I have to prove the simplicity of $A_n$ for a homework but I can't understand one of the steps that I have to follow. It says:
Show that if $N$ is a normal subgroup of ${ A }_{ n }$ and it has an element of type $({ m }_{ 1 }, { m }_{ 2 }, \dots ,{ m }_{ r })$ with ${ m }_{ 1 } = 2$, then $N$ = ${ A }_{ n }$.
Help: Let $\sigma $ be an element of $N$ with that type. Show that ${ m }_{ r-1 }$=2. Let $\sigma$ = $({ a }_{ 1 } { a }_{ 2 })({ a }_{ 3 }, { a }_{ 4 }) \tau $ where $({ a }_{ 1 } { a }_{ 2 })$ and $({ a }_{ 3 }, { a }_{ 4 })$ are disjoints and $\tau$ is product of disjoint cycles of $({ a }_{ 1 } { a }_{ 2 })$ and $({ a }_{ 3 }, { a }_{ 4 })$. Let $\delta$ = $({ a }_{ 1 } { a }_{ 2 } {a}_{3})$.
Show that $({ a }_{ 1 } { a }_{ 3 })( {a}_{2} {a}_{4}) = \sigma ^{ -1 }\delta \sigma \delta ^{ -1 }$.
I don't understand how to work with $\tau$.
Thank you.
Working with $\tau$ is straightforward....It's a product of cycles none of which involve $a_1,a_2,a_3,a_4$. When computing $\sigma^{-1}\delta\sigma\delta^{-1}$, $\delta^{-1}$ leaves any such element $b$ fixed, $\sigma$ moves it according to $\tau$, $\delta$ leaves it fixed again since it wasn't moved to $a_1,a_2,a_3,a_4$, and $\sigma^{-1}$ moves it according to $\tau^{-1}$. Thus, for any such $b$, $\sigma^{-1}\delta\sigma\delta^{-1}b=\tau^{-1}\tau b=b$, which is an important part of showing that $\sigma^{-1}\delta\sigma\delta^{-1}=(a_1~a_3)(a_2~a_4)$.
However, the first claim in the help isn't quite true. For example, (12)(345678) is in $A_8$ and has a cycle of length 2 but not a second cycle of length 2. What is true is that some power of $\sigma$ will have more than one cycle of length 2. Why? Because there must be some other cycle of even length or it would not be in $A_n$. If that has length $2$, you are done. If it has length $2m$ for $m>1$, then $\sigma^m$ has at least $m$ 2-cycles, so just substitute $\sigma^m$ for $\sigma$ and proceed.