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Speak With a Friendly Mentor. ###### 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 1 Answer(s) Answer Now 0 Likes 1 Comments 0 Shares Likes Share Comments A 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). Likes(0) Reply(0) T

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