# Finding all left cosets of the given group generated by $\overline a$

I have just learned about cosets and meet with the following question.

Find all left cosets of the subgroup generated by $\overline a$ in $\mathbb Z_{12}$. I know $\mathbb Z_{12} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. I would like to know (for example if $a=2$). What does $\overline a$ mean here? Thanks.

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Do you want to know how to find the cosets? Or are you just confused about the notation? – Dylan Moreland Dec 18 '11 at 3:28
yes am confused with the notation – neema Dec 18 '11 at 3:53
I see. It seems like Brad's answer should clear things up, then. Just a note: later on you'll learn that $\mathbf Z_{12}$ is the (quotient) group formed by the cosets of $12\mathbf Z$ in the group $\mathbf Z$, so you're looking at cosets of cosets here! – Dylan Moreland Dec 18 '11 at 3:54

The elements of $\mathbb{Z}_n$ are not really integers, but rather equivalence classes of integers. The equivalence class of an integer $a$ is often denoted $\bar a$. It is the set of all integers of the form $a+nk$, where $k$ is an integer. Using this notation,
$$\mathbb{Z}_n = \{\bar 0, \bar 1, \ldots, \overline{n-1}\}.$$ For example, in $\mathbb{Z}_{12}$, we have $\overline{12} = \bar 0$ and $\overline{21} = \bar 9$. However, we often abuse notation and write things like "21 = 9" in $\mathbb{Z}_{12}$ instead.
@neema: In $\mathbb{Z}_n$, $\overline{a}+\overline{b}=\overline{a+b}$. So in $\mathbb{Z}_{12}$, $\overline{0}+\overline{12} = \overline{0+12} = \overline{12} = \overline{0} = \overline{24} = \cdots$ – Arturo Magidin Dec 18 '11 at 4:59