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For the composition table of a cyclic group shown below:

 * a b c d a a b c d b b a d c c c d b a d d c a b

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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Corrected previous table .

Now it is correct table.
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This might help ....

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 can't 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. Then (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 it is not a generator.
Now for c, c*c = b. Then (c*c)*c = b*c = d. Then (c*c*c)*c = d*c = a. Then (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. Then (d*d*d)*d = c*d = a. Then (d*d*d*d)*d = a*d = d. So we have generated all elements of group from d, so d is a generator.
So c and d are generators. So option (C) is correct.

http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2009.html
edited
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can you please tell me how did u get c*c as b...we get d here as per the table given.
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Is this a part of DMGT subject?
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Yes, it is from Group Theory , from Discrete Math subject.
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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.

$c^1 = c, c^2=c^1*c^1=d, c^3=c^2*c^1=d^c=a, c^4=c^3*c^1=a*c\\ d^1 = d, d^2=d^1*d^1=b, d^3=d^2*d^1=b^d=c, d^4=d^3*d^1=c*d=a\\$