$ to $0$.  Both of these occur, the trivial map is always a homomorphism and the other map corresponds to quotienting by the normal subgroup $\langle \tau\rangle$ and the first isomorphism theorem.  
We can also see that the second map is a homomorphism by seeing that the products of pairs of elements are taken to their sum:
\begin{align*}
0=\phi(\sigma^2)&=2\phi(\sigma)=5+5=0\\
0=\phi(\tau^2)&=2\phi(\tau)=0+0=0\\
5=\phi(\sigma\tau)&=\phi(\tau^4\sigma)=5.
\end{align*}
Alternately, we can look at this question on subgroups of $D_5$ to see that there are only three normal subgroups of $D_5$, $\{e\}$, $\langle \tau\rangle$ and $D_5$.  Of these, the quotienting by the trivial group gives $D_5$, which is not a abelian, so it is not a subgroup of $\mathbb{Z}/10$.  The other two quotients give groups isomorphic to $\mathbb{Z}/2$ and the trivial group, both of which are subgroups of $\mathbb{Z}/10$.
