+ /*
+ * To verify that we have a valid group table, it suffices to
+ * test latin-square-hood and associativity only. All the other
+ * group axioms follow from those two.
+ *
+ * Proof:
+ *
+ * Associativity is given; closure is obvious from latin-
+ * square-hood. We need to show that an identity exists and that
+ * every element has an inverse.
+ *
+ * Identity: take any element a. There will be some element e
+ * such that ea=a (in a latin square, every element occurs in
+ * every row and column, so a must occur somewhere in the a
+ * column, say on row e). For any other element b, there must
+ * exist x such that ax=b (same argument from latin-square-hood
+ * again), and then associativity gives us eb = e(ax) = (ea)x =
+ * ax = b. Hence eb=b for all b, i.e. e is a left-identity. A
+ * similar argument tells us that there must be some f which is
+ * a right-identity, and then we show they are the same element
+ * by observing that ef must simultaneously equal e and equal f.
+ *
+ * Inverses: given any a, by the latin-square argument again,
+ * there must exist p and q such that pa=e and aq=e (i.e. left-
+ * and right-inverses). We can show these are equal by
+ * associativity: p = pe = p(aq) = (pa)q = eq = q. []
+ */
+