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What does $\langle\overline{4}\rangle$ mean?

Context: Find the right cosets of $H<G$ and $[G:H]$ where $G=\mathbb{Z}_{12}$ and $H=\langle\overline{4}\rangle$

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    $\begingroup$ H is the group generated by $4\in \mathbb Z_{12}, $ modulo $12$. $\endgroup$
    – amWhy
    May 7, 2018 at 14:47
  • $\begingroup$ @amWhy What is the difference between $\langle 4 \rangle$ and $\langle \overline{4} \rangle?$ $\endgroup$
    – Cure
    May 7, 2018 at 14:48
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    $\begingroup$ Technically, the elements of $\mathbb Z_{12}$ under modulo arithmetic are the equivalence classes: $\{\bar 0, \bar 1, \bar 2, \ldots, \bar {10,} \bar {11}\}$. More casually, we refer to the group elements by the least non-negative integer representing their equivalence class. $\endgroup$
    – amWhy
    May 7, 2018 at 14:51
  • $\begingroup$ If you were working in $\mathbb{Z}$ only, then $\langle{4}\rangle$ would be the set $\{...,-8,-4,0,4,8,...\}$. $\endgroup$ May 7, 2018 at 14:52
  • $\begingroup$ @amWhy I "think" I see it. Would it be right to say that $gH$ is $\{ \overline{4}, \overline{8}, \overline{12}, \overline{11\cdot4} \dots \overline{12}\dots \overline{11\cdot12} \}$? $\endgroup$
    – Cure
    May 7, 2018 at 14:55

1 Answer 1

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Here $\bar{4}=\{4+12k\mid k\in\mathbb{Z}\}$ is an element of $\Bbb Z_{12}$ and so $\langle \bar{4}\rangle$ is the subgroup of $\Bbb Z_{12}$ generated by $\bar{4}$; that is, the group of all powers of $\bar{4}$ in $\Bbb Z_{12}$, which, since the operation is additive, are all multiples of $\bar{4}$.

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