Homework Help: 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!