Youtube Clip: Quotient Groups Part 2

The video shows that "the operation on quotient group is well-defined" by taking two elements from two different cosets and showing their composition leads to some element in one of the other cosets but I don't understand the reason why it needs to be shown since any element is in one of the cosets including $a \circ b$ (cosets partitions the group). I know that to show a function is well-defined I need to show every element can be mapped to exactly one element. But how about this. Are they conceptually the same?


marked as duplicate by Matthew Towers, MJD, Watson, Ethan Bolker, Lee Mosher Nov 6 '16 at 15:58

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I think this will explain things well enough (if not, feel free to ask followup questions). https://en.wikipedia.org/wiki/Well-defined

Edit: This part:

In particular, the term well-defined is used with respect to (binary) operations on cosets. In this case one can view the operation as a function of two variables and the property of being well-defined is the same as that for a function. For example, addition on the integers modulo some $n$ can be defined naturally in terms of integer addition.

$[a]\oplus [b]=[a+b] [a]\oplus [b]=[a+b]$

The fact that this is well-defined follows from the fact that we can write any representative of $[a]$ as $a+kn$, where $k$ is an integer. Therefore,

$[a+kn]\oplus [b]=[(a+kn)+b]=[(a+b)+kn]=[a+b]=[a]\oplus [b]$ and similarly for any representative of $[b]$.

  • $\begingroup$ The article mentions cosets and gives an explicit example from group theory under Independence of Representative, Operations. $\endgroup$ – Justin Benfield Nov 4 '16 at 7:11
  • $\begingroup$ You should probably quote the relevant parts of the Wikipedia article. The content of the target of a link, especially Wikipedia, may change over time. $\endgroup$ – Taladris Nov 4 '16 at 7:12
  • $\begingroup$ For a binary function $f:A\times B \rightarrow C,a\in A ,b\in B$, isn't that we might want to take $a,a',b,b'$ and show they produces the same output in $C$? $\endgroup$ – user241043 Nov 4 '16 at 7:23
  • $\begingroup$ Yes, what they are showing is that the choice of representative element from the cosets in question is unimportant, what matters is only which coset they came from (as far as determining what coset the result will lie in). $\endgroup$ – Justin Benfield Nov 4 '16 at 7:53

Let $G/H$ be a quotient group which automatically implies that $H$ is normal subgroup of $G$.
Suppose $aH\in G/H$.
$aH$ at the same time can be written as $bH$ as long as $b^{-1}a\in H$.
So if $aH=bH$, and $cH=dH$, we need to ensure that $aHcH=bHdH$.

You can take $G=S_3$ and $H=\{1,(12)\}$ which is not a normal subgroup of $G$ as an example where the operation is not well-defined.

$G=D_3=\{1,x,x^2,y,xy,x^2y\}$ where $x^3=1,y^2=1,yx=x^2y$
Recall that the operation is defined by $HaHb=Hab$.
Clearly $Hx\neq H$, so $HxHx^2\neq Hx^2yHxy$.
So the choice of representative of coset affect its image under the operation defined, so the operation is not well-defined.

  • $\begingroup$ Can you give tow elements that contradicts the fact? (do we need only two elements)? $\endgroup$ – user241043 Nov 4 '16 at 7:10
  • $\begingroup$ @Student I have updated my idea about the concept. Check whether it can help you. $\endgroup$ – Alan Wang Nov 4 '16 at 7:51

Well-defined in mathematics means that the choice of representation does not matter. That is if I have $$\frac{1}{2}+\frac{1}{3}$$ then that should give an element that is equal to $$\frac{2}{4}+\frac{3}{9}$$

For group multiplication (or any function we deal with) we must always show that our choice of representation doesn't effect the outcome. For example $$f(\frac{p}{q})=q$$ is not well-defined on rational numbers as different rational numbers that are equal will give different values.

In quotient groups we want the choice of element in a coset to be irrelevant, as otherwise we get some wierd structure that isn't a group.