Suppose the number of homomorphisms from a group $G$ to $G'$ is $n$. Find the number of homomorphisms from $G$ to $G' \times G' \times \cdots \times G'$ ($k$ terms).
I think answer is $n^k$. For this we need to prove:
Let $f_1 \ f_2 , \dots , f_n$ are the homomorphism from $G$ to $G'$ . I want to prove that any homomorphism from $G$ to $G' \times G' \times \cdots \times G'$ (k terms) is of the form $\phi : G \rightarrow G' \times G' \times \cdots \times G'$ (k terms) such that $\phi(g) =( f_1(g) ,f_2(g) , \cdots , f_k(g))$ , where $ 1 \leq i \leq n$, where $f_i = \phi_j$for some $1\leq j \leq n$ for all $1 \leq i \leq k$. I think it could be show by contradiction, but i do not know how to proceed.
I would be thankful for giving me your valuable time.
Hint: Let $\phi : G \to (G')^k$ be a homomorphism. The projection $\pi_j: (G')^k \to G'$ onto the the $j$th coordinate (i.e. $\pi_j((g'_1,\ldots, g'_k)):= g'_j$) is a homomorphism. Then the composition $\pi_j \circ \phi : G \to G'$ is a homomorphism. Then $\phi = (\pi_1 \circ \phi, \ldots, \pi_k \circ \phi)$, and you have your claim.