Suppose $G$ is a group and let $H \triangleleft G$. Consider the factor group $G/H$ where the relation is $aHbH = abH$ for all $a,b \in G$.
Suppose we wanted to show that the above relation is well-defined. Then we would pick some elements $a,b,c,d \in G$ such that $aH = cH$ and $bH = dH$. I then would have to show that $aHbH = cHdH$, which is the same as showing that $abH = cdH$.
Question. Couldn't we begin with the left hand side of the equation $aHbH = cHdH$ and substitute $cH$ for $aH$ and $dH$ for $bH$? This would result in
$$aHbH = (cH)(dH) = cHdH$$
The proof in the book doesn't proceed in this way, and although I understand the author's proof, I don't understand why the above method wouldn't work.
I have looked at other questions regarding this, but my question is addressing why we can't substitute our assumed equations $aH=cH$ and $bH=dH$ into the expression $aHbH$?
Thanks!
Best Answer
Your question is a common theme that arises whenever we define expressions that depend on non-unique representations of the same thing.
Imagine a similar scenario. Suppose $\text{Units}(x)$ is an operation that picks out the units digit in the decimal expansion of $x$. So for e.g. $\text{Units}(10.01)$ outputs $0$ and $\text{Units}(3.45)$ outputs $3$.
Now certainly $42 = 41.99\bar{9} \cdots$ as real numbers. But $$ \text{Units}(42) = 2 \neq 1 = \text{Units}(41.99\bar{9} \cdots) $$ So substituting equal things into the same expression did not result in the same outcome!
Why did this happen? Because in essence $42 = 41.99\bar{9} \cdots$ as real numbers but they are not equal as decimal expansions. That is, decimal expansions are not unique. The same number can have two different decimal expansions unless we are careful about disallowing a continued series of $9$'s at the end. And unfortunately the operation $\text{Units}(x)$ outputs values based on the non-unique representation of $x$ that we know as decimal expansions.
The exact same issue can happen with an expression like $(aH)(bH)$. Sure $aH$ may equal $cH$ and $bH$ may equal $dH$ as sets. But those sets $aH$ and $cH$ may have potentially different representatives $a$ and $c$. And if the operation $(aH)(bH)$ depends on non-unique representatives, which it does as the operation $(aH)(bH)$ picks out representatives $a, b$ of $aH,bH$ and outputs $(ab)H$, then merely substituting equal things may not result in equal things.
This is why proving the well-defined-ness of the definition $(aH)(bH) := (ab)H$ is necessary. You realize that our definition depends on non-unique representatives $a, b$; therefore you realize that mere substitution will not work. And that is why you get your hands dirty by literally verifying that the definition of $(aH)(bH)$ as $(ab)H$ outputs the same result even when you choose potentially different representatives $cH$ and $dH$ for $aH$ and $bH$ respectively.