The first trick here is to think of the multiplication table as the transition table of an automaton $A$ with each state representing a letter in your multiplication table, but not worrying about acceptance yet. So the letters on the left and in the body of the table are actually states — it would be more accurate to write them as $q_a, q_b, q_c$, but I won’t. The letters across the top are inputs. Then construct the automaton $A_T$ (“$T$” for transpose) for reverse multiplication by transposing $A$: $qquad displaystylebegin{array}{c|ccc} A_T & a & b & c hline a & a & c & b b & a & a & c c & c & b & a end{array}$ So $A(abc)$ takes you to state $c$, and likewise $A_T(cba)$ moves into state $a$ of $A_T$, as you note. However, $A_T$ assumes you are going right-to-left, and we still want to go left-to-right. So the second trick is to reverse the automaton (not the multiplication, which would just get us back were we started), by reversing all the arrows, which leads to a non-deterministic automaton $A_{TR}$ given by the transition table below, with subsets indicated by concatenated letters to keep the chicken scratching down, so $ac$ is really ${a,c}$. (hope I got it all right — seems to work). $qquad displaystylebegin{array}{c|ccc} A_{TR} & a & b & c hline a & ab & b & c b & ∅ & c & a c & c & a & b hline ab & ab & bc & ac bc & c & ac & abc ac & abc & ab & bc abc & abc & abc & abc ∅ & ∅ & ∅ & ∅ end{array}$ You can interpret this as a non-deterministic automaton with only the three rows above the line or a determinized version with all 8 rows. Finally, the machine to solve the problem is the cross-product automaton of the original $A$ and $A_{TR}$, that is $A times A_{TR}$ to perform the intersection behavior of the two automata (we don’t need $A_T$ any more). $A times A_{TR}$ has states that are pairs like $langle a,acrangle$. The transition function runs $A$ and $A_{TR}$ independently. A single start state $langle 1,1 rangle$ goes into $langle a,a rangle$ under input $a$, into $langle b,b rangle$ under input $b$, etc. Accepting states in the non-deterministic version are $langle a,a rangle$ etc. In the deterministic version, accepting states are pairs in which the first component is $in$ of the second component set, such as $langle a,a rangle$ or $langle b,bc rangle$. $A times A_{TR}$ augmented and determinized as shown has $25=3cdot 8+1$ states, so forgive me if I don’t write it out in detail. But the non-deterministic version has only $10=3cdot 3+1$ states.
