1
$\begingroup$

"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 $\langle 3 \rangle$.

$\endgroup$
3
  • 2
    $\begingroup$ What makes you think that $H = \{0, 1, 2 \}$? $\endgroup$
    – Calvin Lin
    Jul 19 '15 at 21:47
  • $\begingroup$ Is $<3>$ the same as $\mathbb{Z}_3$? $\endgroup$ Jul 19 '15 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
    Jul 19 '15 at 21:49
2
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ 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? $\endgroup$ Jul 20 '15 at 1:53
  • $\begingroup$ Yes, that's the idea. $\endgroup$ Jul 20 '15 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. $\endgroup$ Jul 20 '15 at 1:57
  • $\begingroup$ Yes, This is correct. $\endgroup$ Jul 20 '15 at 1:58
2
$\begingroup$

$H$ 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 $$H=\{\ldots,-9,-6,-3,0,3,6,9,\ldots\}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.