About group multiplication table 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?
 A: 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.
A: First of all if this structure wants to be a group of order $5$ then it is assumed to be cyclic and so abelian. This fact helps you to have $$ab=ba=d,~ca=ac=e,~dc=cd=b,$$ If $b$ be our identity so from above $a=d$ which is wrong becuse $a\neq d$. If $a=id$ then $c=e$ which is also wrong. The same is true for $c$ and $d$. So your $e=e_G$ the identity one. $G=\{e,a,b,c,d\}$ is a group so every element powered by $5$ would be $e$, so $$ac=e\to a=c^{-1}\to a^2=c^{-2}=c^3$$. You can easily fill the blank considering that there is an element, say $c$, which every other element in $g$ can be written as a power of $c$.

A: Now, I think that the answer is 
$\begin{array}{c|ccccc}
\cdot & a & b & c & d & e \\ 
\hline
a & b & d & e & c & a \\ 
b & d & c & a & e & b \\ 
c & e & a & d & b & c \\ 
d & c & e & b & a & d \\
e & a & b & c & d & e
\end{array}$
Is this true?
