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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
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9 Answers

Best answer
45 votes
45 votes

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

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15 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.
9 votes
9 votes

$1.$ Let $(G,*)$ be a multiplicative group and $e$ is identity.  The smallest postive integer $m$ such that $x^m=e$ is called order of x

$2.$ An element $x\in G$ such that $O(x)=O(G)$ is called generator of group $(G,*)$

$3.$ A group having atleast one generator is called Cyclic Group

$4.$ The number of generators in a cyclic group of order $n=\phi(n)$

$5.$ $\phi(n)=$ No. of positive integers less than n and relatively prime to n.

$6.$ $\phi(n)=n(1-\frac{1}{p1})(1-\frac{1}{p2}).....(1-\frac{1}{pk})$ where $p1,p2...pk$ are distinct prime factors of n.


$G=\{\{a,b,c,d\},*\}$

$\phi(4)=4(1-\frac{1}{2})=2$ $\{1,3\}$

$a^2=a|$ $O(a)\neq O(G)$

$b^2=a|$ $O(b)\neq O(G)$

$c^2=b|c^3=c^2*c=b*c=d|c^4=c^2*c^2=b*b=a|$ $O(c)= O(G)$

Now, either you can continue finding other generators using brute force or after finding one generator you can easily find other generators by putting that generator to the power of $\{1,3\}$  for this question.

$c^1=c$ (Which we already find out as one of the generator)

$c^3=c^2*c=b*c=d$ (We have found another generator which is d)

Correct Answer: C

Answer:

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