Well-definedness of coset multiplication

Question

I am trying to prove this following theorem about multiplying left cosets.

Let $H \subset G$ a subgroup and $G/H$ the set of left cosets of $H$ in $G$. We can define a group structure on $G/H$ by setting $aH \cdot bH = abH$, which is well-defined if and only if $H$ is a normal subgroup of $G$.

Here is my attempt.

Suppose that $H$ is normal in $G$. Then $aHa^{-1} = H$ for all $a \in G$, i.e., $aH = Ha$ for all $a \in G$. We then have \begin{align*} aH \cdot bH = (Ha)(bH) = H(ab)H = (ab)HH = (ab)H \end{align*} for all $a \in G$. Conversely, if $H$ is not normal, there exists $a \in G$ and $h \in H$ such that $aha^{-1} \not \in H$. This operation, if it were well-defined, would give \begin{align*} (aH)(a^{-1} H) = (aa^{-1})H = eH = H. \end{align*} Taking $ah \in aH$ and $a^{-1} e = a^{-1} \in a^{-1} H$, we get $aha^{-1} \in (aH)(a^{-1} h)$ but $aha^{-1} \not \in H$, a contradiction.

I worry I could be waving my hands a bit too much. How does this look? I would appreciate any criticisms.

Answer 1

I don’t much like your first argument, to be honest...

Proving that multiplication “is well defined” means proving that if $aH=a’H$ and $bH=b’H$, then $abH = a’b’H$. I’m not sure your argument establishes that; certainly not as written.

The second is correct, but you wrap it around a “proof by fake contradiction” making it more obscure than it needs to be. (That’s what I call a direct proof in which you essentially add an “assume not” on top, and a “a contradiction” at the bottom). To prove that if coset multiplication is well defined then $H$ is normal, let $a\in G$ and $h\in H$. Then $$aha^{-1} = aha^{-1}e\in aHa^{-1}H = (aa^{-1})H = eH = H,$$ proving that $H$ is normal. No need to argue by contradiction, when you have a perfectly valid direct proof there in the middle of it.

To prove well-definedness, assume $H$ is normal, and that $aH = a’H$, $bH=b’H$. Then: $$\begin{align*} abH &= a(bH) = a(b’H) = a(Hb’) = (aH)b’ \\ &= (a’H)b’ = a’(Hb’) = a’(b’H)\\ &= (a’b’)H. \end{align*}$$

You may find this extensive discussion on this interesting; or you may not.

Answer 2

This isn't what they're looking for. Recall that $S = \{aH | a \in G\}$ is the set of left cosets, where $aH := \{ab | b \in H\}$. We are trying to show that the multiplication operator $\cdot : S^2 \to S$ defined by $aH \cdot bH = (ab) \cdot H$ is in fact a well-defined function.

What does it mean to be well-defined in this case? It means each input corresponds to exactly one output. The problem is that the same coset $s \in S$ may be represented both as $aH$ and as $cH$ where $a \neq c$. Thus, our definition of $\cdot$ may a priori assign $s \cdot g$ to different values depending on whether we write $s = aH$ or $s = cH$, even though both are true.

To illustrate this point further, consider the group $\mathbb{Z} = G$ (under addition) and the subgroup $2\mathbb{Z} = H \subseteq G$. Let's say we tried to define a function $f : S \to \mathbb{Z}$ by $f(n \mathbb{Z}) = n$. This would not be a well-defined function because we would have $0 = f(0H) = f(2H) = 2$, since $0H = 2H$.

What we must show is the following: suppose $H$ is normal. Then for all $a, b, c, d \in G$, if $aH = cH$ and $bH = dH$ then $(a \cdot b) H = (c \cdot d) H$.

And we must also show that if for all $a, b, c, d \in G$ such that $a H = cH$ and $bH = dH$, we have $(a \cdot b) H = (c \cdot d) H$, then $H$ is normal.