Describing nontrivial Homomorphisms between $\mathbb{Z}/10\mathbb{Z}$ and $D_5$

Question

I was wondering if someone could figure out a way to describe the homomorphisms $\phi: \mathbb{Z}/10\mathbb{Z} \rightarrow D_5$ and $\varphi: D_5 \rightarrow \mathbb{Z}/10\mathbb{Z}$. Assume that $D_5 := \langle r,s : r^5 = s^2 = 1, rs = sr^-1\rangle$. I know that for any homomorphism $\varphi: G \rightarrow H$, for $a,b \in G$, we need $\varphi(ab) = \varphi(a)\varphi(b)$. Obviously the identities need to map to one another, $r$ in $D_5$ should correspond with an order 5 element in $\mathbb{Z}/10\mathbb{Z}$ (or the image of r should be the identity), and $s$ in $D_5$ should correspond with an element of order 2 in $\mathbb{Z}/10\mathbb{Z}$ (or again, should correspond with the identity). Am I approaching this correctly?

Answer 1

To figure out $\phi:\mathbb Z/10\mathbb Z\rightarrow D_5$ is easier. As $\mathbb Z/10\mathbb Z$ is cyclic, each $\phi$ is totally determined by the image of the generator $1+\mathbb Z/10\mathbb Z$, and such a $\phi$ exists iff the order of $\phi(1+10\mathbb Z)$ divides $10$. By Lagrange, the order of every element of $D_5$ is a factor of $10$. Therefore there are exactly $|D_5|$ distinct homomorphisms from $\mathbb Z/10\mathbb Z$ to $D_5$.

As for $D_5$ to $\mathbb Z/10\mathbb Z$, since $D_5$ has no element of order $10$, there is no isomorphism between them, and the kernel must be a nontrivial normal subgroup of $D_5$. To gain some intuitions, we recognize $D_5$ as the set of the identity, $4$ rotations of order $5$ and $5$ reflections of order $2$.

The kernel can be $D_5$ itself, in which case only one $\phi$ can realize it, i.e. the trivial homomorphism. Or $C_5$, in which case the image of the nontrivial element in $D_5/C_5$ can be mapped to any element in $\mathbb Z/10\mathbb Z$ of order $2$, and there is also only one of them. As all the reflections are conjugate to each other, there is no normal subgroup of order $2$. Hence there are only $2$ homomorphisms for this direction.