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

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.

• First isomorphism theorem? Commented Jun 19, 2019 at 16:15

Hint:

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

• Thank you, I wrote this exactly but for some reason decided that it's not good. Commented Jun 19, 2019 at 16:18
• I really don't see why. Commented Jun 19, 2019 at 16:51

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

• Can you elaborate? Is there a way to solve this without applying the first isomorphism theorem? Commented Jun 19, 2019 at 16:19
• Let $h\in\mathbb Z_{15}$ be an element whose order is $3$. If there is a surjective homomorphism $\phi$ from $G$ onto $\mathbb Z_{15}$, then there is a $g\in G$ such that $\phi(g)=h$. But the order og $g$ will then be a multiple of the order of $h$, which is $3$. This is impossible, since the order of $g$ must divide $20$ and no multiple of $3$ does that, Commented Jun 19, 2019 at 16:25

$$\ker\phi$$ 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 $$10$$.