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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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Please correct the table

9
Corrected previous table .

Now it is correct table.
2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Here the identity element is $'a '$

------------------------------------------------------------------------

Note

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 .

 

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

6
Let you find a generator "x" for given group then x^m  will be also a generator but condition is GCD(m,n) =1 , where n is order of group (means total number of element in the group).

Also note that inverse of "x" will also be a generator of group.
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7 Answers

37 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

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7 Comments

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?
0
Yes, it is from Group Theory , from Discrete Math subject.
3
I have doubt about cyclic group and  generator.After reading this answer my doubt is clear.Nice explanation!
0
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
5
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.
9
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
0
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.
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\\

6 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

2 votes

Let G be a group and let $a\in G$

The " subgroup generated by a " is the subgroup $\{a^n|n\in Z\}$

$\{...(a^{-1})^2,a^{-1},e,a^1,a^2,a^3...\}=<a>$

Now,

A group is called cyclic if there exists an element $a\in G$ s.t  $<a>=G$ i.e

$G= \{...(a^{-1})^2,a^{-1},e,a^1,a^2,a^3...\}$


$<c>=\{e,c^1,c^2,c^3,c^4...\}=\{e,c,b,d,a\}$

$<d>=\{e,d^1,d^2,d^3,d^4...\}=\{e,d,b,c,a\}$

 

Correct answer is (C)

0 votes

a*a=a, a*b=b, a*c=c, a*d=d
Clearly, from this first row of the table, a is an identity element, and identity element can't be a generator

Eg: 
Suppose we have set-> {1,2,3} and operation is multiplication modulo 4, so here identity element is 1, now see:


(1^1)mod 4=(1) mod 4=1
(1^2)mod 4=(1*1) mod 4=1
(1^3)mod 4=(1*1*1) mod 4=1


So, identity element keep on repeating the same number again & again, hence it can't be a generator.

So, options A and D are ruled out.

Now, from option B and C, c is common in both, so definitely, c is a generator, and now go to that row which contains c, there, c*d=a, which means, d is inverse of c, or c and d are inverses of each other and there is a property that, if an element is a generator then its inverse is also a generator, hence d is also a generator.
 

So, we are left with option C only.

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0 votes

In a group each element has a unique inverse , also inverse of identity element is itself. Now we observe from table that is the identity element . How ? Find an element x such that x when operated with any element , gives that element itself (Here first row/column remains as it is after operation with a ).

Order of group is 4 , apart from identity element 3 elements are left. Now since remaining elements are odd in number , we can infer that one such element exists (apart from identity) such that the inverse of the element is that element itself. Why ? Since each element has unique inverse so elements and their respective inverses come in pair and if we form pairs(of the from { element,inverse(element) } ) then one element will be left alone and that element’s inverse is the element itself. Now check from table which element x when operated with itself gives identity element i.e find a x such that x * = a. We find b is that element , so now the remaining 2 elements are inverse of each other i.e. c * a . Also if an element is a generator its inverse will also be a generator so option C is correct.

Answer:

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