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So I'm very new to group theory and I'm trying to understand the difference between a coset and a quotiënt group. To me it seems that a quotient group is always also a coset but not the other way around. But I can't seem to grap the exact difference betweem them.

Can anyone give a simple example or intuitive explanation?

Bobby
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    A coset is an element of the quotient group. For example take $G = \mathbb{Z}$ and $H = 3\mathbb{Z} \subset G$. Then the quotient group $\mathbb{Z}/3\mathbb{Z}$ is, as a set, ${3\mathbb{Z}, 1 + 3\mathbb{Z}, 2 + 3\mathbb{Z}}$. The element $1 + 3\mathbb{Z}$ is an example of a coset. – Sean Haight Jun 10 '19 at 16:26
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    A coset is generally not a group, for one thing. And what makes you say that "a quotient group is always also a coset"? A quotient group is not, generally, even a subset of the original group. – lulu Jun 10 '19 at 16:26
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    @SeanHaight Ooh okey. So if I understand correctly. A quotient group is made up of multiple cosets? – Bobby Jun 10 '19 at 16:28
  • Yes exactly. The collection of left (or right) cosets of a given subgroup form the quotient group, provided the subgroup is normal. – Sean Haight Jun 10 '19 at 16:30
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    Okey the example SeinHaight gave with the line from lulu: "A quotient group is not, generally, even a subset of the original group" made me realize what it exactly is! Thanks a lot! – Bobby Jun 10 '19 at 16:30
  • You will get good number of examples in Contemporary algebra by Gallian..PDF is available online..it may help. – ogirkar Jun 10 '19 at 16:40
  • Do you already understand equivalence relatiions and quotient sets? You should master that simpler case before moving on to quotients of algebraic structures. – Bill Dubuque Jun 10 '19 at 16:48
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    @Sean The elements of a quotient group are not necessarily cosets. What they are depends on how one actually constructs the quotient group (there are various ways). But the representation of the elements in the quotient group is not what matters algebraicually. Rather what matters is how they relate to each other under the group operations. This is one way that algebra is abstract, i.e. it abstracts away from any particular representations of the elements. For more on this see the Remark here.. – Bill Dubuque Jun 10 '19 at 18:27
  • @BilDubuque Thanks for the remark! I already know what equivalence relations are and was quite sure what quotient gruops were and now are. Anyway, it's all blowing my mind right now and after some good practice it'll all fall into place. Thanks again! – Bobby Jun 11 '19 at 08:25

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So you start with a group $G$ and a subgroup $H$.

Then the (left) coset of $H$ for $g\in G$ is a subset of $G$ of the form $\{gh\ |\ h\in H\}$. Also denoted by $gH$. Let me emphasize this one more time: cosets are certain special subsets of $G$.

The collection of all cosets is denoted by $G/H=\{gH\ |\ g\in G\}$. So elements of $G/H$ are cosets, not elements of $G$.

If $H$ is additionally a normal subgroup then $G/H$ can be naturally turned into a group as well and in this situation we call $G/H$ the quotient group.

Example:

Consider $G=\mathbb{Z}$ integers with addition and $H=5\mathbb{Z}$ i.e. all integers divisible by $5$. So every element of $H$ can be written as $5k$ for some $k\in\mathbb{Z}$. Therefore every coset can be written as $$x+H=\{x+5k\ |\ k\in\mathbb{Z}\}$$

So how many cosets are there? Since every coset is $x+H$ for some integer $x\in\mathbb{Z}$ then you may think that as many as there are integers, right? Wrong. For example

$$\{1+5k\ |\ k\in\mathbb{Z}\}=\{6+5t\ |\ t\in\mathbb{Z}\}$$

These are equal sets and so these two cosets: $1+H$ and $6+H$ are equal. You can verify that there are only 5 cosets of $H$, namely $H$, $1+H$, $2+H$, $3+H$ and $4+H$. They correspond to remainders to division by $5$. So $G/H$ has exactly $5$ elements even though $G$ is infinite.

freakish
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    So $G/H$ does not always per se represent a quotient group. That wasn't stated in my book! Big thanks and also for the explanation! – Bobby Jun 11 '19 at 08:27