1.9k views

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

1. $c\;a\;e\; b$
2. $c\; b\; a\; e$
3. $c\; b\; e\; a$
4. $c\; e\; a\; b$

edited | 1.9k views
+10
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

$a^{1}=e$

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

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

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

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.

by Boss (23.9k points)
edited by
+6
O(g) <= 5  will always be abelian therfore symmetric property can be applied in cayley table
0

@Rajesh Pradhan can we say that every group having some power on prime (say pk, where p is prime and k is some power k>1) is Abelian?

0

@meghna

Did you got any reference for this ??that if order is p^k ...where p is prime...then group is abelian ??

+15
If we don't know the property that "A Group whose order is the square of a prime is abelian group" Then also we can fill the entire table based on the given info.

Here , group is given . So, It will satisfy closure, associativity,identity, inverse properties of a group.

So, Here it is given ,

$1) \;e*e = e$ ,$e*a = a ,\; e*b=b\; , e*c=c$ [Left Identity property]

So,

$2)\;e*e\;,a*e = a ,\; b*e=b\; , c*e=c$    [Right Identity property]

$3)\; a*e = a\; , \; a*a=b,\; a*b=c,\;a*c=e$

Now, based on the given info,

$1)\; b*a = (a*a)*a = a*(a*a) = a*b = c$

$2)\; b*b = (a*a)*b =a*(a*b) = a*c = e$

$3)\; b*c = (a*a)*c = a*(a*c) = a*e = a$

$4)\; c*a= (a*b)*a = (a*(a*a))*a = (a*a)*(a*a) = b*b = e$

$5) \;c*b = (a*b)*b = (b*a)*b =b*(a*b) = b*c = a$

$6)\;c*c=(a*b)*(a*b)=a*(b*a)*b =a*(a*b)*b = (a*a)*(b*b) =b*e=b$
+1

great brother @ankitgupta.1729

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.
by Active (3.8k points)
+15

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.

0
How did you conclude that it is a Cayley table?
+1
Hello mojo-jojo

I'm removing Best Tag from your answer. An answer should be an explanation. exam oriented solution doesn't matter.
0

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.