Return to Answer

Corrected typo and clarified steps

Source Link

edited Feb 18, 2016 at 20:01

I don't know whether you would count this, but the proof of the existence of quotient groups seems to fit your description. Some people define the product of the cosets $gH$ and $g'H$ to be the coset $gg'H$, and then go on to prove that this is well-defined. Another approach is to define two elements $g$ and $h$ of $G$ to be equivalent if $gh^{-1}\in H$, to define cosets to be equivalence classes, and to define the product of two cosets to be ... their product. That is, if $A$ and $B$ are cosets then their product is $AB=\{ab:a\in A, b\in B\}$. Of course, we have to prove that that is a coset. But if $ac^{-1}\in H$ and $bd^{-1}\in H$, then, because elements from $cbd^{-1}c^{-1}\in H$ so$H$ commute, $ab(cd)^{-1}\in H$$(ab)(cd)^{-1} = a(bd^{-1})c^{-1} = (ac^{-1})(bd^{-1})\in H$. Thus, if $a\sim c$ and $b\sim d$ , then $ab\sim cd$.

I can't quite decide whether this is a real example. I did have to take elements, but I wasn't exactly choosing them so much as proving something about all of them. The point of the example is that the product of cosets is defined by multiplying everything by everything -- that is the sense in which it does what the OP asks for.

I don't know whether you would count this, but the proof of the existence of quotient groups seems to fit your description. Some people define the product of the cosets $gH$ and $g'H$ to be the coset $gg'H$, and then go on to prove that this is well-defined. Another approach is to define two elements $g$ and $h$ of $G$ to be equivalent if $gh^{-1}\in H$, to define cosets to be equivalence classes, and to define the product of two cosets to be ... their product. That is, if $A$ and $B$ are cosets then their product is $AB=\{ab:a\in A, b\in B\}$. Of course, we have to prove that that is a coset. But if $ac^{-1}\in H$ and $bd^{-1}\in H$, then $cbd^{-1}c^{-1}\in H$ so $ab(cd)^{-1}\in H$. Thus, if $a\sim c$ and $b\sim d$, then $ab\sim cd$.

I don't know whether you would count this, but the proof of the existence of quotient groups seems to fit your description. Some people define the product of the cosets $gH$ and $g'H$ to be the coset $gg'H$, and then go on to prove that this is well-defined. Another approach is to define two elements $g$ and $h$ of $G$ to be equivalent if $gh^{-1}\in H$, to define cosets to be equivalence classes, and to define the product of two cosets to be ... their product. That is, if $A$ and $B$ are cosets then their product is $AB=\{ab:a\in A, b\in B\}$. Of course, we have to prove that that is a coset. But if $ac^{-1}\in H$ and $bd^{-1}\in H$, then, because elements from $H$ commute, $(ab)(cd)^{-1} = a(bd^{-1})c^{-1} = (ac^{-1})(bd^{-1})\in H$. Thus, if $a\sim c$ and $b\sim d$ , then $ab\sim cd$.

Post Made Community Wiki

Source Link

created Dec 21, 2010 at 15:20

gowers

I don't know whether you would count this, but the proof of the existence of quotient groups seems to fit your description. Some people define the product of the cosets $gH$ and $g'H$ to be the coset $gg'H$, and then go on to prove that this is well-defined. Another approach is to define two elements $g$ and $h$ of $G$ to be equivalent if $gh^{-1}\in H$, to define cosets to be equivalence classes, and to define the product of two cosets to be ... their product. That is, if $A$ and $B$ are cosets then their product is $AB=\{ab:a\in A, b\in B\}$. Of course, we have to prove that that is a coset. But if $ac^{-1}\in H$ and $bd^{-1}\in H$, then $cbd^{-1}c^{-1}\in H$ so $ab(cd)^{-1}\in H$. Thus, if $a\sim c$ and $b\sim d$, then $ab\sim cd$.