I have to show that the group $A_6$ is simple. For the subgroups that have order divisible by $5$ and order of $8,9,18,24,36,$ and $72$ I have shown that those subgroups are not normal. Now we need to show that for the cases of when the order of $H$ is either $2,4,6$ and $12$, that $H$ is also not normal.
We must show H must contain an element of order 2 which in $A_6$ is of the form $\alpha=(ab)(cd)$ Compute $g\alpha g^{-1}$ for $g=(afe)$ So $gag^{-1}=(afe)(ab)(cd)(aef)=(a)(e)(fb)(cd)$.
Then it says argue by repeated conjugation one can generate all pairs of 2-cycles. So by doing $g\alpha g^{-1}$ for $\alpha=(fb)(cd)$. Thus $g\alpha g^{-1}=(afe)(fb)(cd)(aef)=(a)(e)(f)(cd)$? I think this is wrong. But what is the point of the argument here?
Finally argue that any normal subgroup with an order two element must contain all order two elements.
Now for the case when the order of the subgroup is $3$. What two different cycle-structures are possible for elements in $H$?
Two cycle structures are $(abc)$ and $(abc)(def)$.
Show that in either case there will be a conjugate $g\alpha g^{-1}$ such that $g\alpha g^{-1} \in H$. Thus $A_6$ is simple.
This part I dont know where to begin because, but I know that for any product of three cycles we can generate $A_n$ for $n\ge 3$. But I don't think I can use that fact here.
The point is that the cycle-type of an element of order $2$ is $1,1,2,2$. Now I know I can conjugate this element to any other element of the same cycle-type within $S_6$ - because conjugacy classes in $S_n$ are defined by cycle-types for all $n$.
Now suppose I conjugate some permutation $p$ of this type to another permutation $q$. If I am conjugating using an even permutation, I know this is possible also in $A_6$. If the permutation $r$ which conjugates one to the other is odd, I conjugate first by the $2$-cycle $t$ which exchanges the two fixed points, leaving $p$ the same, and then by $r$. Depending on how I am writing my permutations this amounts to conjugating $p$ to $q$ using the permutation $rt$ or $tr$ - either way the permutation is even, and $p$ and $q$ are conjugate within $A_6$.