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.

  • 3
    $\begingroup$ First: can you identify the identity element? $\endgroup$ – Gerry Myerson Nov 19 '12 at 22:45
  • $\begingroup$ Gerry Myerson's question gives you six squares in the table immediately. I got the rest of them quickly by recalling that all groups of order less than $6$ are abelian. $\endgroup$ – Michael Hardy Nov 19 '12 at 22:51

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.

  • $\begingroup$ It's possible that OP is at an early point in the course where the groups of order 4 have not yet been catalogued. It should still be possible to see the problem through, though it will take more work. $\endgroup$ – Gerry Myerson Nov 19 '12 at 22:55
  • $\begingroup$ @Gerry: I agree, but what’s needed to distinguish the two is also what’s needed to identify the group even if one doesn’t know that there are only two, so I thought that it probably wouldn’t hurt to mention the stronger fact. $\endgroup$ – Brian M. Scott Nov 19 '12 at 22:58
  • 2
    $\begingroup$ The classification of groups of order 4 is not necessary. For example to compute $bd$ we reject $a$ and $d$ since they already appeared. If $bd=b$ then by cancellation we get $d=e$ which is not possible. Therefore $bd=c$ and $ad=b$. The rest is done similarly. $\endgroup$ – PAD Nov 19 '12 at 23:00
  • $\begingroup$ @Pantelis: I know; I pointed out as much in my previous comment. And while what you’re suggesting works, it’s unnecessarily complicated. $\endgroup$ – Brian M. Scott Nov 19 '12 at 23:03
  • 1
    $\begingroup$ @jmi4 : If $ac=a$ then $a^{-1}(ac)=a^{-1}(a)$. That certainly tells you whether $ac=a$ is sufficient to identify $c$ as the identity. $\endgroup$ – Michael Hardy Nov 19 '12 at 23:07

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.

  • $\begingroup$ You mean that $b^2 = a$ right? Because then $ | b | = 4$, and so this is $\Bbb Z_4$? $\endgroup$ – Moderat Nov 25 '12 at 20:57
  • $\begingroup$ Fixed. ${{{{{{}}}}}}$ $\endgroup$ – Michael Hardy Nov 26 '12 at 3:14

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 $=c$.


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