# Constructing Multiplication Table for Multiplication Modulo

I have this question

Consider G={1,5,7,11,13,17} under Multiplication Modulo 18.Construct Multiplication Table for G.I have constructed the following

Im i correct ?

• The table seems to be correct. You did p.ex. $11\times 17=187$ Then you divided 187 by 18 and the remainder was $7$. I did not check all the boxes though. – zoli Sep 27 '15 at 8:36
• @zoli Yeah, i did the same x18-Multiplication modulo 18- (a*b)mod 18 – techno Sep 27 '15 at 8:41
• There should not be two the same in a given column, yet you have two copies of 5 in the column under the 5. (That column is also missing an 11, and each number should appear in each row/column somewhere.) – coffeemath Sep 27 '15 at 8:49
• @coffeemath Thanks,but i really did not get what you have meant.Can you point out where i have gone wrong? – techno Sep 27 '15 at 8:58
• @coffeemath means that 13x5=65 and 65/18 = 3 and the remainder is 11. So In the box (13,5) the 5 is wrong the correct result is 11. – zoli Sep 27 '15 at 9:05

When just multiplying in the reals I get:

$$\begin{array}{c|cccccc} \times & 1 & 5 & 7 & 11 & 13 & 17\\ \hline1 & 1 & 5 & 7 & 11 & 13 & 17\\ 5 & 5 & 25 & 35 & 55 & 65 & 85\\ 7 & 7 & 35 & 49 & 77 & 91 & 119\\ 11 & 11 & 55 & 77 & 121 & 143 & 187\\ 13 & 13 & 65 & 91 & 143 & 169 & 221\\ 17 & 17 & 85 & 119 & 187 & 221 & 289 \end{array}$$

Then $\mod 18$ we get

$$\begin{array}{c|cccccc} \times & 1 & 5 & 7 & 11 & 13 & 17\\ \hline 1 & 1 & 5 & 7 & 11 & 13 & 17\\ 5 & 5 & 7 & 17 & 1 & 11 & 13\\ 7 & 7 & 17 & 13 & 5 & 1 & 11\\ 11 & 11 & 1 & 5 & 13 & 17 & 7\\ 13 & 13 & 11 & 1 & 17 & 7 & 5\\ 17 & 17 & 13 & 11 & 7 & 5 & 1 \end{array}$$

So only the $5$ on the bottom left is wrong, this should be $11$. As it is a commutative group, you should have a symmetry relative to the diagonal, this would provide a quick way that finds this error.

• Thanks.. so for Abelian Groups the diagonal elements will have symmetry,right? – techno Sep 27 '15 at 11:24
• I mean a symmetry relative to the diagonal: $13 \times 5 = 11 = 5 \times 11$ in your example here. – flawr Sep 27 '15 at 12:03
• Okay..To be clear if i have 17 in the top right part of the diagonal i will have a 17 in the bottom left part and so on.. right? – techno Sep 27 '15 at 12:31
• Exactly, because the group is abelian e.g. $7\times 5$ must be the same as $5 \times 7$ when looking it up in the table, so the top right triangle is just mirrored along the main diagonal to the bottom left triangle. – flawr Sep 27 '15 at 13:55

Calculating these things are really neat to do in for instance Matlab / Octave.

Two lines:

ps = [1,5,7,11,13,17];

table = mod(ps'*ps,18)

First line defines the numbers as a row-vector.

Second line does outer product the vector with itself followed by an element-wise modulo.

My output is $$\text{table} = \left[\begin{array}{cccccc} 1&5&7&11&13&17\\ 5&7&17&1&11&13\\ 7&17&13&5&1&11\\ 11&1&5&13&17&7\\ 13&11&1&17&7&5\\ 17&13&11&7&5&1 \end{array}\right]$$

• Thanks for the input on matlab. – techno Sep 27 '15 at 14:05