GATE CSE 2009 | Question: 22

Question

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?

• A. $a,b$ are generators
• B. $b,c$ are generators
• C. $c,d$ are generators
• D. $d,a$ are generators

Comments

Good reads: http://sites.millersville.edu/bikenaga/abstract-algebra-1/cyclic-groups/cyclic-groups.pdf

---

Without checking the Cayley table,we can answer it directly by inspecting the options. Here, we see a is an identity element,so it can't be generator,hence throw option A and D right now. Generators can't exist in isolation,so if g is generator,so is g inverse. We see C and D are inverses of each other ,so it has to right  

---

$\color{Red}\text{Understand the solution:}$

https://youtu.be/83GuHnqT8UA?list=PLIPZ2_p3RNHhXves0XVa8d5O6F4rUi3KR&t=1786

Answer 1

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

Comments

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

Answer 2

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.

Answer 3

$1.$ 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 4

Answer is C.