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I'm currently watching Scoratica's YouTube series on Group Theory. On one of the videos, the following argument is made ($N$ is a normal subgroup of $G$): Condition for the cosets to act like a group

I don't understand why $xy \in (xN)(yN)$ is a necessary condition. We haven't even defined what $(xN)(yN)$ means yet. It could be the set formed by multiplying every element in $xN$ by each element of $yN$, it could be the set formed by the union of the sets $xN$ and $yN$, etc. Is there only one definition of the operation $(xN)(yN)$ that allows the cosets to behave like a group? If so why does it include $xy$?

Sammy Black
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Guilherme Mendonça
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  • All group operations are derived from the operation that defines the original group. So in the quotient (where the cosets get multiplied), we have to use that operation. – Sammy Black May 21 '22 at 03:14
  • Possible duplicate https://math.stackexchange.com/a/2426231/977780 – Sourav Ghosh May 21 '22 at 03:17
  • See this question. – Sammy Black May 21 '22 at 03:17
  • I haven't watched the video, so I can't comment on what its maker means by $(xN)(yN)$. But I can tell you that in Group Theory it is generally defined as in your first suggestion; it is the set of all $rs$, where $r$ is in $xN$ and $s$ is in $yN$. – Gerry Myerson May 21 '22 at 03:33
  • Given a group $G$ and a subgroup $H\le G$, a more general way to try getting from the cosets $xH$ and $yH$ another coset of $H$ (closure), by using the representatives $x,y$ and group's operation, only, is to set: $$xHyH=x^my^nH$$ for some positive integers $m,n$. But for $(m,n)\ne(1,1)$, the operation turns out to be not even associative, while for $(m,n)=(1,1)$, but $H$ non-normal, it turns out to be not well-defined. –  May 21 '22 at 12:18

1 Answers1

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There are two natural approaches to studying groups (and algebraic objects more generally). The first is to study subgroups, and the second is to study their operation-preserving maps (homomorphisms).

Quotients arise naturally from the study of homomorphisms, in that the cosets are the non-empty preimages of the homomorphism. That is, let $\varphi : G \to H$ be a homomorphism. Now let: $\{ \varphi^{-1}(h) : |\varphi^{-1}(h)| > 0 \}$ be the set of preimages that are non-empty.

Take $x \in \varphi^{-1}(h_{1}), y \in \varphi^{-1}(h_{2})$. So $\varphi(xy) = \varphi(x)\varphi(y) = h_{1}h_{2}$.

Note that if $N = \text{ker}(\varphi)$, then $x \in xN, y \in yN, xy \in xyN = xN \cdot yN$.

There are technical tools that need to be developed to equate these concepts, but this is the high level idea as to why cosets appear naturally and why the operation $xN \cdot yN = xyN$ is the correct operation.

ml0105
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