# Checking inverses axiom from Cayley table

I'm checking whether a multiplication table (Cayley table) satisfies the inverses axiom. Suppose I've already confirmed the identity axiom (but nothing else).

Which (if either) of the following statements are correct?

a) A multiplication table satisfies the inverses axiom if and only if the identity element appears once in each row and in each column and its occurrences are symmetrical with respect to the main diagonal.

b) A multiplication table satisfies the inverses axiom if and only if the identity element appears at least once in each row and in each column and its occurrences are symmetrical with respect to the main diagonal.

I feel that b) is correct, but some textbooks seem to suggest a).

Any help would be greatly appreciated,

Jack

• I'm not entirely sure about where you are heading, thus I only answer in a comment: Are you trying to show group axiomes, one at a time? If yes, you might want to remember that in a group, every element has a unique inverse. To show that you need associativity of the group, so depending on where you are heading, an element might actually have multiple inverses (however, that would be a really strange structure...). – Dirk Apr 11 '17 at 8:51
• Thanks for your reply - I simply want to know how to check whether a Cayley table satisfies the inverses axiom (alone) at this stage - even if it doesn't (ultimately) define a group. – Jack Apr 11 '17 at 8:56
• That depends on the definition of said axiom. Some authors might say "every element has an inverse", then you should go with b). Some might (already having the group they are aiming for in mind) say "every element has a unique inverse", then you should go with a). – Dirk Apr 11 '17 at 9:00
• Thanks - I naively (?) assumed that the group axioms were set in stone. – Jack Apr 11 '17 at 9:05

If you're talking about groups then yes, the inverse axiom often states that there exists an inverse, but doesn't state that the inverse is unique. However with the other axioms the inverse can be shown to be unique. Suppose that $b$ and $c$ are inverse to $a$ then we have:
$$b = b1 = b(ac) = (ba)c = 1c = c$$