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"A Book of Abstract Algebra" presents the following exercise:

Describe the cosets of the subgroups:

The subgroup $\langle 3 \rangle $ of $\mathbb{Z}$

Given that $H=\langle 3 \rangle $, I tried to figure out the cosets:

$$H=\lbrace 0,1,2 \rbrace$$ $$H+1=\lbrace 1,2,0 \rbrace$$ $$H+2=\lbrace 2,0,1 \rbrace$$

Since $H=H+1=H+2$, it seems to me that there's a single coset.

Please confirm, clarify or correct my attempt at figuring out the cosets of $\langle 3 \rangle$.

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    $\begingroup$ What makes you think that $H = \{0, 1, 2 \}$? $\endgroup$
    – Calvin Lin
    Commented Jul 19, 2015 at 21:47
  • $\begingroup$ Is $<3>$ the same as $\mathbb{Z}_3$? $\endgroup$
    – Kevin Meredith
    Commented Jul 19, 2015 at 21:48
  • $\begingroup$ $\langle 3 \rangle$ in $\Bbb Z$ is the subgroup of all integer multiples of 3. It is not a finite set. $\endgroup$
    – coldnumber
    Commented Jul 19, 2015 at 21:49

2 Answers 2

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$H\ne\{0,1,2\}$, $H=\{3n:n\in\Bbb Z\}$. That should help.

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    $\begingroup$ So a sample of $H=\lbrace 0, 3, 6, ... \rbrace$. So, $H+1 = \lbrace 1,4,7,... \rbrace = \lbrace 1 \rbrace$. $H+2 = \lbrace 2, 5, 8 \rbrace = \lbrace 2 \rbrace$. At $H+3=H$. So, the cosets of $H$ are $H, H+1, H+2$. Is that right? $\endgroup$
    – Kevin Meredith
    Commented Jul 20, 2015 at 1:53
  • $\begingroup$ Yes, that's the idea. $\endgroup$
    – Tim Raczkowski
    Commented Jul 20, 2015 at 1:56
  • $\begingroup$ Thank you. And that's the idea means it's correct? Could you please let me know if I'm not properly and fully answering the exercise? It's self-study, not homework for school. $\endgroup$
    – Kevin Meredith
    Commented Jul 20, 2015 at 1:57
  • $\begingroup$ Yes, This is correct. $\endgroup$
    – Tim Raczkowski
    Commented Jul 20, 2015 at 1:58
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$H$ is the subgroup generated by $3$, that is, the subset of $\Bbb Z$ of the numbers that can be obtained adding or substracting $3$. This is $$H=\{\ldots,-9,-6,-3,0,3,6,9,\ldots\}$$

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