How fill in this multiplication table? The following multiplication table was given to me as a class exercise. I should have all the necessary information to fill it completely in. However, I'm not sure how to take advantage of the relations I am given to fill it in?
The Question
A group has four elements $a,b,c$ and $d$, subject to the rules $ca = a$ and $d^2 = a$. Fill in the entire multiplication table. 
\begin{array}{c|cccc}
\cdot & a & b & c & d \\ \hline
 a&  &  &  & \\ 
 b&  &  &  & \\ 
 c& a &  &  & \\ 
 d&  &  &  & a
\end{array}
I imagine I might proceed like this: 
To find $ab$, write $a = d^2$ and thus $ab = d^2b = db\cdot b$....but my chain of reasoning always stops around here. 
 A: As Gerry Myerson said, first identify the identity element; you have enough information to do this. Identifying it will let you fill in its row and it column in the table. Then use the fact that up to isomorphism there are only two groups of order $4$, the cyclic group of order $4$ and the Klein $4$-group; the fact that $d^2=a$ is important for deciding which one you have.
A: I didn't even need anything as strong as the fact that up to isomorphism there are only two groups of order $4$.  Once you've identified the identity element, you fill in one row and one column of the table.  Then suppose you ask what $bd$ is.  Since neither $b$ nor $d$ is the identity, it can't be $b$ and it can't be $d$, and it can't be either of the two elements that we already know appear in that column, so that leaves only one thing.  And when you've filled in three squares in a column, there's only one thing left to fill in the fourth square.  So you've got $bd$ and $db$ and $ad$ and $da$.  That leaves four squares still blank.  So $b^2$ must be $a$ or $d$.  But it can't be $d$, since that would leave only $a$ to fill in the $ba$ square, and that would imply $b$ is the identity, which is wrong.  So $b^2=a$, and then there's only one square to complete the second row and second column, so that's got to be the only thing left.  And then there's only one more square to fill in, and again it's the only thing left.
A: Since each element has an inverse, each row and each column of the table must contain all four elements. 
After filling in the row and the column of the identity element $c$, we have three $a$'s in the table, and it follows that the only place for the last $a$ is given by $b^2=a$. Trying to distribute four $d$'s in the table, we are led to $ab=ba=d$. The only place for the remaining $b$'s is now given by $ad=da=b$, and the remaining products are then $=c$.
