Is there a group G of order 20 such that there exists a surjective homomorphism $\phi: G \rightarrow \mathbb{Z}_{15}$?

Question

Is there a group G of order 20 such that there exists a surjective homomorphism $\phi: G \rightarrow \mathbb{Z}_{15}$?

I am not sure how to approach this.

$\mathbb{Z}_{15}$ is a cyclic group, and if $\phi$ is surjective then there is $g\in G$ such that $\phi(g)=1$.

By the definition of homomorphism, I get that for all $1\leq m \leq 14$, $\phi(g^m)=m$, and that $\phi(g^{15})=0$.

I know that $G$ doesn't have an element of order $15$, but not sure how to use this.

Answer 1

Hint: $\mathbb{Z}_{15}$ has an element whose order is $3$.

Answer 2

Hint:

If $f:G \twoheadrightarrow H$ is a surjective homomorphism, then $\;|H|=[G:\ker f]$ is a divisor of $|G|$.

Answer 3

$\ker\phi$ is either trivial or has at least $2$ elements. In the first case, we have an isomorphism, in the second the image has order at most $10$.