# Find the Cosets of Subgroup $\langle 3 \rangle$ of $\mathbb{Z}$

"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 $$\langle 3 \rangle$$.

• What makes you think that $H = \{0, 1, 2 \}$? Jul 19 '15 at 21:47
• Is $<3>$ the same as $\mathbb{Z}_3$? Jul 19 '15 at 21:48
• $\langle 3 \rangle$ in $\Bbb Z$ is the subgroup of all integer multiples of 3. It is not a finite set. Jul 19 '15 at 21:49

$H\ne\{0,1,2\}$, $H=\{3n:n\in\Bbb Z\}$. That should help.
• 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? Jul 20 '15 at 1:53
• 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. Jul 20 '15 at 1:57
$H$ 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 $$H=\{\ldots,-9,-6,-3,0,3,6,9,\ldots\}$$