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Vaishnavi Mishra posted an Question
January 05, 2022 • 20:47 pm 30 points
  • IIT JAM
  • Mathematics (MA)

Choose the correct statements for every positive integer m with 1msn.s has eycfic subgroup of order m. for every positive integer n and for every m with n km

Choose the correct statements For every positive integer m with 1mSn.S has eycfic subgroup of order m. For every positive integer n and for every m with n km

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  • Anonymous User best-answer

    (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).

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