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I'm watching this video on quotient groups and came across the following(you don't have to click the link):

$G$ is a group and $N$ is its subgroup.
Using $N$ to make cosets : $N, g_1N, g_2N, g_3N, \ldots$

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I get why $x\in xN$ and $y\in yN$, but I'm not able to figure out the last two lines. Any help understanding why they say $x\cdot y \in (xN)(yN)$ for cosets to act like a group? It is obvious right? (Since $x\in xN$ and $y\in yN$, it is obvious $x\cdot y \in (xN)(yN)$. I don't get how this makes the cosets a group...)

AgentS
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    I think if you want to understand normal subgroups and quotients better, you should just read Arturo's answer here: https://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only – rschwieb Jul 29 '19 at 15:22

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We want to define the product pointwise, i.e $(xN)(yN)=\{gh:g\in xN, h\in yN\}$. From this definition obviously $xy$ must be in the product. Now, the set we get from multiplying two cosets is not necessary a coset itself. But since $xy$ belongs there we know that if it is a coset (which will happen if the cosets act like a group) then it must be the coset $xyN$.

Mark
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  • Ohk $(xN)(yN)$ is like the cartesian product of the sets. I get this may not be identical to a coset. But if it is identical to a coset, then the cosets act like a group. I understand this far. I still don't get why the cartesian product must equal to the particular coset $xyN$ ? – AgentS Jul 29 '19 at 15:30
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    If $(xN)(yN)$ is a coset and $xy$ belongs there then what coset can it be? There is only one left coset to which $xy$ belongs and that is $xyN$. – Mark Jul 29 '19 at 15:31
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    @rsadhvika It is not at all like a cartesian product. We usually call this "pairwise product." – rschwieb Jul 29 '19 at 15:31
  • @Mark That's another question I have. Do you mean any element in a coset can generate its parent? – AgentS Jul 29 '19 at 15:34
  • For example, if $a,b,c \in aN$, then are the $3$ cosets $bN$, $cN$, and $aN$ identical? – AgentS Jul 29 '19 at 15:35
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    Yes. Different left cosets are disjoint. Let $x\in bN$. Then we can write $x=bn_1$ when $n_1\in N$. But also $b\in aN$ and hence $b=an_2$ when $n_2\in N$. Hence we get $x=bn_1=a(n_2n_1)\in aN$. This shows $bN\subseteq aN$. For the other direction note that if $b=an_2$ then $a=bn_2^{-1}\in bN$. And as I showed this implies $aN\subseteq bN$. – Mark Jul 29 '19 at 15:37
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    @Mark That looks beautiful! May be because I'm seeing for the first time. I totally get it now. Thank you so much XD – AgentS Jul 29 '19 at 15:40
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    I think group theory is beautiful in general. – Mark Jul 29 '19 at 15:41
  • @rschwieb let me think some more... ok I agree its not same as cartesian product. It's more than that. After taking cartesian product $(xN) \times (yN)$, we perform the group operation on the data in each element of the cartesian product to get the pairwise product. Is my understanding ok? – AgentS Jul 29 '19 at 15:51
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    @rsadhvika I would say: "if $(G,\cdot)$ is a group, then $\cdot$ is a function from the Cartesian product $G\times G\rightarrow G$ with special properties." – rschwieb Jul 29 '19 at 15:59