Can anyone help me in this proof ?
If A and B are in S_n, then $\displaystyle ABA^{-1}$ is the permutation that has the same cycle structure as B and that is obtained by applying A to the symbols in B.
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.
Let $\displaystyle \sigma = (a_1, ... , a_k)$ (where $\displaystyle k\geq 1$). Let $\displaystyle \tau$ be a permutation. Prove that $\displaystyle \tau \sigma \tau^{-1} = (\tau(a_1), ... , \tau(a_k))$. Now let $\displaystyle \theta$ be any permuation. Then we can write $\displaystyle \theta = \sigma_1\sigma_2 ... \sigma_m$ where $\displaystyle \sigma_i$ are disjoint cycles. Therefore, $\displaystyle \tau \theta \tau^{-1} = (\tau \sigma_1 \tau^{-1})(\tau \sigma_2 \tau^{-1}) ... (\tau \sigma_m \tau^{-1})$. But each $\displaystyle \tau \sigma_i \tau$ has the same structure as $\displaystyle \sigma_i$ by above. Therefore, $\displaystyle \theta$ and $\displaystyle \tau \theta \tau^{-1}$ both have the same number of disjoint cycles in their cyclic decompositition. Therefore $\displaystyle \theta$ and $\displaystyle \tau \theta \tau^{-1}$ have the same structrure.