 Solution

A cyclic group is a group generated by a single element: there is $g\in G$ such that every element is $g^n$ for some integer $n$. For a finite group of prime order $p$, choose $g\ne e$. Its cyclic subgroup $⟨g⟩$ has order dividing $p$ by Lagrange's theorem, and its order is greater than one. Hence $|⟨g⟩|=p$, and $⟨g⟩=G$. Thus every group of prime order is cyclic.

A group homomorphism from $C_{11}=⟨a⟩$ is determined by $h=\phi (a)$, since $\phi (a^j)=h^j$. Its image generator must satisfy $h^{11}=e$, so its order of a group element divides $11$. As an element of $C_{14}$ its order also divides $14$. These integers are coprime, so the only possible order is one. Thus $h=e$, andThis map is indeed a group homomorphism, so there is exactly one, the trivial homomorphism.