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From "A First Course In Abstract Algebra" by John B. Fraleigh

Theorem
If $G$ is a group with binary operation $\ast$, and if $a$ and $b$ are any elements of $G$, then the linear equations $a \ast x = b$ and $y \ast a = b$ have unique solutions $x$ and $y$ in $G$.

Proof
First we show the existence of at least one solution by just computing that $a' \ast b$ is a solution of $a \ast x = b$. Note that $$\begin{align*} a \ast (a' \ast b) &= (a \ast a') \ast b & &\text{associative law} \\ &= e \ast b & &\text{definition of $a'$} \\ &= b & &\text{property of $e$} \\ \end{align*}$$ Thus $x = a' \ast b$ is a solution of $a \ast x = b$. In a similar fashion, $y = b \ast a'$ is a solution of $y \ast a = b$.
To show uniqueness of $y$, we use the standard method of assuming that we have two solutions, $y_1$ and $y_2$, so that $y_1 \ast a = b$ and $y_2 \ast a = b$. Then $y_1 \ast a = y_2 \ast a$, and by Theorem 4.15, $y_1 = y_2$. The uniqueness of $x$ follows similarly. $\blacksquare$

My question. Why didn't Fraleigh simply solve the equation to show that a solution exists? I mean: given the equation $a \ast x = b$ we can do as follows $$\begin{align*}a \ast x &= b \\ a' \ast (a \ast x) &= a' \ast b \\ (a' \ast a) \ast x &= a' \ast b \\ e \ast x &= a' \ast b \\ x &= a' \ast b\end{align*}$$ and this shows two things: that a solution exists and also that the solution is unique.

Is it a matter of style or in this context my way is wrong?

rookie_of_math
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    Why does that prove uniqueness? What if there was another way to concoct a solution, and it gave a different one? – Randall Jun 08 '21 at 12:48
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    What you can do: supposing a solution exists, show that it must necessarily be $a'*b$. That does show uniqueness. – Randall Jun 08 '21 at 12:50
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    a list of equations isn't an argument - you need to explain why each one is true if and only if the next is true (Fraleigh's chain of equalities does not have this problem). In the real numbers, for example, $a=b$ implies $ca=cb$ but $ca=cb$ doesn't imply $a=b$. – Matthew Towers Jun 08 '21 at 12:51
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    It is a matter of style. Yes, your argument does show both statements. – plop Jun 08 '21 at 12:55
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    I think this is really a question of habits; the usual song is in two verses: 1/ show that a solution exists, 2/ show it is unique. It has the advantage of clarity. – Picaud Vincent Jun 08 '21 at 13:35
  • @MatthewTowers I can 't undestrand. Given the equation $a \ast x = b$ I solved it like we usually do with linear equations in $\mathbb{R}$. I simply used the properties of operation $\ast$ in a group. – rookie_of_math Jun 08 '21 at 13:36
  • @PicaudVincent So do you think my way is right? Do you think solving the equation $a \ast x = b$ like I did is OK to show that a solution exists? – rookie_of_math Jun 08 '21 at 13:38
  • @rookie_of_math I personally think your solution is right in this peculiar case. But I also think it is a good habit to split the problem into two parts 1/ & 2/. It improves clarity and this is helpful when tackling trickier problems. One thing at a time is the safer way to get correct results :) – Picaud Vincent Jun 08 '21 at 13:41
  • @Randall I solved the equation $a \ast x = b$ like we usually do with equations in $\mathbb{R}$. When we solve equations in $\mathbb{R}$ the procedure to solve the equation shows existence and uniquity of solution. Why in this case does the procedure fail? – rookie_of_math Jun 08 '21 at 13:41
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    It is a subtle point, but I would again argue that ONE method of solving equation might not be the ONLY method. As it turns out, in groups in your particular equation this turns out to be the "only sensible" way to solve it, but that is precisely what Fraleigh is proving. – Randall Jun 08 '21 at 13:47
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    It think yours shows that if a solution exist it must be $a'*b$ which, if you have already proven inverses are unique, must be unique. But it doesn't show a solution exists in the first place. Claiming "$ax =b$" is (similar to claiming "$0\cdot x = 7$") simply assumes without justification that a solution exists. Admittedly we can just work backwards and show that if $x=a'b$ then it is a solution but... you didn't. – fleablood Jun 08 '21 at 15:17
  • Consider the following proof that there is a unique positive integer, $n$ so that $n^2 = -4$: Let $n^2 = -4\implies n^4 = 16 \implies n = -2,2,2i, -2i$ but of those only $2$ is positive. So $n=2$ is the unique positive integer solution to $n^2 = -4$. Do you see the error. – fleablood Jun 08 '21 at 15:23
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    ... and so your proof that given I solution exists that it must be unique is valid, I think, IMO, that Fraileigh's is more instructive to a first time student as to what issues we are supposed to be addressing and aware of and the need to nail them down. – fleablood Jun 08 '21 at 15:27
  • Your proof is correct only if you use bidirectional arrows between all lines of the proof. This subtlety is discussed here and here, where the confusion caused by this subtlety is clearly illustrated in the first linked post (cf. claim that there is an error in Rosenlicht's textbook). – Bill Dubuque Sep 09 '24 at 16:14

2 Answers2

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The difference between "a sequence of equations" and a proof are: clearly expressed logical relations between steps; and clearly expressed justification of each step.

Regarding your comment in reponse to @MatthewTowers, you may have "simply used the properties of operation $*$ in a group", but you did not justify your use of those properties, nor did you explain the logical relations amongst your equations.

If you include those justifications and implications, then your list of equations can be converted into a true proof of uniqueness, something like this (which I did by simply cutting and pasting your align* environment and adding some stuff): $$\begin{align*} a \ast x &= b \\ \implies a' \ast (a \ast x) &= a' \ast b \quad\text{(well-definedness of $(*)$)} \\ \iff (a' \ast a) \ast x &= a' \ast b \quad\text{(associative law applied on the LHS of the equation)}\\ \iff e \ast x &= a' \ast b \quad\text{(inverse law applied on the LHS of the equation)} \\ \iff x &= a' \ast b \quad\text{(identity law applied on the LHS of the equation)} \end{align*} $$ and therefore, following all the implication arrows forwards, we have $a*x=b \implies x = a'*b$, which proves uniqueness.

But, has existence been proved??!?

Look carefully at the implications that I have written out. The first one is only $\implies$; the remainder are $\iff$. In each $\iff$ step, the group law I am applying allows me to rewrite one side of the equation by substitution, leaving the other side unchanged. But the first implication is different: well-definedness of $(*)$ gives only the forward direction of implication $\implies$.

Notice that if the converse of the first implication is true then you can follow all of the implication arrows backwards to prove existence.

So, is the converse of the first implication true?

Yes, of course it is, and one can prove it like this: \begin{align*} a' * (a * x) &= a' * b \\ \implies a * (a' * (a * x)) &= a * (a' * b) \quad\text{(well-definedness of $(*)$)}\\ \end{align*} and so on applying the associative law and inverse law and identity law...

However, that leads to a rather gloppy proof of existence. Fraleigh's proof of existence is much shorter and better.

Lee Mosher
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  • The OP's proof is correct if written using bidirectional arrows. This subtlety is discussed here and here, where the confusion caused by this subtlety is clearly illustrated in the first linked post (cf. claim that there is an error in Rosenlicht's textbook). – Bill Dubuque Sep 09 '24 at 16:16
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I'll type some of the things that have been said in the comments here because it can be helpful to have them in one place.

As Matthew says: we can't just have a list of equations, we need to be precise about the relation between one equation and the rest. If, like here, nothing is written about how the equations are related some readers will assume that you mean that each line follows naturally from the previous one, while some others will assume that each line follows naturally from the previous one AND MOREOVER that conversely the previous one follows from the current one. But these are two very different things. So we better are clear about which one we mean.

Now suppose that we go with only the first interpretation and you would have annotated your solution as follows:

Suppose that for some $x$ we have:

$$\begin{align*}a \ast x &= b \textrm{, then it follows that} \\ a' \ast (a \ast x) &= a' \ast b \textrm{, so that}\\ (a' \ast a) \ast x &= a' \ast b \textrm{, and hence}\\ e \ast x &= a' \ast b \textrm{, from which it follows that}\\ x &= a' \ast b\end{align*}$$

Now what does this prove? Obviously it shows that IF such an $x$ exist, THEN it must be equal to $a' \ast b$. What it doesn't prove is THAT such an $x$ exist, so you need to type a separate proof of that fact.

Of course, giving such a proof is easy: since we know that IF $x$ exist it must be equal to $a' \ast b$, all we need to do is check that $x = a' \ast b$ does indeed 'do the job'.

AND in a sense you are right that you already typed such a proof: it is just your sequence of equations, but now with the silent assumption that each equation follows from the one below it, rather than the one above it. In annotated form:

$$\begin{align*} \textrm{define } x &= a' \ast b \textrm{, then clearly} \\ e \ast x &= a' \ast b \textrm{, from which it follows that} \\ (a' \ast a) \ast x &= a' \ast b \textrm{, and hence} \\ a' \ast (a \ast x) &= a' \ast b \textrm{, so that clearly obviously trivially} \\ a \ast x &= b \\ \end{align*}$$ showing that this $x$ is a solution to the desired equation.

In summary: your sequence of statements does indeed show both things (existence and uniqueness) but the distinction is still very much 'there': you get uniqueness by reading from top to bottom and existence by reading from bottom to top.

In this context it is not clear that each step can be taken in both directions just because you typed them above each other, so at least you should make that more explicit.

I hope this answers your question. Apart from that there is a second issue with your proof, also pointed out by Matthew:

I type 'clearly obviously trivially' in the last proof step above because this step (and only this step) is in fact not clear, obvious and trivial at all. In fact it is valid in groups but not, e.g. in $\mathbb{R}$ so adding some additional reasoning why this step holds here is highly desired.

(E.g. we could add some extra lines where we first multiply both sides of the equation by $a$, then simplify the resulting $a \ast a'$ to $e$ and finally removing the $e$'s. But these extra lines are only needed for existence and not for uniqueness, further illustrating the point that in fact there is a difference between the two even in your approach.)

Vincent
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  • I don't think your writing the proof backward works at all. The more straightforward backwards proof would be $$x=a'b\
    a    x =a(a'b)\ax = (aa') b\ax=eb\ax=b$$. The proof the op provided isn't a complete backwards forward proof.     – fleablood Jun 08 '21 at 15:33
  • @fleablood yes I also thought of the backwards proof you mention, which is obviously better then using the original backwards proof and adding three lines, but I wanted to stay as close the the original proof as possible. – Vincent Jun 08 '21 at 15:44
  • Also, of course, following the example of the book, instead of adding extra lines we could appeal to the Theorem 4.15 mentioned in the quote to justify the step that I say needs additional reasoning. But something is needed here. That is of course also what @fleablood says when stating that the proof is not complete. – Vincent Jun 08 '21 at 15:52
  • @fleablood The proof written using bidirectional arrows yields both existence and uniqueness of roots, cf. this old answer. – Bill Dubuque Sep 09 '24 at 16:10