Final answer:
There are two homomorphisms from \(\mathbb{Z}_{25}\) to \(\mathbb{Z}_{10}\), but neither of them is an epimorphism because they do not map onto all elements of \(\mathbb{Z}_{10}\).
Step-by-step explanation:
To determine how many homomorphisms there are from \(\mathbb{Z}_{25}\) to \(\mathbb{Z}_{10}\), we can use the property that in a homomorphism between two groups (\phi: G \rightarrow H), if g is an element of group G, then the order of \(\phi(g)\) must divide the order of g.
Given that 25 and 10 are not coprime (share a common factor of 5), there are homomorphisms possible, because any homomorphism must map the generator of \(\mathbb{Z}_{25}\) to an element of \(\mathbb{Z}_{10}\) with order dividing 25. In \(\mathbb{Z}_{10}\), the orders of elements divide 10, which means they are 1, 2, 5, or 10.
Since 5 is the only one of these which also divides 25, the generator of \(\mathbb{Z}_{25}\) can only be mapped to elements of order 1 (the identity element) and 5 in \(\mathbb{Z}_{10}\). The elements of order 5 in \(\mathbb{Z}_{10}\) are 0 and 5, yielding two possible homomorphisms.
An epimorphism is a surjective homomorphism, meaning it maps G onto H. In our case, since \(\mathbb{Z}_{10}\) has more than just the two elements 0 and 5, the only homomorphism that could potentially be an epimorphism would be the one that maps the generator to 5. However, this homomorphism doesn't hit all elements in \(\mathbb{Z}_{10}\), so neither of the two homomorphisms are epimorphisms.