vsage
I was given this problem to work out but I'm still a little bad when it comes to proofs, but here's the question. I have given it a little thought but I can't seem to prove what I feel is the correct answer without brute forcing the answer in such an ugly way.
Let G = S_7, where S_7 is the group of permutations of the cyclic group (1, 2, 3, 4, 5, 6, 7) (for example (7, 6, 5, 4, 3, 2, 1)). Determine the centralizer C_G(\sigma) where \sigma is the cycle (1, 2, 3, 4, 5, 6, 7), where the centralizer by definition is any element C of the given group such that C\sigma = \sigma C over the given operation. Also, prove that your answer is correct.
Part 1 wasn't too bad: C_G(\sigma) = k\sigma (mod 7), or (k, 2k, 3k, 4k, 5k, 6k, 7k)(mod 7) where k is any integer not divisible by 7, and 0 is taken to be equivalent to 7. Obviously any integer k = 7p + r would produce the same results as k = r for integer p, so I only have to deal with k = 1-6. I hope that made sense: I'm not sure my first way of writing what the centralizer is was correct. However, I'm having trouble producing a proof that I think is acceptable. I think I could easily show that each k satisfies commutativity but it seems so brute-forced. Is there a more elegant solution I can employ? Thanks
Let G = S_7, where S_7 is the group of permutations of the cyclic group (1, 2, 3, 4, 5, 6, 7) (for example (7, 6, 5, 4, 3, 2, 1)). Determine the centralizer C_G(\sigma) where \sigma is the cycle (1, 2, 3, 4, 5, 6, 7), where the centralizer by definition is any element C of the given group such that C\sigma = \sigma C over the given operation. Also, prove that your answer is correct.
Part 1 wasn't too bad: C_G(\sigma) = k\sigma (mod 7), or (k, 2k, 3k, 4k, 5k, 6k, 7k)(mod 7) where k is any integer not divisible by 7, and 0 is taken to be equivalent to 7. Obviously any integer k = 7p + r would produce the same results as k = r for integer p, so I only have to deal with k = 1-6. I hope that made sense: I'm not sure my first way of writing what the centralizer is was correct. However, I'm having trouble producing a proof that I think is acceptable. I think I could easily show that each k satisfies commutativity but it seems so brute-forced. Is there a more elegant solution I can employ? Thanks
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