# Number of homomorphisims from $C_5\times C_4\times C_4$ onto $C_{10}$

I know that any homomorphism between groups is determined by it's action a generating set of the group and that the kernel of such homomorphism must be of order 8 by the first isomorphism theory. By the Chinese remainder theorem the group $C_5\times C_4\times C_4$ isomorphic to $C_{20}\times C_4$ so if $a$ is a generator of $C_{20}$ and $e$ is the identity element in $C_4$ an epimorphism $f$ must map $(a,e)$ to an elememnt of order $10$ or $5$. I would like a hint on how to continue from here.

The first move is fine, i.e. looking at $C_{20}\times C_4$. This is generated by $x$ and $y$ of order $20$ and $4$ respectively. You know the order of their images divides the original order and divides the order of $C_{10}$, thus the order of $f(x)$ divides $(20,10)=10$ and the order of $f(y)$ divides $(4,10)=2$. This last means $f(y)$ has either order $1$ or $2$ in $C_{10}$, so things have become a little easier. It suffices you take care of the possiblities $f(x)$ has, and when they produce an onto homomorphism or not. Suppose then $C_{10}$ is generated by $z$.
The homomorphism $f(x)=z$ and $f(y)=1$ will be onto, while $f(x)=z^2$, $f(y)=1$ will be not. However, $f(x)=z^2$ and $f(y)=z^5$ will be onto, since $f(yx^{-2})=z$
Hint The point, maybe, is the following: for which choices of $f(x)$ and $f(y)$ can you find $m,n$ such that $f(x^my^n)=z^k$ is a generator of $C_{10}$, i.e. $(k,10)=1$?