# Centralizer problem help

1. May 15, 2006

### JFo

Find $C_{S_9}( \sigma) = \{ \tau \in S_9 | \sigma \tau = \tau \sigma \}$, where $\sigma = (2 \ 7)(3 \ 1 \ 8)$

I know I'm supposed to show my work to get help, but I am completely lost on this one. Any hints would be great.

I have proved that for $\tau \in C_{S_9}( \sigma)$, $\tau(\mbox{Fix} (\sigma)) = \mbox{Fix}(\sigma)$, where $\mbox{Fix}(\sigma) = \{ i | \sigma(i) = i \}$. I have been able to come up with subsets of $C_{S_9}$ but I have not been able to find the entire set.

2. May 15, 2006

### AKG

You want to find the group of $\tau$ such that $\tau \sigma \tau ^{-1} = \sigma$. Observe that:

$$\tau \sigma \tau ^{-1} = \tau (2\, 7)(3\, 1\, 8)\tau ^{-1} = \tau (2\, 7)\tau ^{-1}\tau (3\, 1\, 8)\tau ^{-1}$$

Conjugation preserves cycle structure, so for any $\tau$, the above will result in a product of a transposition and a 3-cycle, and since the transposition and the three-cycle in $\sigma$ are disjoint, you need the conjugate of (2 7) to be (2 7), and the conjugate of (3 1 8) to be (3 1 8). In other words:

$$C_{S_9}(\sigma ) = C_{S_9}((2\, 7))\, \cap \, C_{S_9}((3\, 1\, 8))$$

You should be able to figure it out from here.

3. May 15, 2006

### JFo

Thanks for your reply AKG. I don't understand what you mean when you say you need the conjugate of (2 7) to be (2 7), and the conjugate of (3 1 8) to be (3 1 8). Could you elaborate?

4. May 15, 2006

### AKG

Do you follow the part that says that conjugation preserves cycle structure? It means that if you have any permutation, and expressed as a product of disjoint cycles it has the form, say, xyz where x is a 3-cycle, y is a 4-cycle, and z is a 2-cycle, then any conjugate of this permutation will, when expressed as a product of disjoint cycles, be the disjoint product of a 3-cycle, 4-cycle, and 2-cycle. In particular, any conjugate of (2 7) will be a 2-cycle, and any conjugate of (3 1 8) will be a 3-cycle.

If

$$\tau (2\, 7)\tau ^{-1}\tau (3\, 1\, 8)\tau ^{-1} = (2\, 7)(3\, 1\, 8)$$

then I claim that $\tau (2\, 7)\tau ^{-1} = (2\, 7)$ and likewise for (3 1 8). To prove this, take what I just said (that the conjugate of a 2-cycle is a 2-cycle, and the conjugate of a 3-cycle is a 3-cycle) and prove that if xy = (2 7)(3 1 8) where x is a 2-cycle and y is a 3-cycle, that x = (2 7) and y = (3 1 8). This is easy to prove - start by proving that if x is a 2-cycle, y is a 3-cycle, and xy = (2 7)(3 1 8), that x and y must be disjoint.

5. May 15, 2006

### JFo

Ah I see. Thanks alot for your help!