1
$\begingroup$

I'm struggling to find the cosets of $\langle\mathbb{R}, +\rangle / \langle 360\mathbb{Z}, +\rangle$.

Usually I only find cosets of finite groups, in which case I know the cardinality of the subset and then divide the groups cardinality by that to get the number of cosets. Then it is just a matter of finding what elements are in the other cosets.

But I can't figure out what it is for the infinite sets above. How many cosets are there and how do you know? At the very least, I can imagine the subgroup would be $\{0, 360, 720, ...\}$ (because it has to be a subgroup and have the identity).

$\endgroup$
1
  • 1
    $\begingroup$ "I can imagine the subgroup would be $\{0,360,720,\dots\}$". Be careful: You need inverses in the subgroup! $\endgroup$
    – Shaun
    Nov 21, 2019 at 1:29

1 Answer 1

Reset to default
1
$\begingroup$

The cosets are of the form $r+H$ for $r$ in the interval $[0, 360)$ and $H:=360\Bbb Z$. Here's an example for illustration:

$$\pi+H=\{\pi+h\mid h\in 360\Bbb Z\}.$$

That's about as concise as I can get it, prima facie.

Further: $$\pi+H=\{\dots , \pi-720, \pi-360, \pi, \pi+360, \pi+720, \dots\}.$$

It might help to note that two cosets $a+H$ and $b+H$ are the same if and only if $a$ and $b$ differ by an element of $H$; that is, if and only if $a-b\in 360\Bbb Z$.

$\endgroup$
2
  • $\begingroup$ ah, thanks. what i wasnt getting is how there are infinite cosets because the cosets have an element of the group added to them, which seems clear now because you are adding an element of the group to the subgroup to generate the cosets. pleas correct me if im wrong. $\endgroup$
    – tau
    Nov 21, 2019 at 1:33
  • 1
    $\begingroup$ You're welcome, @tau. Yes, you're right (if I understand you properly: you were just confused with how there can be infinitely many cosets, etc., right?). Please don't forget to upvote & accept this answer if it works for you! $\endgroup$
    – Shaun
    Nov 21, 2019 at 1:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .