Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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?

share|cite|improve this answer
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
@JL90: I didn’t check every single entry, but I checked enough to be pretty sure that it is, yes. – Brian M. Scott Nov 18 '12 at 23:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.