TIFR GS 2010 Complete Solution
Here is fully solved paper of TIFR GS 2010 with justification of each and every step, hope you'll enjoy it. For any query feel free to write at sushantkala786@gmail.com . PART - 'A' Q.1 Ans. (c). phi(60) = 16 Justification -: We know that in a cyclic group G of order n generated by a. The order of an element a^(k) is n/(n,k)=n iff (n,k)=1. So the number of generators of G are phi(n). Q.2 Ans. (a). (a) False, Consider the direct product Z_3*Z_3*Z_3. (b) True, Any Abelian group of order 14 must-have elements a and b with respective order 2 and 7 ( Cauchy Theorem ). Then a*b generates the group. (c) True, Same logic as in option (b). (d...
Comments
Post a Comment