Question

A partial operation table for a group $G=\{e,a,b,c,d,f\}$is shown below.Complete the table. [Hint: Exercises 27 and 28.]

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& & \\ b& b& & & & & \\ c& c& f& & & & a\\ d& d& & & & & \\ f& f& & & & & \end{array}$

Short answer

The rest of the table is:

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& f& c\\ b& b& e& a& f& c& d\\ c& c& f& d& e& b& a\\ d& d& c& f& a& e& b\\ f& f& d& c& b& a& e\end{array}$

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 the 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:

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& & \\ b& b& & & & & \\ c& c& f& & & & a\\ d& d& & & & & \\ f& f& & & & & \end{array}$

Since, from the table we have, $ab=e\Rightarrow ba=e$,

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& & \\ b& b& e& & & & \\ c& c& f& & & & a\\ d& d& & & & & \\ f& f& & & & & \end{array}$

Step 4:Apply the property to complete the table

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& & \\ b& b& e& & & & \\ c& c& f& & & & a\\ d& d& c& & & & \\ f& f& d& & & & \end{array}$

Step 5:Apply the property to complete the table

Similarly,

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& f& c\\ b& b& e& & & & \\ c& c& f& & & & a\\ d& d& c& & & & \\ f& f& d& & & & \end{array}$

Step 6:Apply the property to complete the table

Since $df=\left(ac\right)f=a\left(cf\right)=a\cdot a=b$,

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& f& c\\ b& b& e& & & & \\ c& c& f& & & & a\\ d& d& c& & & & b\\ f& f& d& & & & \end{array}$

Step 7:Apply the property to complete the table

By using the property in Step 2:

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& f& c\\ b& b& e& & & & d\\ c& c& f& d& & & a\\ d& d& c& & & & b\\ f& f& d& & & & e\end{array}$

Step 8:Apply the property to complete the table

Now, by using property in Step 2: and,$b=df=\left(cb\right)f=c\left(bf\right)=cd$ we get,

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& f& c\\ b& b& e& & & & d\\ c& c& f& d& e& b& a\\ d& d& c& & & & b\\ f& f& d& & b& & e\end{array}$

Step 9:Apply the property to complete the table

Again, by using the property in Step 2, and$bc=\left(fc\right)c=f\left(cc\right)=f\cdot e=f$

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& f& c\\ b& b& e& & f& & d\\ c& c& f& d& e& b& a\\ d& d& c& f& a& e& b\\ f& f& d& & b& & e\end{array}$

Step 10:Apply the property to complete the table

Using,$bb=b\left(fc\right)=\left(bf\right)c=dc=a$ , by using property in Step: 2:

$\begin{array}{ccccccc}& e& a& b& c& d& f\\ e& e& a& b& c& d& f\\ a& a& b& e& d& f& c\\ b& b& e& a& f& c& d\\ c& c& f& d& e& b& a\\ d& d& c& f& a& e& b\\ f& f& d& c& b& a& e\end{array}$