Prove that $A_5$ is simple.
The solution presented is as follows:
Let $N$ be a non trivial normal subgroup of $A_5.$ If $(12)(34)\in N,$ then $$(25)(34)(12)(34)(34)(25) = (15)(34)$$ and $$(12)(34)(15)(34)=(152).$$ Again, if $(12345)\in N,$ then $$(23)(45)(12345)(45)(23) = (13254)$$ and $(12345)(13254) = (142).$ Now we know that, if a normal subgroup of $A_n$ contains even a single $3-$cycle it must be all of $A_n$, it follows that $N=A_5.$
In this solution I don't get, why only they consider these two cases $(12)(34)$ and $(12345),$ are present in $N.$ There are so many more cases possible for example, say if $(13)(24)$ may belong to $N$, then what? Obviously, we can verify, that we again find a three cycle in $A_5$, but I want to know how does the author make it so obvious that, there will be a three- cycle in $A_5$ for every other possible cases by just considering these two examples. As this seems not at all obvious to me. I dont get this at all...
I know there are tons of questions on this site concerning the same topic. But I want an explanation for this particular proof.
Any two permutations with the same cycle structure are conjugate in $S_n$.
So suppose your normal subgroup $N$ contains some permutation that is a product of two disjoint transpositions: $(ab)(cd)$. Then there is a permutation $\sigma\in S_5$ such that $\sigma(ab)(cd)\sigma^{-1}=(12)(34)$. Now consider the subgroup $\sigma N\sigma^{-1}\triangleleft \sigma A_5\sigma^{-1}=A_5$. If $\sigma N\sigma^{-1}=A_5$, then $N=A_5$. So considering the case of $(12)(34)$ covers all cases in which $N$ contains a permutation that is a product of two disjoint transpositions.
Likewise, if $N$ contains a $5$-cycle, then conjugating by a suitable $\sigma\in S_5$ you may assume that $N$ contains $(12345)$. Thus, it suffices to consder that case.