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!
 A: 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.
A: 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?
A: 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 $K$.
