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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!

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  • $\begingroup$ Although not directly your question, you might find math.stackexchange.com/a/14315 useful. It defines well defined (that is, if $xH = x'H$ and $yH=y'H$ then $xyH = x'y'H$). $\endgroup$ Aug 2, 2019 at 11:29

5 Answers 5

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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.

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    $\begingroup$ Nice analogy, gets at the heart of the problem. +1 for you. $\endgroup$
    – MPW
    Aug 2, 2019 at 0:02
  • $\begingroup$ Excellent explanation. I also like how it illuminates well-defined-ness in general. Very nice. $\endgroup$
    – J. Dunivin
    Aug 2, 2019 at 0:44
  • $\begingroup$ @BenedictVoltaire Thank you. Glad to be of help. $\endgroup$
    – balddraz
    Aug 2, 2019 at 0:45
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Interesting question.

This works if you have previously proved that $aHbH$, thought of as the set of all products of elements one from the each coset, is in fact a coset of $H$. (Once you know that it is clearly the coset of $ab$, since $e\in H$.)

Without knowing that and thinking of $aHbH = abH$ as defining the product to be the coset of $ab$ you are just asserting what you want to prove.

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  • $\begingroup$ I think this is the best answer, as it highlights that $(aH)(bH) = abH$ can be either taken as a definition or as something we need to prove (a theorem or lemma). The two turn out to be equivalent (for normal subgroups) so it ends up not really mattering, but when people are still building their basic understanding it's good to be painfully explicit about what your assumptions are. $\endgroup$
    – JonathanZ
    Aug 2, 2019 at 15:31
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This is an example of "proof by notation", which is where one use symbols to denote concepts in such a way that a claim appears to simply fall out of how objects are represented. A more direct proof by notation for this claim would be "Given $aH$ and $bH$, let $abH$ be the coset generated by $ab$ and $H$. I just defined $abH$, so clearly $abH$ is well-defined". That's clearly fallacious. In your proof, the fallacy isn't as blatant, but it's the same basic principle. You write $aHbH=(cH)(dH)$ = $cHdH$, but what does that even mean? You're relying on implicitly understanding what "two things next to each other means": after years of mathematics classes, you've gotten used to $(a)(b)$ and $ab$ both representing the same thing, namely $a$ multiplied by $b$, but do $(aH)(bH)$ and $aHbH$ both represent "multiplication", and what does "multiplication" even mean? Remember, "$aH$" and $bH$" represent sets. What does it mean to multiply to sets by each other?

The notation $aH$ relies of a basic understanding of cosets and is shorthand for the more complicated notation $\{ah: h\in H\}$. Once you have a solid understanding of cosets, you may be able to easily work with these shorthands, but this question is asking you to prove a basic property of cosets, so using a high-level shorthand is a bit inappropriate. The fallacy should be more clear if we dispense with this shorthand. Instead of writing $aH$, we have $\{ah: h\in H\}$. Instead of $bH$, we have $\{bh: h\in H\}$. And so on. Instead of $aHbH$, we have $\{ah_1bh_2: h_1,h_2 \in H \}$. And instead of $aH=cH,bH=dH$, we have $\exists h_3,h_4 \in H:c=ah_3, d = bh_4$. Now the claim that $aHbH=cHdH$ expands to the claim that $\{ah_1bh_2: h_1,h_2 \in H \} = \{(ah_3)h_1(bh_4)h_2: h_1,h_2 \in H \}$. And that, in turn, is equivalent to $\forall h_1,h_2 \in H, \exists h_1' ,h_2'\in H:ah_1bh_2=(ah_3)h_1'(bh_3)h_2'$.

That is, when you say $aHbH=(cH)(dH)$, you're invoking the entire complicated statement of $\forall h_1,h_2 \in H, \exists h_1' ,h_2'\in H:ah_1bh_2=(ah_3)h_1'(bh_3)h_2'$, and just saying "Well, I've written this claim in a form that makes it look like it's obviously true, so I'm going to just take it as true".

The problem with simply substituting one expression in for another is that $aHbH$ is not defined directly in terms of $aH$ and $bH$. We have $aHbH :=\{ah_1bh_2: h_1,h_2 \in H \}$, so $aHbH$ is defined in terms of $a$ and $b$. So just because $cH=aH$, that doesn't mean that substituting $cH$ in for $aH$ will give the same result: when you replace $a$ with $c$ in the formula $\{ah_1bh_2: h_1,h_2 \in H \}$, you're not guaranteed to get the same answer.

Suppose we had an operation $*$ on fractions that said "the numerator of the result is the sum of the input numerators, and the denominator is the sum of the input denominators". Then $\frac 1 2 = \frac 2 4$, but $\frac 1 2 * \frac 1 3 \neq \frac 2 4 * \frac 1 3$. This shows that $*$ is not well-defined on the rational numbers, because it looks at not the fraction as a whole, but the individual parts of its representation. Similarly, for non-normal subgroups, $aHbH$ is not well-defined, because it is defined not in terms of the cosets as a whole, but rather in terms of the individual parts of its representation ($a$ and $b$). Once you define an operation in terms of the representation of an object, then if the object has more than one presentation, you are no longer guaranteed to have a well-defined operation.

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The hypotheses $aH=cH$ and $bH=dH$ do not ensure in an obvious way that $ab H=cd H$, in other words that the equivalence relation on $G$ associated to the subgroup $H$ is compatible with the group operation. You have to prove it.

However this isn't very hard: $abH=cdH$ means that $\;(cd)^{-1}ab\in H$. Now $$(cd)^{-1}ab=d^{-1}\underbrace{c^{-1}a}_{\in H}\,b\in d^{-1}Hb= d^{-1}b H\quad\text{since }H\text{ is normal},$$ and $d^{-1}b\in H$, so $\;d^{-1}b\,H\subseteq H.$

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    $\begingroup$ I think the OP knows your proof (or an equivalent one) but wants to know why he can't just use the fact that $aH$ and $cH$ are the same set and substitute in the set product $aHbH$. $\endgroup$ Aug 1, 2019 at 22:30
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    $\begingroup$ That's what I (hopefully) answer to in the prologue of the proof. Do you think it's not clear enough? $\endgroup$
    – Bernard
    Aug 1, 2019 at 22:32
  • $\begingroup$ That prologue does indeed help. My answer addresses what I think confused the OP. In any case the OP has enough to go on now. $\endgroup$ Aug 2, 2019 at 1:47
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It might help for you to calculate what can go wrong with this operation when $H$ is not a normal subgroup. Some good examples can be found by considering order-two subgroups of $S_3$.

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    $\begingroup$ Welcome to MSE. This should be a comment, not an answer. $\endgroup$ Aug 2, 2019 at 13:37

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