Quotient Group Isomorphic to $\mathbb{Z}_{3}$?

I have a question that I'm tripped up on for an exam review. The problem is as follows:

Prove that $\mathbb{Z}_{18}/\left\langle\right [3]\rangle \approx \mathbb{Z}_{3}$.

I think we need to use the Fundamental Homomorphism Theorem for groups, but I'm not entirely sure. Thanks for any help!

Here is one way that might work for you. The group $\mathbb{Z}_{18}$ has order $18$. The subgroup generated by $\langle 3\rangle = \{0, 3, 6, 9, 12, 15\}$ has order $6$, so the quotient has order $18/6 = 3$. Now you might now that (up to isomorphism) there is only one group of order $3$, namely $\mathbb{Z}_3$. So the quotient has to be isomorphic to $\mathbb{Z}_3$.

Another way would be to find a surjective homomorphism $\phi: \mathbb{Z}_{18} \to \mathbb{Z}_{3}$ with kernel $\langle 3 \rangle$. Then the First Isomorphism Theorem will give you your isomorphism.

Define $f:Z_{18}\to Z_3$ through $f(\overline{x}) = x \textrm{ mod } 3.$ Check that this is a homomorphism onto $Z_3.$ What is $\ker f$? Note that you can't do the same for, say, $Z_{19}.$ What fails?

The subgroup $\langle[3]\rangle$ is actually $3\mathbb{Z}/18\mathbb{Z}$ and one of the homomorphism theorems says $$(\mathbb{Z}/18\mathbb{Z})\big/(3\mathbb{Z}/18\mathbb{Z}) \cong \mathbb{Z}/3\mathbb{Z}$$ More generally, if $H\subseteq K$ are normal subgroups of $G$, then $$(G/H)\big/(K/H)\cong G/K$$ The statement can be proved by considering the two canonical maps $$p\colon G\to G/H,\qquad q\colon G/H\to (G/H)\big/(K/H)$$ The kernel of $q\circ p$ is $K$.