Question

The following is the incomplete operation table of a 4-element group.
$\begin{array}{ccccc}\ast & \text{e}& \text{a}& \text{b}& \text{c}\\ \text{e}& \text{e}& \text{a}& \text{b}& \text{c}\\ \text{a}& \text{a}& \text{b}& \text{c}& \text{e}\\ \text{b}& & & & \\ \text{c}& & & & \end{array}$
The last row of the table is

• A. $\text{c a e b}$
• B. $\text{c b a e}$
• C. $\text{c b e a}$
• D. $\text{c e a b}$

Answer 1

The correct option is D $\text{c e a b}$

Step 1:
By looking at the row for $e$, we see that it is a copy of the column headers.
So $e$ must be the identity element. Since right identity and left identity element must both be same. The column corresponding to $e$ must be a copy of the row headers.
We can now say that the operation table is:
$\begin{array}{ccccc}\ast & e& a& b& c\\ e& e& a& b& c\\ a& a& b& c& e\\ b& b& -& -& -\\ c& c& -& -& -\end{array}$

Step 2:
From table above we see that $a\ast c=e$
$\therefore c\ast a$ must also be $=e$ (if $a$ is the inverse of $c$, then $c$ is the inverse of $a$)
Now the operation table looks like
$\begin{array}{ccccc}\ast & e& a& b& c\\ e& e& a& b& c\\ a& a& b& c& e\\ b& b& -& -& -\\ c& c& e& -& -\end{array}$

Step 3:
The blank in second column must be $c$ (since in a cayley table, every row and every column is a unique permutation of the row and column headers).
Now the operation table looks like
$\begin{array}{ccccc}\ast & e& a& b& c\\ e& e& a& b& c\\ a& a& b& c& e\\ b& b& c& -& -\\ c& c& e& -& -\end{array}$

Step 4:
Now the blanks in third row can be filled as $a,e$ or $e,a$. Let us try each one in turn.
If we fill $a,e$ in third row the operation table will look like
$\begin{array}{ccccc}\ast & e& a& b& c\\ e& e& a& b& c\\ a& a& b& c& e\\ b& b& c& a& e\\ c& c& e& -& -\end{array}$

Now the blank in fourth row and third column must be filled by $e$
However this is not possible since $e$ is already entered in fourth row and second column. Therefore filling $a,e$ in third row blanks is wrong. So let us try filling the third row blanks with $e,a$
Now the operation table looks like
$\begin{array}{ccccc}\ast & e& a& b& c\\ e& e& a& b& c\\ a& a& b& c& e\\ b& b& c& e& a\\ c& c& e& -& -\end{array}$
Now the blanks in fourth row has to be filled with $a,b$
The final operation table looks like
$\begin{array}{ccccc}\ast & e& a& b& c\\ e& e& a& b& c\\ a& a& b& c& e\\ b& b& c& e& a\\ c& c& e& a& b\end{array}$
which is consistent with all the rules of a Cayley Table
The last row of this table is $c,e,a,b$
Therefore the correct answer is (d)