# Me, myself and conjugate permutations

1. Apr 22, 2005

### Chen

Hi,

Is there a general method, given $$\sigma$$ and $$\rho$$ in Sn, for finding a permutation $$\tau$$ in Sn such that $$\rho = \tau ^{-1} \sigma \tau$$? I know how to do it when $$\sigma$$ and $$\rho$$ are made of a single k-cycle, but what happens when they are more complex?

For example, for:
$$\sigma = (1, 2)(3, 4)$$
$$\rho = (5, 6)(1, 3)$$
In S6.

Thanks,
Chen

Last edited: Apr 22, 2005
2. Apr 22, 2005

### Palindrom

Do you know you've just solved me a problem I've been working on all day? (By reminding me of conjugation).

Now to current business: I don't remember it precisely, but isn't the conjugation lemma suppose to do the trick? Or does it work only with k-circles?

3. Apr 22, 2005

### matt grime

As a hint you could try thinking of change of basis in a vector space where we realize the permutations as a permutation of basis elements. That ought to work.

4. Apr 23, 2005

### Chen

I figured it out, by trial and error. I needed to find $$\tau \in S_6$$ such that:
$$\tau (1, 2)(3, 4) \tau ^{-1} = (5, 6)(1, 3)$$
Which can be written as:
$$\tau (1, 2)\tau ^{-1} \tau (3, 4) \tau ^{-1} = (5, 6)(1, 3)$$
So I assumed that:
$$\tau (1, 2) \tau ^{-1} = (5, 6)$$
$$\tau (3, 4) \tau ^{-1} = (1, 3)$$
Which means that:
$$\tau (1) = 5$$
$$\tau (2) = 6$$
$$\tau (3) = 1$$
$$\tau (4) = 3$$
So I get:
$$\tau = (4, 3, 1)(1, 5)(2, 6)$$

Hopefully though this wasn't a fluke and this method will work all the time. Matt, unfortunately I don't really know what you're talking about... or maybe I know it by a different name. Thanks thought.

5. Apr 23, 2005

### Palindrom

It should work every time, if I remember my first modern algebra course correctly.

Oh, and Hag Sameah :)

6. Apr 24, 2005

### Chen

Thank you very much, Happy Passover.

7. Apr 24, 2005

### Palindrom

Pessah my friend, I'm from Haifa.

8. Apr 24, 2005

### Chen

Yeah, I thought so. I think I know you from ASAT.

9. Apr 24, 2005

### Palindrom

It's a small world after all...