need help in this permutation proof

**deniselim17** · September 21st 2008, 08:11 PM

Can anyone help me in this proof ?

If A and B are in S_n, then $ABA^{-1}$ is the permutation that has the same cycle structure as B and that is obtained by applying A to the symbols in B.

**wisterville** · September 21st 2008, 10:08 PM

Hello,

Prove it when the length of A is 2 and the B is cyclic:
A=(i j), B=(m_1 m_2... m_k).
Consider several cases depending on whether i, j are included in B.
For example, if i=m_n and j doesn't occur in B,
ABA^{-1}=(m_1 m_2... m_{n-1} j m_{n+1} ... m_k).

Bye.

**ThePerfectHacker** · September 22nd 2008, 10:09 AM

Originally Posted by deniselim17

Can anyone help me in this proof ?

If A and B are in S_n, then $ABA^{-1}$ is the permutation that has the same cycle structure as B and that is obtained by applying A to the symbols in B.

Let $\sigma = (a_1, ... , a_k)$ (where $k\geq 1$ ). Let $\tau$ be a permutation. Prove that $\tau \sigma \tau^{-1} = (\tau(a_1), ... , \tau(a_k))$ . Now let $\theta$ be any permuation. Then we can write $\theta = \sigma_1\sigma_2 ... \sigma_m$ where $\sigma_i$ are disjoint cycles. Therefore, $\tau \theta \tau^{-1} = (\tau \sigma_1 \tau^{-1})(\tau \sigma_2 \tau^{-1}) ... (\tau \sigma_m \tau^{-1})$ . But each $\tau \sigma_i \tau$ has the same structure as $\sigma_i$ by above. Therefore, $\theta$ and $\tau \theta \tau^{-1}$ both have the same number of disjoint cycles in their cyclic decompositition. Therefore $\theta$ and $\tau \theta \tau^{-1}$ have the same structrure.

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