How to show that $3\mathbb{Z}/15\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z}$?

Question

How to show that $3\mathbb{Z}/15\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z}$ as $\mathbb{Z}$-module over $\mathbb{Z}$?

My proof:

Define surjective function $f:3\mathbb{Z}/15\mathbb{Z} \rightarrow \mathbb{Z}/5\mathbb{Z}$ by $f(3n+15\mathbb{Z})=n+5\mathbb{Z}$.

Then $$f(3a+15\mathbb{Z}+3b+15\mathbb{Z})=f(3a+3b+15\mathbb{Z})=(a+b)+5\mathbb{Z}=f(3a+15\mathbb{Z})+f(3b+15\mathbb{Z})$$ $$f((3a+15\mathbb{Z})n)=f(3an+15\mathbb{Z})=an+5\mathbb{Z}=(a+5\mathbb{Z})n=f(3a+15\mathbb{Z})n$$

So $f$ is a surjective homomorphism.

Besides, $f(3n+15\mathbb{Z})=5\mathbb{Z}$ iff $n\in 5\mathbb{Z}$, which implies $3n\in 15\mathbb{Z}$. Hence $ker(f)=15\mathbb{Z}$.

Hence we conclude that $f$ is an isomorphism.

Do we really need to prove it like this? Or there are simpler ways so that the statement is just sort of trivial?

Answer 1

Use the third isomorphism theorem.

If $G$ is a group (or ring, or module) and $H$ and $K$ are normal subgroups (or ideals, or submodules, respectively) of $G$, with $H\subseteq K$, then there is a natural isomorphism $$(G/H)/(K/H)\cong G/K.$$

Answer 2

Consider $f\colon\mathbb{Z}\to3\mathbb{Z}$ defined by $f(x)=3x$; then compose it with the canonical projection $\pi\colon 3\mathbb{Z}\to3\mathbb{Z}/15\mathbb{Z}$.

The kernel of $\pi\circ f$ consists of the integers $x$ such that $$ 3x+15\mathbb{Z}=15\mathbb{Z} $$ that is, $3x\in15\mathbb{Z}$.

Can you prove that $\ker(\pi\circ f)=5\mathbb{Z}$? Is $\pi\circ f$ surjective?