GATE CSE 2004 | Question: 72

Question

The following is the incomplete operation table of a $4-$element group.

$$\begin{array}{|l|l|l|l|l|} \hline \textbf{*} & \textbf{e}& \textbf{a} &\textbf{b} & \textbf{c}\\\hline \textbf{e} & \text{e}& \text{a} & \text{b} & \text{c} \\\hline \textbf{a} & \text{a}& \text{b} & \text{c} & \text{e}\\\hline \textbf{b}\\\hline \textbf{c} \\\hline\end{array}$$

The last row of the table is

• A. $c\;a\;e\; b$
• B. $c\; b\; a\; e$
• C. $c\; b\; e\; a$
• D. $c\; e\; a\; b$

Comments

Although Rajesh's answer is best way to approach this problem , yet these is another approach.

We can observe given group is cyclic group , where 'a' is generator , and every cyclic group is abelian too. So fill the missing entries on the bases of commutative property of abelian groups

$a^{1}=a$

$a^{2}=a*a=b$

$a^{3}=a*(a*a)=a*b=c$

$a^{4}=a*a*(a*a)=a*(a*b)=a*c=e$

---

Another approach :-

By observing above Cayley  table, you can see that "e" is right identity. As identity is unique in group, so "e" also should be left identity. So, using this concept, we able to get 1st element of last row as "c".

Now lets see, how to find the 2nd element of last row :-

By observing above Cayley table, you can find that inverse of "a" is "c" (because in row of heading "a", we have identity element in last column and that column heading is element "c"). By using properties of group, we can say that inverse of  "c" is "a" . So, fill "e" in the cell, which is intersection of row reading "c" and column heading "a". That cell is 2nd cell of last row. So, value of 2nd element of last row is "e". Our result is matching only with option (D), so option (D) is true.

---

@Aalok8523 I used the same method as yours while solving for the first time and then in the second thought to fully get the last row, @Rupendra Choudhary, I followed the same approach as you did...

---

Its a well known group

Answer  : (D)

Answer 1

1. Group of order Prime Square $(p^2)$ is always abelian. See here 
2. If the group is abelian, then $x * y = y * x$ for every $x,y$ in it (Commutative). Therefore, the $(i,j)$ entry is equal to the$( j,i)$ entry in the Cayley table making the table is symmetric. $($Informally, $1^{st}$ row is same as $1^{st}$ column, $2^{nd}$ row is same as $2^{nd}$ column and so on$)$

Here, order $4=p^2=2^2  (p=2).$ Hence, it is abelian group.

Now abelian group's Cayley  table is symmetric. So, $1^{st}$ row will be same as $1^{st}$ col and $2^{nd}$ row will be same as $2^{nd}$ column.

Matches with option D only.

Comments

great explanation

---

actually how i got this question is its following order e a b c in rows as well as in columns we just have tofollow that order if first element is e then in row next 3 elements respectively are a b c similarly if first element is a then next three elements in row are b c e like that it will solve in 10 seconds same will apply for columns also

---

 

ankitgupta.1729 Nice Approach !!

Answer 2

From First row you can conclude that e is the identity element.

=> Using the above fact, from second row you can conclude that a and c are inverses of each other.

=> In fourth row:

First element : c*e = c (e is identity)

Second element : c*a = e (inverse)

Option 4 matches this.

Comments

Row1 tells e is left identity, to be an identity element it should be right as well as left identity.

in first column we see a*e = e, this means e is right identity. therefore e is identity.  and identity element is unique hence e is the only identity.

This gives - 

*	e	a	b	c
e	e	a	b	c
a	a	b	c	e
b	b
c	c

Property of Cayley Table:-  it can not contain any element twice in any row or column. (see here)

using these two properties we can fill every entry above.

---

How did you conclude that it is a Cayley table?

---

Hello mojo-jojo

I'm removing Best Tag from your answer. An answer should be an explanation. exam oriented solution doesn't matter.

---

yes by looking at first row we can say that e is an identity element because it's already given that table is a 4 Element Group and Group satisfies Identity Property and Identity is Unique.

---

This I think should be the proper way of solving this problem without any facts real to be known.

Answer 3

__ __ __ __

 

Inverse of a is c. So, inverse of c is a.

So, we have

__ e __ __

Only Option D satisfies.

---

Alternatively

 

In some Cayley tables, you'd notice this pattern that the elements are being rotated by 1 position to the left.

First row: $e,a,b,c$

Second row: $a,b,c,e$

Third row: $b,c,e,a$

Fourth row: $c,e,a,b$

Answer 4

It is given that the given set of 4 elements is group.

The element ’e’ is clearly identity as the row corresponding to it has all values same as the other operand.

 

Also, since a*c is e, c*a should also be e which is only the case in option D

Answer 5

Given -> Cayley Table of Group ,

1. In Cayley Table of Group we always  have distinct element in a row and in a column.

2. In this table e is the identity element (as there should be exactly one identity element for Group and only e satisfies that property).

Now the easiest way is to elliminate the options or we can just try the combination of different element which satisfies above 2 points and property of a Group.

Correct Answer (D)

Answer 6

.

Comments

what is Fui Uki Sudoko ?? @Shaikh_hasib

---

fill like sudoku

in a row or column no element should repeat