The thing I don't understand about the proof is that we are using the fact that $A_5$ contains all its three cycles. I can't see why this doesn't prove $A_4$ is simple. How can we use this argument when it should also work for $A_4$?
Also, why can we just break down $A_5$ into three types?
- $2^2$
- $1^2 3^1$
- $5^1$
Surely you have to include $4^1 1^1$ and $3^1 2^1$ too. However, in written notes it doesn't do this.
Also, does $A_4$ contain all three cycles? Maybe I'm incorrectly assuming this.
I'm trying to understand this proof. It's written notes that I have scanned.