I have a doubt regarding one of my textbook problems regarding homomorphisms.

My textbook says:

"Suppose that we wish to determine all possible homomorphisms $\phi$ from $\mathbb{Z}_7$ to $\mathbb{Z}_{12}$.

Since $\ker\phi$ must be a subgroup of $\mathbb{Z}_7$, there are only two possible kernels, $\ker\phi=\{0\}$ or $\ker\phi=\mathbb{Z}_7$.

The image of a subgroup of $\mathbb{Z}_7$ must be a subgroup of $\mathbb{Z}_{12}$. Hence, there is no injective homomorphism; otherwise $\mathbb{Z}_{12}$ would have a subgroup of order 7, which is impossible.

Consequently, the only possible homomorphism from $\mathbb{Z}_7$ to $\mathbb{Z}_{12}$ is the one mapping all elements to zero."

I don't understand how the last sentence follows from the preceding ones. Just because there is no injective homomorphism, why does it mean that everything in $\mathbb{Z}_7$ is mapped to zero? Why can't $\mathbb{Z}_7$ be mapped to $\mathbb{Z}_4$, for instance?


Let $\phi$ be a homomorphism from $\mathbb{Z}_7$ to $\mathbb{Z}_{12}$. Then as you said, $\ker\phi=0$ or $\ker\phi=\mathbb{Z}_7$. In the first case, $\ker\phi=0$ means $\phi$ is injective since $\ker\phi$ is by definition all elements that go to $0$ under $\phi$. So the image of $\phi$ is a subgroup of order $7$ inside of $\mathbb{Z}_{12}$. But since $7$ does not divide $12$, this is not possible. So only the second case happens, i.e. $\ker\phi=\mathbb{Z}_7$. This says all elements go to $0$, i.e. $\phi$ is the $0$ map.


That is because, as $\phi$ cannot be injective, i.e. $\ker\phi\neq\{0\}$, necessarily $\;\ker\phi=\mathbf Z_7$.


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.