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

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

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
+3
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

This might help ...

1. Group of order Prime Square $(p^2)$ is always abelian. See
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.

edited by
+2
O(g) <= 5  will always be abelian therfore symmetric property can be applied in cayley table
+1

Following theorem says the group is abelian if and only if order of the group is prime.

In this example order is 4 which is not prime .

https://yutsumura.com/a-simple-abelian-group-if-and-only-if-the-order-is-a-prime-number/

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?

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.
+11

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?
0
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.