Homomorphic Images of $\mathbb{Z}_{6}$? I had a question that's stumping me from an exam review for my abstract algebra class. The problem is as follows:
Find all of the homomorphic images of $\mathbb{Z}_{6}$.
From what I recall, this should be pretty basic, but it's been a while since I did something like this. I'm not sure where to even start. Thanks in advance for any help!
 A: By the first isomorphism theorem, any homomorphic image of a group is isomorphic to a quotient by the kernel. So, you should find all normal subgroups of $\mathbb{Z}_6$ (there will be 4 of them), and the quotients will classify all possible homomorphic images.
A: This approach is in some sense more elementary than lisyarus's answer, but this does not mean the first isomorphism theorem is not an important tool in your algebraic arsenal.
For any homomorphism $h$ from $\mathbb{Z}_6$ onto a group a $G$ with identity $e$, it must be that $(h(1))^6 = e$. More importantly, because $\mathbb{Z}_6$ is cyclic, its image under $h$ will also be cyclic and entirely determined by the order of $h(1)$, its generator.
How many possibilities do we have for the order of $h(1)$? The only requirement is that the order divide $6$, for otherwise we would not have $(h(1))^6=e$. Since $6$ has four divisors, there are four homomorphic images of $\mathbb{Z}_6$. 


*

*Divisor $1$: $\mathbb{Z}_{1} = \{e\}$, the group with a single element.

*Divisor $2$: $\mathbb{Z}_{2}$; $h$ is the remainder modulo $2$.

*Divisor $3$: $\mathbb{Z}_{3}$; $h$ is the remainder modulo $3$.

*Divisor $6$: $\mathbb{Z}_{6}$; $h$ is the 'identity'.

