Is there any group $G$ with $ord(G)=20$ so that $\varphi:G\rightarrow \mathbb{Z}_{10}$ is epimorphism?

I thout about $\mathbb{2\cdot Z}_{40}$, is it right?


Sure! For example the group $\Bbb{Z}_2 \times \Bbb{Z}_{10}$, where $\varphi$ is the projection on the second component.

If by $2 \cdot \Bbb{Z}_{40}$ you mean the image of the multiplication by $2$ homomorphism in $\Bbb{Z}_{40}$ then that works, too. Indeed, we know that $$ \Bbb{Z}_{40} \simeq \Bbb{Z}_8 \times \Bbb{Z}_5 $$ and since $\Bbb{Z}_5$ has no elements of order $2$ it follows that $$ G := 2 \cdot \Bbb{Z}_{40} \simeq \Bbb{Z}_4 \times \Bbb{Z}_5 \simeq \Bbb{Z}_{20} $$ which clearly has cardinality $20$. Similarly, we see that multiplication by $2$ in $G$ gives a surjective homomorphism $$ \Bbb{Z}_4 \times \Bbb{Z}_5 \to \Bbb{Z}_2 \times \Bbb{Z}_5 \simeq \Bbb{Z}_{10} $$

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