Coset of $[18]$ in $\Bbb Z_{55}/11\Bbb Z_{55}?$

Question

Q: Let $G=\mathbb{Z}_{55}$, $H=11\mathbb{Z}_{55}$ and $g=18\pmod{55}.$ Write down the elements of the left coset $gH$ as a comma-separated list of elements of $G$.

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My thoughts: First find $G=\mathbb{Z}_{55}$, then $H=11\mathbb{Z}_{55}$, then $gH=18+11\mathbb{Z}_{55}$. Thus, $H=\mathbb{Z}_{55}=\{0,11,22,33,44,55\equiv0\}\implies gH=\{18,29,40,51,7\} (\mod55)$, whereby I simply multiplied all elements of of $H$ by $g=18\mod55$.

However, I also realise that since $aK=K\iff a\in K$ for some group $K$, so $\dots$ $11\in\mathbb{Z}_{55}$ and $18\in H\implies\mathbb{Z}_{55}=11\mathbb{Z}_{55}\implies gH=H=\mathbb{Z}_{55}$??? (Set of 55 elements!??)

Answer 1

Hint: The coset is $$[18]+11\Bbb Z_{55},$$ and not $11\Bbb Z_{55}$ (nor $18\cdot 11\Bbb Z_{55}$) because the operation of $\Bbb Z_{55}$ is addition, not multiplication. Note that $11\mid 55$.

It looks like you got it right the first time. It's just that you wrote it mutliplicatively instead of additively. Also, $18\notin H$.