The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+19 votes
1.2k views

For the composition table of a cyclic group shown below:
 

* a b c d
a a b c d
b b a d c
c c d b a
d d c a b


Which one of the following choices is correct?

  1. $a,b$ are generators
  2. $b,c$ are generators
  3. $c,d$ are generators
  4. $d,a$ are generators
asked in Set Theory & Algebra by Boss (18.1k points)
retagged by | 1.2k views
+6

Please correct the table

+1
Corrected previous table .

Now it is correct table.
+1

This might help ....

3 Answers

+22 votes
Best answer

An element is a generator for a cyclic group if on repeated applications of it upon itself, it can generate all elements of group.
For example here: $a*a = a,$ then $(a*a)*a = a*a = a,$ and so on. Here, we see that no matter how many times we apply $a$ on itself, we cannot generate any other element except $a.$ So, $a$ is not a generator.

Now for $b, b*b = a.$ Then, $(b*b)*b = a*b = b, (b*b*b)*b = b*b = a,$ and so on. Here again, we see that we can only generate $a$ and $b$ on repeated application of $b$ on itself. So, $b$ is not a generator.

Now for $c, c*c = b.$ Then, $(c*c)*c = b*c = d, (c*c*c)*c = d*c = a, (c*c*c*c)*c = a*c = c.$ So, we see that we have generated all elements of group. So, $c$ is a generator.

For $d, d*d = b.$ Then $(d*d)*d = b*d = c, (d*d*d)*d = c*d = a, (d*d*d*d)*d = a*d = d.$ So, we have generated all elements of group from $d.$ So, $d$ is a generator.

$c$ and $d$ are generators. Option (C) is correct.

http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2009.html

answered by Loyal (5.9k points)
edited by
0
can you please tell me how did u get c*c as b...we get d here as per the table given.
0
Is this a part of DMGT subject?
+2
Yes, it is from Group Theory , from Discrete Math subject.
0
I have doubt about cyclic group and  generator.After reading this answer my doubt is clear.Nice explanation!
+1
i want to add an additional information here that,,,if a group is cyclic then it will be abelian group but vice-versa need not to be true
0
When we have found c as a generator, note that in a cyclic group if g is a generator so $g^{-1}$ is also a generator.

Since a is the identity element of the group, $c^{-1}$ is d, so d is also a generator.
+10 votes
a * a = a. So a can not generate anything other than a. In fact a is identity so a can not be generator.

So this eliminates Option A & D.

b*b = a. Also a is just identity, so b*a=a*b = a. So b can not generate all elements . So Option B is out.

So Answer is C.
answered by Boss (42.6k points)
+5 votes

Answer is C.

c^1 = c, c^2=c^1*c^1=d, c^3=c^2*c^1=d^c=a, c^4=c^3*c^1=a*c\\ d^1 = d, d^2=d^1*d^1=b, d^3=d^2*d^1=b^d=c, d^4=d^3*d^1=c*d=a\\

answered by Active (4.2k points)


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

38,174 questions
45,676 answers
132,605 comments
49,562 users