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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.

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

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  • $\begingroup$ Can you elaborate? Is there a way to solve this without applying the first isomorphism theorem? $\endgroup$ – איתן לוי Jun 19 at 16:19
  • $\begingroup$ 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, $\endgroup$ – José Carlos Santos Jun 19 at 16:25
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Hint:

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

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  • $\begingroup$ Thank you, I wrote this exactly but for some reason decided that it's not good. $\endgroup$ – איתן לוי Jun 19 at 16:18
  • $\begingroup$ I really don't see why. $\endgroup$ – Bernard Jun 19 at 16:51
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$\ker\phi$ 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 $10$.

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