I have the following question on group theory:
A. show that there are 745 different homomorphisms from $\mathbb Z_5 \times \mathbb Z_5 \times \mathbb Z_5$ to $S_5$
B. How many different homomorphisms are from the group $S_5$ to $\mathbb Z_5 \times \mathbb Z_5 \times \mathbb Z_5$ ?
I have done a similar question about the number of homomorphisms from $D_4$ to $Q_8$, which uses the generating set of $D_4$. I assume that for A I need to do the same.
However, I don't think I have a generated set for $S_5$, and it appears I need to look at the kernel and image of the homomorphism.
I don't really have a strong grasp on group theory, so I could use some help with this. Thank you for answering!
alex.jordan's answer has already addressed part (a), but I think there is an error in their solution to part (b).
Let $f \colon S_{5} \to \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ be an arbitrary homomorphism. Let $\sigma$ be an arbitrary transposition in $S_{5}$. Since $\sigma$ must have order $2$, we have $$f(\sigma) + f(\sigma) = f(\sigma \cdot \sigma) = f(e) = (0, 0, 0),$$ hence $f(\sigma)$ has order dividing $2$ in $\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$.
However, since $\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ is a group of odd order, there are no elements of order 2, hence the only element with order dividing $2$ in $\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ is the identity.
Therefore, for any transposition $\sigma \in S_{5},$ $f$ must send $\sigma$ to the identity. However, we know that $S_{5}$ can be generated by transpositions, hence $f$ must send all of $S_{5}$ to the identity.
So, any homomorphism from $S_{5}$ to $\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ must be trivial, hence there is only 1 such homomorphism.