Can we calculate the order of $\hom (G,G')$ in terms of $|G| $ and $ |G'|$ , when $G,G'$ are finite abelian groups?

Question

Let $G,G'$ be abelian groups and let $\hom (G,G')$ be the set of all homomorphisms from $G$ to $G'$.

We define an operation $\ast$ on $\hom (G,G')$ as: for $f,g \in \hom(G,G') \space , (f\ast g)(x)=f(x)g(x) , \forall x \in G$.

Then $\hom (G,G')$ becomes a group w.r.t. the operation $\ast$.

If $G, G'$ are finite then can we calculate the order of $\hom (G,G')$ in terms of $|G|$ and $ |G'|$?

Answer 1

No, you cannot, as this does not only depend on the order of the respective groups. You can have $|G_1|=|G_1'|$ and $|G_2|=|G_2'|$ while $|\hom (G_1,G_1')| \neq |\hom (G_2,G_2')|$.

For a specific example, consider $G_1$ a cyclic group of order $p^2$ and $G_2$ a direct product of two cyclic group of order $p$.

An element $f$ of $\hom (G_1,G_1)$ is determined uniquely by $g(e)$ where $e$ a generating elemnt of $G_1$ thus there are (at most) $p^2$.

However, an element $g$ of $\hom (G_2,G_2)$ is not determined by $g(e)$ for some nonzero element of $G_2$. By contrast, you still have $p^2$ choices for this, yet then can still "choose" the value on an element indepent of $e$.