Number of group homomorphism from $ \mathbb{Z}_{2} * \mathbb{Z}_{2} \to D_{8}$. How many group homomorphism are there from $ \mathbb{Z}_{2} * \mathbb{Z}_{2} \to D_{8}$?. 
There are $5$ elements of order $2$ in $D_{8}$ and any non-trivial element in $\mathbb{Z}_{2} * \mathbb{Z}_{2}$ either goes to an element of order $2$ or to an trivial element of $D_{8}$. I cannot go further from here.
Any help would be appreciated. Thanks in advance.
 A: The group $\mathbb{Z}_2 \star \mathbb{Z}_2$ is generated by two elements $(r, s)$ subject to the relation $r^2 = s^2 = 1$ and no other relations.
So the homomorphisms $\mathbb{Z}_2 \star \mathbb{Z}_2 \to G$ are in bijective correspondence with ordered pairs of elements in $G$ of order dividing $2$.
Since $D_8$ has $6$ elements of order dividing $2$, there are $6^2$ possible such ordered pairs and hence $36$ homomorphisms.
(EDIT: a little more detail on the second paragraph: the "universal property of the free product" says that to give a map from $K \star H$ to a third group $G$ is the same as to give a map from $H$ to $G$ and a map from $K$ to $G$. I am not sure what definition you are working with of the free product but this should have been proven as a theorem immediately after the definition. Now in the case that $K = H = \mathbb{Z}_2$, to give a map from $\mathbb{Z}_2$ is just to pick out an element of order dividing two, where you will send the generator. So we're just combining these two properties.)
