Question

Here is part of the operation table for a group whose $G$elements are $a,b,c,d$. Fill in the rest of the table. [Hint: Exercises 27 and 28.]

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& & \\ c& c& & a& \\ d& d& & & \end{array}$

Short answer

The elements of the rest of the tables are:

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& d& c\\ c& c& d& a& b\\ d& d& c& b& a\end{array}$

Step-by-step solution

Operation table/composition table of a group

The composition table is the representation of a binary operation in a finite set. This table helps us to verify all the properties of a group.

In this table, the elements of a finite group are written in rows and columns and the composition of these elements are placed at the intersection of the respected row and column.

For example, let$G=\{e,a,b,\}$ form a group under binary composition$\ast$ , then the operation table will be given as follows:

$\begin{array}{cccc}\ast & e& a& b\\ e& e& a& b\\ a& a& b& e\\ b& b& e& a\end{array}$, where$e$ is the identity element,$a$ is the inverse of b and vice-versa.

Formation of table

Each element of a finite group$G$ appears exactly once in each row and exactly once in each column of the operation table.

Apply the property to complete the table

The given table is:

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& & \\ c& c& & a& \\ d& d& & & \end{array}$

Then, at the intersection of $4^{th}$row and$4^{th}$ column, we must have the element

Using the result stated in Step 2:

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& & \\ c& c& & a& \\ d& d& & & a\end{array}$

Step 4:Apply the property to complete the table

Again, using the same property, we have, $d\cdot b\ne 0$since, from the table$a\cdot b=b$

So we get,

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& & \\ c& c& & a& \\ d& d& & b& a\end{array}$

Step 5:Apply the property to complete the table

Now, each row contains all elements of the group, so the empty position in the last row is$c$

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& & \\ c& c& & a& \\ d& d& c& b& a\end{array}$

Step 6:Apply the property to complete the table

Again, each column contains all elements of the group, so the empty position in the second column is$d$

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& & \\ c& c& d& a& \\ d& d& c& b& a\end{array}$

Step 7:Apply the property to complete the table

Now, each row and column contains all elements of the group, so the empty position in the third column is $d$and third row is$b$.

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& d& \\ c& c& d& a& b\\ d& d& c& b& a\end{array}$

Step 8:Apply the property to complete the table

Again, each column contains all elements of the group, so the empty position in the last column is$c$

$\begin{array}{ccccc}& a& b& c& d\\ a& a& b& c& d\\ b& b& a& d& c\\ c& c& d& a& b\\ d& d& c& b& a\end{array}$