# Trouble with this multiplication table

I am having trouble filling out this multiplication table for an unspecified group of order 6. The table is

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

Where I was given that $ae = a$ and thus that $e = \bf 1$. I mananged to surmise that $fb = a$ by noticing that $a$ had appeared in every row thus far and that was the only column left, however I now am stuck. How can I proceed?

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Note: I also know that since $df = a = fb$, this table is not abelian. –  Mike Nov 25 '12 at 21:39
There are not many non-abelian groups of order 6. –  Daan Michiels Nov 25 '12 at 21:45
So it is $S_3$. Now to go through and identify each element with the cycle. Is there a better way to do this? I feel like it should be possible in some other manner? –  Mike Nov 25 '12 at 21:53
Yup, you beat me to posting that $S_3$. There will be subgroups of the group that are abelian, as you know. –  amWhy Nov 25 '12 at 21:55

Once you know that the group is nonabelian, you can fill the mult table in, if you’ve stared at the mult table of $S_3$ long enough. The facts to know about the symmetric group are that there are three elements of order $2$, two of order $3$. These two, with the identity, form the only proper normal subgroup; I’ll call it $H$. The other coset of $H$ I’ll call $X$: it contains the three elements of order $2$. Also, the elements of order $3$ form a conjugacy class, and so do the elements of order $2$. Finally, the center is trivial: the only thing that commutes with everything is the identity element, $e$ in our situation; and the centralizer of an element $x$ of order $2$ is $\{e,x\}$, while the centralizer of an element of order $3$ is $H$.
Now let’s move: Since $c^2=a$, $c$ is one of the elements of order $3$, and $a$ is the other. So you can fill in $a^2=c$, and $ac=ca=e$, and also $b^2=d^2=f^2=e$.
Next, consider $df=a$. Conjugating with $d$, you get $ddfd=fd=dad\in H$, but unequal to $a$ because $a$ doesn’t commute with $d$ (remember I’m using $d^{-1}=d$). Thus $fd=c$. Similar arguments give you $fb=a\Rightarrow bf=c$, $df=a\Rightarrow da=f$, and you go on filling in products using all this that you now know.