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?
 A: The only non-abelian group of order six is the dihedral group of order six. Here is its multiplication table.
A: 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.
