Let $\sigma, \tau \in S_n$. Show that if $\sigma$ is a product of disjoint cycles of lengths $l_1,...,l_r$, then so is $\tau \cdot \sigma \cdot \tau^{-1}\\\\\\$.
Is it correct to do it in this way:
Suppose $\sigma = (a_1...a_{l_1}) \cdot ... \cdot (x_1...x_{l_r})$ where all cycles are disjoint, then
$\tau \cdot \sigma \cdot \tau^{-1} = \tau \cdot (a_1...a_{l_1}) \cdot ... \cdot (x_1...x_{l_r}) \cdot \tau^{-1} = \tau \cdot (a_1...a_{l_1}) \cdot e \cdot ... \cdot e \cdot (x_1...x_{l_r}) \cdot \tau^{-1} = \\\tau \cdot (a_1...a_{l_1}) \cdot \tau^{-1} \cdot \tau \cdot ... \cdot \tau^{-1} \cdot \tau \cdot (x_1...x_{l_r}) \cdot \tau^{-1} =\\ (\tau \cdot (a_1...a_{l_1}) \cdot \tau^{-1}) \cdot \tau \cdot ... \cdot \tau^{-1} \cdot (\tau \cdot (x_1...x_{l_r}) \cdot \tau^{-1}) = \\ (\tau(a_1) ...\tau(a_{l_1})) \cdot ...\cdot(\tau(x_1) ...\tau(x_{l_r}))\\$
$\sigma, \tau \in S_n$ are bijections, therefore cycles $(\tau(a_1) ...\tau(a_{l_1})), ... ,(\tau(x_1) ...\tau(x_{l_r}))$ are disjoint with lengths $l_1, ..., l_r$ respectively, as cycles $(a_1...a_{l_1}), ... ,(x_1...x_{l_r})$ are disjoint.