# Homework Help: Abstract Algebra - Centralizers

1. Feb 24, 2005

### 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

Last edited by a moderator: Feb 24, 2005
2. Feb 24, 2005

### Hurkyl

Staff Emeritus
Nitpick: S7 is the group of permutations on any 7 distinct objects.

It's clear that each k produces an element of the centralizer... that cycle is simply the k-th power of the given cycle!

Frankly, I think the most straightforward approach is to write down an arbitrary permutation:

1->a
2->b
...
7->g

plug into the equation of commutativity, and then solve.

(Though, I haven't tried it)...

3. Feb 24, 2005

### vsage

Helpful as always, thanks! Yeah I meant your definition of $$S_7$$, I was trying to paraphrase the question from my notes and it always comes out half-conceived. I think I completely solved it now though with your suggestions.