# Group(Non-abelian) Multiplication Table

Given Group(Non-abelian) Multiplication Table.
Find $ca,bb\text{ and }df$.

$$\begin{array}{ l | c r r r r r } * & e & a & b & c & d &f \\ \hline e & e & a & b & c & d & f \\ a & a & b & e & d \\ b&b\\ c&c&&&e&&a\\ d&d\\ f&f\\ \end{array}$$

My question:

Is there an algorithm for filling up the empty cells? if there isnt, what is then the best approach?

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Hint: Each row and each column must contain each of $a,b,c,d,e,f$ exactly once. –  MJD Nov 18 '12 at 22:11
There may not be enough information to fill in the whole table, but there’s enough to answer the question. For instance, $ca=c(cf)=(cc)f=ef=f$. Note that $bb=(aa)b$ and $df=(ac)f$; can you take it from there?
I suspect there is enough information to fill in the whole table (though of course as you show there is no need to do so). From $ca=f$ and $ac=d$ we know it's not abelian, so it must be $S_3$. $a,b$ have order 3, $c,d,f$ must be transpositions, and so on. –  Gerry Myerson Nov 18 '12 at 22:44
@Gerry: True; I wasn’t thinking about the fact that it has to be $S_3$. I’m too lazy at the moment to try, so I’ll just change the wording slightly. –  Brian M. Scott Nov 18 '12 at 22:49
@BrianM.Scott I came up with this: $$\begin{array}{ l | c r r r r r } * & e & a & b & c & d &f \\ \hline e & e & a & b & c & d & f \\ a & a & b & e & d&f&c \\ b&b&e&a&f&c&d\\ c&c&f&d&e&b&a\\ d&d&c&f&a&e&b\\ f&f&d&c&b&a&e\\ \end{array}$$ is this correct? –  Onur Nov 18 '12 at 22:56