Find the order $\bar{5} + H$

Question

Let $G=\mathbb{Z}/18\mathbb{Z}$ and $H=\langle 6+ 18\mathbb{Z}\rangle= \langle \bar{6}\rangle $ a subgroup of $G$. Consider the group $G/H$. What is the order of $(5 + 18 \mathbb{Z}) + H \in G/H$?

Attempt:

We know that as $G$ is a commutative group, then $H$ is a normal subgroup of $G$. So all we have to do is to find the smaller integer $n$ such that $n(5 + 18 \mathbb{Z}) + H = H$. The smaller $n$ here is 18. So $ord(\bar{5} + H) = 18$.

I am just a bit confused here. According to Lagrange Theorem, the order of an element divide the order of the group. However, the order of the group is 6, but 18 does not divide 6. Where am I wrong?

Answer 1

Here is an alternative answer, trying to bring structure in the game. We have the short exact sequence: $$ 0 \longrightarrow \underbrace{6\Bbb Z/18\Bbb Z}_{H} \longrightarrow \underbrace{\Bbb Z/18\Bbb Z}_{G} \longrightarrow \underbrace{\Bbb Z/6\Bbb Z}_{G/H} \longrightarrow 0 \ . $$ The image of that complicated element $5 + 18 \mathbb{Z}\in G$ inside $G/H$, so $$ (5 + 18 \mathbb{Z}) + H = (5 + 18 \mathbb{Z}) + (6\Bbb Z+18\Bbb Z) = (5 + 6\Bbb Z) \in G/H\ , $$ is simply $5+6\Bbb Z$, an element of order six.

Answer 2

Observe, that $H=\{-6+18\mathbb Z, 18\mathbb Z, 6+18\mathbb Z\}$. Hence, $|G/H|=[G:H]=18:3=6$. Now: \begin{align*}\overline{5}&=5+H=\{-1+18\mathbb Z, 5+18\mathbb Z, 11+18\mathbb Z\}\\\overline{5}+\overline{5}&=10+H=\{4+18\mathbb Z, 10+18\mathbb Z, -2+18\mathbb Z\}\\\overline{5}+\overline{5}+\overline{5}&=15+H=\{9+18\mathbb Z, 15+18\mathbb Z, 3+18\mathbb Z\}.\end{align*} None of these is $H$. But you can easily calculate $\overline{5}+\overline{5}+\overline{5}+\overline{5}+\overline{5}+\overline{5}$. Is it $H$ or not?