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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.2012-11-18

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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?

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    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.2012-11-18
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    @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.2012-11-18
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    @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?2012-11-18
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    @JL90: I didn’t check every single entry, but I checked enough to be pretty sure that it is, yes.2012-11-18