Well definedness of factor group multiplication

Question

I have been reading on this site and the internet, and I do not quite understand the comments which are being made with respect to the problem of proving that factor group multiplication is well defined. According to John Fraleigh, A First Course in Abstract Algebra, 7th ed (page 137), if we make the following multiplication rule for factor groups, $$ (xH)(yH) = (xy)H, $$ we must check whether it is well defined. I have searched on this website and the internet to try and find out what the concern is, and it seems that the problem is that we must make sure that if $xH = x'H$ and $yH = y'H$, then $(xy)H = (x'y')H$.

First question: Is this Fraleigh's concern?

If the answer to this question is indeed yes. How can it be a problem, since if $xH = x'H$ and $yH = y'H$, then $$(x'y')H = (x'H)(y'H)=(xH)(yH) = (xy)H$$ is automatic? This is my second question.

Answer 1

That is indeed the concern. It is always a concern when you define an operation on equivalence classes by only saying what the operation does to a representative of the equivalence class. You have to check that picking different representatives gives the same result.

The elements $xy$ and $x'y'$ are generally distinct elements of the group, but you want them to represent the same coset, assuming that $x$ represents the same coset as $x'$, and similarly for $y$ and $y'$. That's what you need to prove. To see why this is not automatic, try doing this when $H$ is not assumed to be normal in $G$ (better still: write down a counterexample).

Answer 2

First question: Is this Fraleigh's concern?

Yes.

How can it be a problem, since if $xH=x′H$ and $yH=y′H$, then $(x′y′)H=(x′H)(y′H)=(xH)(yH)=(xy)H$ is automatic? This is my second question.

The argument that you give here relies on the multiplication $(xH)(yH) \mapsto (xy)H$ to be a well-defined function in the first place. A priori this multiplication operator could have "multiple possible outputs", depending on whether you write the "input" element as $xH$ or as $x'H$. Thus you cannot reason with it as if it is a function before you have proved that it actually is a function.

Consider the following similar problem: there are some numbers $a, b$ (which have been set beforehand). I define a function $$ f(x) = \begin{cases} a + x & \text{if $x \geq 0$},\\ b + x^2 & \text{if $x \leq 0$}. \end{cases} $$ Now, you claim that this might not be valid definition; after all, if we have $x = 0$, then the output could be either $f(0) = a + x$ or $f(0) = b + 0^2$, and we don't actually know what the output is. If I told you that of course you don't have to worry, because actually $$ a + 0 = f(0) = b + 0^2, $$ I hope you'd agree that that was silly!

Answer 3

Yes, this is Fraleigh's concern.

For the second question, you must be very careful. There are two ways to define $(xH)(x'H)$ and you are mixing them up to cook a fallacious proof of well-definedness.

• You can define $(xH)(x'H)$ to be $(xx'H)$, which is done by Fraleigh. In that case you are saying "to compute the product of two cosets, just take an element from the first (say $x$), an element from the second (say $x'$), and consider the coset of their product. But What if you had taken two other elements? Would the coset of their product still be the same? You need to check that for well-definedness.

$x$ and $x'$ are not canonical choices of elements of their own cosets, that is the point here.

With this definition, the left and right equalities are correct by definition, but the middle one has no reason to stand if you don't prove it: $$(x'y')H = (x'H)(y'H)\color{red}=(xH)(yH) = (xy)H$$

• You can define $(xH)(x'H)$ to be the set of all possible products $ab$ where $a\in xH$ and $b\in x'H$. In that case you still need to prove that with this definition, $(xx'H)$ is a coset! The fact that it contains $xx'$ is obvious.

With this definition, the middle equality is correct by definition, but the left and right one have no reason to stand if you don't prove them: $$(x'y')H \color{red}= (x'H)(y'H)=(xH)(yH) \color{red}= (xy)H$$

Answer 4

(x′y′)H=(x′H)(y′H)=(xH)(yH)=(xy)H

This is known as "proof by notation". It's a trap you have to be careful about falling into. Simply because you declare some expression to represent a value, doesn't mean that the value exists or is well-defined.

For instance, suppose we define f(p) to be the area of triangles with perimeter p. We can "prove" that given a perimeter, all triangles with that perimeter have the same area with the following argument:

Given triangle T1 and T2 with the same perimeter p, the area of T1 is f(p), and the area of T2 is also f(p). Thus, their areas are the same.