Given the following groups, what is the maximum possible order for an element for (a) S5 S 5 (b) S6 S 6 (c) S7 S 7 (d) S10 S 10 (e) S15 S 15 My book justifies the answer as (a) The greatest …

There are many ways to find the order of an arbitrary element of Sn S n, it's sometimes useful to write σ σ as a product of disjoint cycles. Say σ = μ1μ2 …μj σ = μ 1 μ 2 … μ j.

Step 1 : If you have a k cycle τ in Sn then, what can you say about the number of elements in the centralizer of τ ? Answer: k ⋅ (n − k)! is the number of elements in the centralizer of a k cycle in …

This is Gallian 5.22. According to the manual the answer is 10 10, but it's not clear what combination of cycles from S10 S 10 would give an lcm l c m of 30 30 (and as such an …

Aug 29, 2013 · The order of an element of Sn S n is the least common multiple of the size of its cycles. So if n n is a prime power, then the only way this is possible is if one cycle is of length …

Mar 15, 2018 · Since #σ(S) = k # σ (S) = k, (1) (1) defines an action of Sn S n on the set Xk X k. The kernel of this action is:

Mar 9, 2020 · An element of Sn S n is a permutation, i.e., a function from a set to itself, and function composition can easily be shown to be associative. For simplicity, take f, g, h f, g, h to …

Example For S10 S 10 we have that the maximal order of an element consists of 3 cycles of length 2,3, and 5 (or so I think) resulting in an element order of lcm(2, 3, 5) = 30 lcm (2, 3, 5) = …

Jul 2, 2022 · The hint as written is somewhat misleading: Notice that all of the cycles in your permutation in S21 S 21 have order 2 2 or 5 5, so the element has order lcm(2, 5) = 10 lcm ⁡ …

I need to find how many elements of order k k are in Sn S n (where k ≤ n k ≤ n). So if k k is prime, it's easy: k k can't be the lcm l c m of any integers besides itself and one's (which we're …