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This should be simple, but for some reason I get stuck on this.

Let $G = \{x_1, \ldots x_n\}$ be a set equipped with operation $*$ satisfying the following :

1) $G$ has an identity element $e$ with $e *x = x = x * e$ for all $x \in G$.

2) All elements are self-inverse (i.e. $x * x = e$ for all $x \in G$).

Prove that $*$ is associative if and only if for all $i,j,k$, $(x_i * x_j) * (x_j * x_k) = x_i * x_k$.

It's the $\Leftarrow$ part that's bugging me. From 2) I can show that the multiplication matrix of $G$ is a Latin Square. And I think $*$ is commutative since $(x_i * x_j) * (x_j * x_i) = x_i * x_i = e \Rightarrow x_i * x_j = x_j * x_i$.
But I can't see how this leads to associativity.

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  • $\begingroup$ I'm afraid you are implicitly assuming the inverse is unique. Is it in the problem? Because from $(x_i*x_j)*(x_j*x_i)=e$, as you correctly stated, it only follows that $x_i*x_j$ is the left inverse of $x_j*x_i$ and the latter is the right inverse of the former. But maybe I'm just forgetting a theorem telling me the inverse is unique in some conditions satisfied by the problem? $\endgroup$ – MickG Sep 25 '14 at 16:30
  • $\begingroup$ The inverse is indeed unique. Suppose that $x y = e$. Then $y = e y = (e x) (x y) = x (x y) = x e = x$. $\endgroup$ – polmath Sep 25 '14 at 16:42
  • $\begingroup$ @polmath hang on a sec. $y=ey$ OK. But how did you get $ey=(ex)(xy)$? I can tell $ey=e(xx)y$ or $ey=(xx)y$, but to go on I seem to need associativity, which we do not have. And my passages assume $xx=e$ as per the post, which is not necessarily true outside the post's problem. So how do I get $ey=(ex)(xy)$? $\endgroup$ – MickG Sep 25 '14 at 16:49
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    $\begingroup$ $ey = (ex)(xy)$ because we have that $(x_ix_j)(x_jx_k) = x_ix_k$ $\endgroup$ – Manuel Lafond Sep 25 '14 at 16:51
  • $\begingroup$ @manuellafond : that G be a finite set does not seem to be necessary, see below. $\endgroup$ – polmath Sep 25 '14 at 16:55
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I use multiplicative notation for simplicity ($x y$ instead of $x * y$) Suppose that the property $[x y] [y z] = x z$ holds for all $x, y, z$

Lemma 1 : $x (x y) = y$ and $(x y) y = x$ for all $x, y$

This is because we have $x (x y) = [e x] [x y] = e y = y$.

Lemma 2 : $x y = y x$ for all $x, y$.

This is because $[x y] [y x] = x x = e$ on the one hand, and multiplying on the left by $x y$ gives $y x = (x y) e = x y$ thanks to Lemma 1.

Thus, for all $x, y, z$, $(x y) z = x (y z)$ iff $(x y) z = x (z y)$ iff $[(x y) z] [z y] = x$ iff $(x y) y = x$.

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  • $\begingroup$ That's wonderful ! Thanks. $\endgroup$ – Manuel Lafond Sep 25 '14 at 16:48

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