Complete group multiplication table with 6 elements.

The following group multiplication table shall be completed. How many possibilities are there? (This is an exercise from a textbook).

$\begin{array}{c|cccccc} \cdot & a & b & c & d & e & f \\ \hline a & & & & & & \\ b & &a &e & & & \\ c & & & & & & \\ d & & & & & & \\ e & &d & & & & \\ f & & & & & & \end{array}$

So we have $bb=a, bc=e$ and $eb=d$. What I can derive here is that $$(bc)b=eb=d=b(cb),$$ $$(bb)c=ac=b(bc)=be$$ and $$b(bb)=ba=(bb)b=ab.$$ Of corse when you know that there are only two groups of order 6 ($S_3$ and $\mathbb{Z}6$) you can just check which one is compatible with this table, however I think this is not the idea of this exercise.

• The two groups are $S_3$ and $\mathbb Z_6$ – Jorge Fernández Hidalgo Jan 5 '16 at 14:26
• yes${}{}{}{}{}{}$ – Jorge Fernández Hidalgo Jan 5 '16 at 15:28
• In my view the line with header $a$ is missing because of some slip of the typesetter. In any case, I don't see how you conclude from the given data that $a$ must be the neutral element. – Christian Blatter Jan 5 '16 at 16:03
• I found another argument, since $da = d$ and $ea=e$, $a$ must be the neutral element, agree? – Rudi_Birnbaum Jan 5 '16 at 19:20
• What you have in the table doesn't to match your argument, either in the main text or in the comment. In the table, I only see three entries, corresponding to $bb=a$, $bc=e$, $eb=d$, which is completely different from what you seem to be using. (For example, you write $eb=a$...) Is there something wrong with your typesetting? – verret Jan 6 '16 at 1:52