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Suppose $G$ is a group and $N$ is a subgroup of $G$ show if for all $g\in G$ we have $gNg^{-1}=N$ then the following operation is well defined $$uNvN=uvN$$.

So This is proposition 5 from Dummit and Foote Chapter 3.1 part (1) converse direction. It can be seen here Understanding proof from Dummit's Abstract Algebra Section 3.1. I see how the proof work, but I dont know why does it have to be so convoluted. For example, why take $u,u_1\in uN$?

To show a operation $G \times G \mapsto G$ is well defined dont we only need to show that given an element in $G \times G$ you will get back an element in $G$?

If I prove the claim as follows am I wrong? Can someone check If I understood everything correctly? Thank you!

Suppose $x\in vNuN$ then there exists $n_1,n_2\in N$ such that $x=vn_1un_2$ since we are given everything in $G$ normalizes $N$. We must have some $n_3\in N$ such that $n_1u=un_3$ and so we have $x=vun_3n_2\in vuN$ because $N$ is a subgroup $n_3n_2\in N$.

Remu X
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    The issue here is that the elements of $G/N$ have many "names". For example, $eN$ and $nN$ represent the same element, if $n\in N$, even if $n\neq e$. However, the definition of the operation depends on which name you are using for each element. So suppose you want to multiply $uN$ by $vN$, but you also know that $uN$ has a "different name", namely $wN$. The definition tells you that to compute $(uN)(vN)$ you take $uvN$; but the definition tells you that to compute $(wN)(vN)$ you take $wvN$. They don't look the same. So you need to make sure you just got different names for the same answer – Arturo Magidin Sep 25 '23 at 00:51
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    Because you should get the same answer from $(uN)(vN)$ and from $(wN)(vN)$, because $uN$ and $wN$ are the same element of $G/N$. But it is not immediate that they are the same answer. They look different... but that could just be because they are two different "names" for the same element. So you need to check that the final output is "well defined" and depends only on who the inputs are, and not on which name you picked for them. – Arturo Magidin Sep 25 '23 at 00:53
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    What you are doing, however, does not address that issue in any way. So what you are doing does not check that the operation is "well defined". Instead, your computation shows that the definition is reasonable (it's what you "should" try). But it does not show the definition actually works. – Arturo Magidin Sep 25 '23 at 00:55
  • I think I understand the idea thank you! – Remu X Sep 25 '23 at 01:04
  • @ArturoMagidin Suppose if I extend what I was doing there a little bit and continue by assuming that $u_1\in uN$ and $v_1\in vN$ and conclude that $v_1Nu_1N=vuN$. Would this be a reasonable proof all together for well defineness? THanks – Remu X Sep 25 '23 at 01:08
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    No. You are checking that the element-wise multiplication of left cosets of $N$ is contained in a left coset of $N$. Your "extension" does not change that. But the operation you are given is not defined as being elementwise-multiplication of cosets. It's defined as an operation on cosets: "if you are given two cosets, the result of "multiplying" them will be the coset represented by the product of the representatives". What you are doing can be shown to be equivalent to this, but it is not what you are being told to do, and you have not proven their equivalence anyway. – Arturo Magidin Sep 25 '23 at 01:11

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As I've said before, the meaning of "the function is well-defined" is, ahem, not very well defined.

Essentially it means that what you have really is a function and the outputs are sensible. Usually that means checking one (or both) of two things:

  1. That every input really does give an output and the output lies in the appropriate set (what you seem to be trying to do); and/or

  2. That the "instructions" for how to get the output will yield an unambiguous unique element of the codomain (that is, that it really provides one and only one output for every valid input).

The issue here is the second, not the first. And what you attempt to do in order to establish the first is not really to the point.

Note that, as defined, it is clear that the output of this function $(G/N)\times(G/N)\to(G/N)$ (or at least, what we hope will be a function once we establish item 2) is an element of $G/N$: the output is a left coset of $N$. And you can compute it given any pair of cosets of $N$. So that point 1 is not really an issue: given two left cosets, it tells you to produce a left coset. Your computations aren't checking that. Instead, what you are checking is that $vNuN\subseteq vuN$. Which... okay, sure, one might want to check. But you don't even know if $vNuN$ is a left coset at all, so that this "check" doesn't tell you much. And you don't even know that this element-wise multiplication will correspond to the function you are given. So you are really doing a computation that is orthogonal to the problem at hand. That said, it does suggest that if $vNuN$ is really going to be a coset, it better be the coset $vuN$. So it suggests that if we try to define multiplication of cosets elementwise, that should be our definition.

But here the problem is something else. Remember that left cosets in general have many different "names": you have $uN=wN$ if and only if $w^{-1}u\in N$. But the definition tells you to calculate the answer using whatever name you have on hand. So if $uN=wN$, and $vN=xN$, taking the cosets to be $uN$ and $vN$ and following the instructions give, you compute their product $(uN)(vN)$ to be $uvN$. But if you take the cosets to be $wN$ and $vN$, you get $wvN$. And if you take them to be $uN$ and $xN$ you get $uxN$. And if you take them to be $wN$ and $xN$, you get $wxN$. But you are multiplying the same cosets. So you want the answer to be the same answer (otherwise, you don't have a function). They don't immediately look to be the same answer... but then $uN$ doesn't look the same as $wN$, yet they are assumed to be two different names for the same coset.

So what we need to verify is that if $uN=wN$ as cosets, and $vN=xN$ as cosets, then $uvN$ should be the same as $wxN$ as cosets. That's what we need to verify in order to make sure that our proposed definition of an operation $(G/N)\times (G/N)\to (G/N)$ is really a function (single output for every valid input). That is what we need to do to check that this is "well-defined".

Note also that while one can prove that the result of performing the element-wise product of $uN$ and $vN$ will be the coset $uvN$, that is not what you are being given. Instead you are being given a function on pairs of cosets. The "instructions" of that function are not about multiplying all the elements in $uN$ with all the elements in $vN$. The instructions say: "to multiply two left cosets of $N$, take representatives for each coset, and the result will be the coset of the product of the representatives." That is the function you are trying to check. At this point, you have not established that this amounts to the same thing as multiplying the cosets elementwise, so checking that the set $uNvN$ is contained in the coset $uvN$ does not really accomplish much.

Arturo Magidin
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