Please correct the table

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?

- $a,b$ are generators
- $b,c$ are generators
- $c,d$ are generators
- $d,a$ are generators

4

**^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^**

**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 .**

**^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^**

35 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

0

I have doubt about cyclic group and generator.After reading this answer my doubt is clear.Nice explanation!

5

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

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.

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.

3 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->

(1^1)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.