$(\mathbb{Z}/18\mathbb{Z})/(6\mathbb{Z}/18\mathbb{Z})\cong\mathbb{Z}/6\mathbb{Z}$ Proof

Question

Use the homomorphism theorem or the first or/and second isomorphism theorems to show that $(\mathbb{Z}/18\mathbb{Z})/(6\mathbb{Z}/18\mathbb{Z})\cong\mathbb{Z}/6\mathbb{Z}$.

I was wondering if it is necessary to show that $6\mathbb{Z}/18\mathbb{Z}$ is the image of the ideal $6\mathbb{Z}$ ?

Answer 1

Hint:

Consider the map

$$\mathbb{Z}/18 \mathbb{Z} \to \mathbb{Z}/6 \mathbb{Z}: x+18\mathbb{Z} \mapsto x + \mathbb{6}\mathbb{Z}$$

and use the first isomorphism theorem (i.e. check this is well-defined, that the kernel is $6\mathbb{Z}/18\mathbb{Z}$ and that this is a surjection).

Answer 2

Hint:

Prove that the map $\begin{aligned}[t]f\colon \mathbf{Z} /18 \mathbf{Z}&\longrightarrow \mathbf{Z}/6 \mathbf{Z,}\\ n+18 \mathbf{Z}&\longmapsto n+6 \mathbf{Z}\enspace\end{aligned}$ is well-defined and determine its kernel.