# Efficient way to show the Associativity of $\oplus: E \times E \rightarrow E,\ x\oplus y \equiv x+y \bmod 5$

Let $E:=\{0, 1, 2, 3, 4\}$ be a set and let $\oplus$ be an internal binary operation on $E$ such that $$\oplus: E \times E \rightarrow E,\ x\oplus y \equiv x+y \bmod 5$$

So, we have: \begin{array}{|c|c|c|c|c|c|c|}\hline \oplus & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 & 0 \\ \hline 2 & 2 & 3 & 4 & 0 & 1 \\ \hline 3 & 3 & 4 & 0 & 1 & 2 \\ \hline 4 & 4 & 0 & 1 & 2 & 3 \\ \hline \end{array}

I need to conclude the associativity of $\oplus$ on $E$ by using the above table.

Is there an efficient way to conclude that without going through all $5\cdot5\cdot5= 125$ possibilities?

• Hint Use the definition of $\oplus$ and the associativity of $+$. – Travis Nov 30 '17 at 17:51
• What does [5] mean? – Shine On You Crazy Diamond Nov 30 '17 at 17:57
• Let $a, b, c \in E$. To show that $\oplus$ is associative, show that $(a \oplus b) \oplus c = a \oplus (b \oplus c)$. As @Travis has suggested, using the associativity of $+$ should make this doable – Smeef Nov 30 '17 at 17:57
• @Travis The question is how to conclude the associativity by using the given table not by using the equation of associativity. – Ayoub Falah Nov 30 '17 at 18:10
• I suppose it depends on what exactly you mean by "using the given table" then. Using the symmetry of the table, it only takes $\frac{1}{2}(5)(5 + 1) = 15$ checks to show that $\oplus$ is actually given by the given algebraic rule, after which you can freely use that characterization. – Travis Nov 30 '17 at 18:55

If you want to prove the associativity "using the table" yo have to do all $125$ cases, or else have to give a law describing the buildup of the table.
What I want to say: Your problem can be solved only by doing the $125$ cases using the associativity in ${\mathbb Z}$ individually, or by proving once and for all "abstractly" that the addition in ${\mathbb Z}_5$ (or a similar quotient structure) is again associative. Then your table just has to do with the naming of the equivalence classes.