The following multiplication table was given to me as a class exercise.
The Question
A group has five elements $a$, $b$, $c$, $d$ and $e$, subject to the rules $ab=d$, $ca=e$ and $dc=b$. Fill in the entire multiplication table.
$\begin{array}{c|ccccc}
\cdot & a & b & c & d & e \\
\hline
a & & & & & \\
b & & & & & \\
c & & & & & \\
d & & & & & \\
e & & & & &
\end{array}$
I'm not so sure how to solve this. Any help?
Best Answer
Generally these tables are interpreted as taking an $x$ from the left column and a $y$ from the top row and putting its product $xy$ in the $(x,y)$ position.
You have already been told that $d$ goes in the $(a,b)$ position, and that $e$ goes in the $(c,a)$ position, and that $b$ goes in the $(d,c)$ position. Since groups of prime order are abelian, you can also conclude what $ba,cd$ and $ac$ are.
Since $a,b,c,d,e$ are likely assumed to be distinct, you can also tell from these that the only candidate for the identity is $e$, and so that allows you to fill in the last column and the last row rapidly. At this point you also learn that $a$ and $c$ are inverses of each other.
Using that relationship, you can deduce from $ab=d$ that $b=cab=cd$, so another entry appears in the $(c,d)$ position. As you get further along, you should be able to deduce each position.
Don't forget also that you have another tool at your disposal, namely that all the elements satisfy $x^5=e$. Another thing is that $a,c$ are paired up as inverses, and $e$ is its own inverse... what can you conclude about $b$ and $d$? Also, show that $a^2\in\{b,d\}$: if you try both of them out, you should see immediately that only one is consistent with the relations.
Please update us with your progress.