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I'm having trouble distinguishing between two different methods of proof regarding coset multiplication. I believe they end up being equivalent in the sense that they get us to the same situation, which is that coset multiplication "works" provided that the subgroup is normal. But I just want to be sure.

Method 1: Proving Well-definedness

(a) Let's say $G$ is a group and $H$ is a subgroup. If $H$ isn't normal, there is no guarantee that coset multiplication $(aH) \cdot (bH) = (ab)H$ is well-defined. But if $H$ is normal, I can prove that if $aH = a'H$ and $bH = b'H$ in $G/H$, then $(ab)H = (a'b')H$, so we can safely define this multiplication without issues, and then prove that it gives $G/H$ a group structure.

Method 2: Proving it directly

This is the approach in Lemma 2.12.5 in Artin, and it seems to avoid having to prove well-definedness. It proceeds as follows. Let $H$ be a normal subgroup and $aH, bH$ two cosets of $H$ in $G$. Then elements of $(aH)(bH)$ have the form $ah_1 b h_2$ for $h_1, h_2 \in H$. As $H$ is normal, we have $aH = Ha$ and $bH = Hb$, so $$\begin{align} (aH)(bH) &= (aH)(Hb) \\ &= (a(HH))b \\ &= (aH)b \\ &= a(Hb) \\ &= a(bH) \\ &= (ab)H, \end{align}$$ so this multiplication is in some sense "natural" in the sense that if $H$ is a normal subgroup, then the product of two cosets $aH$ and $bH$ is another coset of $H$ in $G$ with representative $ab$.

I understand both proof strategies individually, but I do not understand how they fit together. Are these proofs mutually exclusive? In other words, if I use method 2, as in Artin, do I no longer need to prove well-definedness of coset multiplication because I already know what the "product" looks like?

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    $\begingroup$ Let's put it this way: there is over 350 different proofs of Pythagoras' Theorem (because there's a book with 350 of them in). Not all of them are related to each other nor do they "fit together". $\endgroup$
    – Shaun
    Commented Jun 26, 2023 at 21:54
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    $\begingroup$ For me they are quite the same inly the way of putting things changes $\endgroup$ Commented Jun 26, 2023 at 22:02
  • $\begingroup$ @Shaun So is it fair to say that these are distinct proofs of the same thing? $\endgroup$ Commented Jun 26, 2023 at 22:12
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    $\begingroup$ I'll leave that to someone more intelligent than me. But there's no reason, given two proofs of a single theorem, to assume that those proofs are related. $\endgroup$
    – Shaun
    Commented Jun 26, 2023 at 22:15
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    $\begingroup$ What does it mean for two proofs to be "mutually exclusive"? $\endgroup$
    – Lee Mosher
    Commented Jun 26, 2023 at 22:51

2 Answers 2

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The proof in Artin shows that $(aH)(bH)=(ab)H$ as sets, so if $aH=a'H$ and $bH=b'H$, then:

$$(ab)H=(aH)(bH)=(a'H)(b'H)=(a'b')H$$

So coset multiplication is well defined.

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  • $\begingroup$ So is it fair to say that this proof of well-definedness is still necessary, even after proving set equality? $\endgroup$ Commented Jun 26, 2023 at 22:26
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    $\begingroup$ I think not, because the second proof defines the product of two cosets to be: "the set of elements of the form xy, where x is in the first coset and y is in the second coset". This definition is representative-independent, and the essence of the proof is essentially just showing that this set we defined is indeed a coset. $\endgroup$
    – Yaneda
    Commented Jun 26, 2023 at 22:30
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    $\begingroup$ I guess one way to think of it is this: The first proof defines the product of two cosets as another coset through coset representatives, and proves well-definedness by showing that this definition is independent of which representative you chose. The second proof defines the product of two cosets to be a SET, but this definition is independent of the coset representatives. The proof shows that this is a well defined product by showing that this set is indeed a coset. $\endgroup$
    – Yaneda
    Commented Jun 26, 2023 at 22:35
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Both methods are identical. Second method is kind of misleading because binary operation is not explicitly stated and it mixes with our usual (informal) notation of $NK=\{nk\mid n\in N, k\in K\}$ where $N,K$ are subsets of $G$. Though it is tempting to prove well definedness as user Yaneda did, it is wrong. If it were a valid proof, then we don’t need to check well definedness, it follows automatically. Product of two cosets is a set but that don’t make binary operator $\cdot$ immune to well definedness. See here for well defined map.

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