Time management is very much important in IIT JAM. The eduncle test series for IIT JAM Mathematical Statistics helped me a lot in this portion. I am very thankful to the test series I bought from eduncle.
Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Anonymous User
(c): Every permutation can split to disjoint cycles and its order is their orders lcm. So it suffices to find such. For n⩽4 and n=6,there is no such mm. But for n=5 and n⩾7 you have always some. You can split all nn elements into two subset of order [n/2],[(n+1)/2] (for odd n) and [n/2],[(n−1)/2](for even n) and then product their cycles that have order n(n±1)/4 which always satisfies the criteria. (d): Since in group, the equation gx=h, for every g,h, has always a unique solution (x), so every element permutes all group elements (including itself), so they behave like permutations on n things (actually themselves). Therefore we can see them as permutation functions, g(x)=gx. Now since h(g(x))=(hg)(x), they respect to group multiplication. Hence group (G), is isomorphic to a subgroup of Sn. (a) is true as I can choose an element of length m which will generate cyclic subgroup of order m (b) Choose n=4and m=19 to contradict (b).