Mapping between two cyclic groups

Let $\displaystyle G$ be a cyclic group of order 5. Let $\displaystyle G'$ be a cyclic group of order 7. Let $\displaystyle f : G -> G'$ be a homomorphism. Why is $\displaystyle f$ the trivial homomorphism?

In my group, the trivial homomorphism means the map that takes every element of the domain and sends it to the $\displaystyle e$ element of the codomain.

I know what $\displaystyle G$ and $\displaystyle G'$ look like.

$\displaystyle G = \{e, a, a^2, a^3, a^4\}$ and $\displaystyle G' = \{e, b, b^2, b^3, b^4, b^5, b^6\}$

But how is it that $\displaystyle f$ must be the trivial homomorphism?