If $n\geq 5.$ Prove for all $(ab)(cde)$ of $S_n$ (with different $a,b,c,d,e$) are conjugated.
My proof is the following:
Given $\alpha=(a_1b_1)(c_1d_1e_1)$ y $\beta=(a_2b_2)(c_2d_2e_2)$, then
\begin{eqnarray*} \alpha(a_1)&=&b_1\\ \alpha(b_1)&=&a_1\\ \alpha(c_1)&=&d_1\\ \alpha(d_1)&=&e_1\\ \alpha(e_1)&=&c_1\\ \end{eqnarray*} and \begin{eqnarray*} \beta(a_2)&=&b_2\\ \beta(b_2)&=&a_2\\ \beta(c_2)&=&d_2\\ \beta(d_2)&=&e_2\\ \beta(e_2)&=&c_2\\ \end{eqnarray*} Given two cycles $(a_1,b_1,c_1,d_1,e_1)$ y $(a_2,b_2,c_2,d_2,e_2)$, consider the following permutation $\sigma:$ \begin{eqnarray*} \sigma(a_1)&=&a_2\\ \sigma(b_1)&=&b_2\\ \sigma(c_1)&=&c_2\\ \sigma(d_1)&=&d_2\\ \sigma(e_1)&=&e_2\\ \end{eqnarray*} then \begin{eqnarray*} \sigma\alpha\sigma^{-1}(a_2)&=&\sigma \alpha(a_1)=\sigma(b_1)=b_2,\\ \sigma\alpha\sigma^{-1}(b_2)&=&\sigma \alpha(b_1)=\sigma(a_1)=a_2,\\ \sigma\alpha\sigma^{-1}(c_2)&=&\sigma \alpha(c_1)=\sigma(d_1)=d_2,\\ \sigma\alpha\sigma^{-1}(d_2)&=&\sigma \alpha(d_1)=\sigma(e_1)=e_2,\\ \sigma\alpha\sigma^{-1}(e_2)&=&\sigma \alpha(e_1)=\sigma(c_1)=c_2. \end{eqnarray*}
What do you think about this proof?
