Determining a finite set is a group from its multiplication table.

At the end of page 4 of this document, it is proved from the multiplication table that $a^{-1}(ab)=b$. The point is to prove associativity, so I don't see how the conclusion that $a^{-1}(ab)=b$ follows. Can someone explain what the author is pointing at?

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The proof of associativity continues in the next page; it is not established until almost all the way through page 5. – Arturo Magidin May 20 '12 at 21:56
I thought the first argument ends there, and what follows on page 5 is an alternative argument to show associativity. – pagan May 20 '12 at 21:59
Sorry: the argument for associativity ends in the top two lines of page 5, looking at the two diagrams at the bottom of page 4. I'll post the answer. – Arturo Magidin May 20 '12 at 22:02
Thank you sir, I appreciate it. – pagan May 20 '12 at 22:04

There is in fact a gap in the first proof. First, he didn’t bother to prove Step 3. For that he should be comparing the tables

$$\begin{array}{c|cc} &1&ab\\ \hline 1&1&ab\\ a^{-1}&a^{-1}&a^{-1}(ab) \end{array}$$

and

$$\begin{array}{c|cc} &a^{-1}&b\\ \hline a&1&ab\\ 1&a^{-1}&b \end{array}$$

in order to conclude that $a^{-1}(ab)=b$.

Then in order to prove that $(ab)c=a(bc)$, he needs not only the table

$$\begin{array}{c|cc} &b&bc\\ \hline b^{-1}&1&c\\ a&ab&a(bc) \end{array}\tag{1}$$

but also the table

$$\begin{array}{c|cc} &1&c\\ \hline 1&1&c\\ ab&ab&(ab)c \end{array}\tag{2}$$

The correctness of $(1)$ follows from Step 3, and the correctness of $(2)$ is obvious. This fills the gaps in his first proof of associativity.

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Thank you, these were precisely the gaps that were giving me trouble. – pagan May 20 '12 at 22:13
@ArturoMagidin I apologize that I was unclear. When I said "I don't see how the conclusion follows" I was referring to the conclusion that $a^{-1}(ab)=b$. – pagan May 20 '12 at 22:16
@pagan: I thought "the conclusion" was "prove associativity", which you mention immediately before. – Arturo Magidin May 20 '12 at 22:17

The problem asks to prove that a set with a binary operation and an identity in which the following two conditions hold is in fact a group:

(i) every row and every column of the multiplication table contains every element of the set; and

(ii) for every pair of elements $x\neq 1$ and $y\neq 1$, if $R$ is a rectangle in the body of the multiplication table that has $1$ in one vertex, $x$ in the corner in the same row as $1$, and $y$ in the corner in the same column as $1$, then the fourth corner of the rectangle depends only on the pair $(x,y)$ and not on the position of $1$.

From the diagram at the bottom of page 4, you get that if you have the corners $$\begin{array}{cc} 1 & c\\ ab& \Box \end{array}$$ then $\Box$ must be $a(bc)$.

But you get the same three corners if instead of having $$\begin{array}{c|cc} &b&bc\\ \hline b^{-1}&1 & c\\ a&ab & a(bc) \end{array}$$ you take $$\begin{array}{c|cc} &1 & c\\ \hline 1&1 & c\\ ab& ab& \Box \end{array}$$ but here the $\Box$ must be $(ab)c$. The assumption (ii) of the problem is that the bottom right corner depends only on the top right and bottom left corners, not on the rows and columns, so we conclude that $(ab)c=a(bc)$.

You used the fact that $x^{-1}(xy) = y$ to construct the first of these two tables, to get the $c$ on the top right corner.

Added. I assumed, from the way you phrased your question, that you already had and agreed that $x^{-1}(xy) = y$ for all $x,y\in G$. This is hinted at with the first diagram at the bottom of page 4: $$\begin{array}{c|cc} &1&ab\\ \hline 1&1 & ab\\ a^{-1}&a^{-1}& a^{-1}(ab) \end{array}$$ which you should compare with $$\begin{array}{c|cc} & a^{-1} & b\\ \hline a & 1 & ab\\ 1 & a^{-1} & b \end{array}$$ i.e., the same argument as used later to prove associativity.

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