# Multiplication of cosets: well-definedness

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?

• 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". Commented Jun 26, 2023 at 21:54
• For me they are quite the same inly the way of putting things changes Commented Jun 26, 2023 at 22:02
• @Shaun So is it fair to say that these are distinct proofs of the same thing? Commented Jun 26, 2023 at 22:12
• 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. Commented Jun 26, 2023 at 22:15
• What does it mean for two proofs to be "mutually exclusive"? Commented Jun 26, 2023 at 22:51

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