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?

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Is $e$ supposed to be identity element? If not, which is the identity? –  Git Gud Jan 26 '13 at 12:54
@GitGud Since, all the letters, are disinct, it is clear from those three relations that $a,b,c, d$ cannot be the identity. –  rschwieb Jan 26 '13 at 13:24
This exercise is surprisingly sudoku-like! –  rschwieb Jan 26 '13 at 13:31
For future reference, these are usually referred to as Cayley tables. –  Sam DeHority Jan 26 '13 at 13:37

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.

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and we have d=dca=ba ? –  izasg Jan 26 '13 at 13:28
Yes, that agrees with what I have! –  rschwieb Jan 26 '13 at 13:37
Why do we have cc=d? –  izasg Jan 26 '13 at 13:56
This should help you derive it: each row and each column must have exactly one of $a,b,c,d,e$. –  rschwieb Jan 26 '13 at 14:11
Thanks for your consideration to my answer. +1 –  Babak S. Jan 26 '13 at 14:26

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 $c$.

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There is a $b$ in the $(c,c)$ position, where you probably meant to put a $d$. –  rschwieb Jan 26 '13 at 13:40
@rschwieb: As you noted this is like a suduko. In fact the only cases we have is to assume $bc=a,bc=d$. Each one has its own story and implication. I assumed $bc=d$ and then we have $bc^2=dc=b$ or $c^2=b$. I don't know if we take abelian away, what we can say about this puzzel. –  Babak S. Jan 26 '13 at 13:51
It's probably analogous to what I did, checking the cases of $a^2\in\{b,d\}$. After that was determined, the rest could be filled in. –  rschwieb Jan 26 '13 at 14:13
@rschwieb: It certainly is. –  Babak S. Jan 26 '13 at 14:26

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?

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Matches what I got. Good work! –  rschwieb Jan 26 '13 at 15:47