I am trying to self study groups, and may be stuck on this definition:

The quotient group has group elements that are the distinct cosets, and a group operation $(g_1 H)(g_2 H) = g_1 g_2 H$
where $H$ is a subgroup and $g_1,g_2$ are elements of the full group $G$.

Let's take this example:
$G$ is the group of integers, with addition.
$H$ is the group of integers divisible by 3 also with addition, -3,0,3,6,9,....

What is the quotient group for this case?

I take the definition to mean that the quotent group elements are not single integers, but infinite sets of integers, the cosets.

The cosets are ${i + H, i \le 3} = 0+H,1+H,2+H = (...,0,3,6...), (,1,4,7...), (,2,5,8…)$

so the quotient group has 3 elements. Is this correct?

Then what is $G/H$?
From the definition above it seems that an element of the quotient group should be

$G/H = g_1 + g_2 + H = g_1 + g_2 + (-3,0,3,6,9,…)$

i.e. it is the subgroup H with two arbitrary integers added to each element.

  • $\begingroup$ What you saying is mainly correct. Indeed there are $3$ elements in your example (the cosets). So the 'underlying' set of $G/H$ is $\left\{ H,1+H,2+H\right\} $. If you define the groupoperation here then you are ready. A group is a set together with a groupoperation. I don't understand why you are still puzzling about (elements of) $G/H$ and mention something like $g_{1}+g_{2}+H$ in the second part. (Merry Christmas) $\endgroup$ – drhab Dec 25 '13 at 11:01
  • $\begingroup$ Merry Christmas @beginer. Don't forget to register soon. It will really help you to track your way here at the site. :-) $\endgroup$ – mrs Dec 25 '13 at 11:07

If $G=\langle \mathbb Z, +\rangle$ so, it is abelian infinite group and every subgroups of $G$ is normal in it so we can speak about $G/H$ freely. Here, you took $H=\langle 3\rangle=3\mathbb Z$, so we have $$G/H=\{0+H,1+H,2+H\}$$ because if we divide any integer numbers by $3$, the reminders are $0,$ $1$ or $2$ and the rest would be put inside $H$. This means that $H$ eats a considerable part of any integer numbers but three numbers $0,$ $1$ and $2$. You may have seen that every finite group of order $3$ can be written as $$\mathbb Z_3=\{0,1,-1\}$$ Now you can set $0\in\mathbb Z_3$ with $0+H\in G/H$ in your mind and think of $G/H$ as $\mathbb Z_3$. Note that this imagination is not a Mathematical way but here you are allowed to do that. I hope you read about isomorphisms later ann then do this final part with a simple map.

  • $\begingroup$ If the identity is written as $0$ (indication for additive group) then it can be confusing to write $a^{-1}$ for the inverse of $a$. I should prefer $\mathbb Z_3=\{0,1,-1\}$ here. Merry Christmas. $\endgroup$ – drhab Dec 25 '13 at 13:08
  • $\begingroup$ Dear @drhab: Yes, exactly! I am fixing it. Thanks. $\endgroup$ – mrs Dec 25 '13 at 13:10
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    $\begingroup$ $\ddot\smile$+1 $\endgroup$ – user63181 Dec 25 '13 at 16:10

$G/H$ is a group with $3$ elements.

A theorem in group theory states that a group $G$ with $|G| = p$, $p$ prime, is cyclic. So $G/H$ is the cyclic group with $3$ elements.


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