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+27 votes

For the composition table of a cyclic group shown below:
$$\begin{array}{|c|c|c|c|c|} \hline \textbf{*} & \textbf{a}& \textbf{b} &\textbf{c} & \textbf{d}\\\hline \textbf{a} & \text{a}& \text{b} & \text{c} & \text{d} \\\hline \textbf{b} & \text{b}& \text{a} & \text{d} &\text{c}\\\hline \textbf{c} & \text{c}& \text{d} & \text{b} & \text{a}\\\hline \textbf{d} & \text{d}& \text{c} & \text{a} & \text{b} \\\hline \end{array}$$
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
in Set Theory & Algebra by Boss (17.5k points)
edited by | 2.2k views

Please correct the table

Corrected previous table .

Now it is correct table.

This might help ....



Here the identity element is $'a '$



If x is a generator the inverse(x) will be generator


So the generators of the cyclic group will give the identity element when we do operation $'* '$ on them 

Here,     $c*d=d*c=a$ 

So $c,d$ are generators .



4 Answers

+34 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.

by Loyal (5.8k points)
edited by
can you please tell me how did u get c*c as b...we get d here as per the table given.
Is this a part of DMGT subject?
Yes, it is from Group Theory , from Discrete Math subject.
I have doubt about cyclic group and  generator.After reading this answer my doubt is clear.Nice explanation!
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
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.
Proof that b is not a generator:

if b's power is even:   b^(2n) = b^2 * b^2 * ........ n times = a * a * ........ n times  = a

if b's power is odd:   b^(2n+1) = b^(2n) * b= a * b = b

Hence b can only generate a and b
+14 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.
by Boss (41.9k points)
+7 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\\

by Active (4.2k points)
0 votes
correct answer is D. i.e c,d
by (41 points)

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