0
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $$ \Bbb{Z}_4 \times \Bbb{Z}_5 \to \Bbb{Z}_2 \times \Bbb{Z}_5 \simeq \Bbb{Z}_{10} $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.