I am working on the following problem:
Consider the symmetric group $S_n$ with $n\ge 5$. Prove the following:
- Show that any $3$-cycle is a commutator.
- Let $G$ be a subgroup of $S_n$ and let $H$ be a normal subgroup of $G$ such that $G/H$ is abelian. If $G$ contains all $3$-cycles then show that $H$ contains all $3$-cycles.
- Deduce that $S_n$ is not solvable.
Now this is fairly simple using facts about symmetric groups, alternative groups, and the commutator subgroup. For instance, here is what I wrote down:
- Let $(a~b~c)$ be a $3$-cycle in $S_n$. We have $(a~b~c) = (a~b)(a~c)(a~b)(a~c) = [(a~b),(a~c)]$ to be a commutator.
- Note that $G/H$ is abelian if and only if $[G,G]\subseteq H$. Recall that $A_n$ is generated by $3$-cycles for $n\ge 5$, so if $G$ contains all $3$-cycles, then $G$ is either $A_n$ or $S_n$. If $G = S_n$, then $[G,G] = A_n$, hence $H$ is either $A_n$ or $S_n$, and either way it contains all $3$-cycles. If $G = A_n$, then we know $A_n$ is non-abelian and simple for $n\ge 5$, therefore the commutator is non-trivial, but $[G,G]\lhd G$, so $G = [G,G] = H$, thus $H$ contains all $3$-cycles as well.
- Suppose $S_n$ is solvable for $n\ge 5$, then $A_n\lhd S_n$ is also solvable, but any non-abelian simple group is not solvable, contradiction.
However, I feel like my solution is overkill to the problem, and it does not give a real connection between different parts of the problem. Is there a better and more natural solution to this question?
We can show that $S_n$ is not solvable for $n\ge 5$ without knowing that $A_n$ is a simple, non-abelian group for $n\ge 5$. It is enough to know that $A_n$ is a perfect subgroup of $S_n$, i.e., a subgroup satisfying $[A_n,A_n]=A_n$. Hence $A_n$ is not solvable by definition (using the derived series). Since subgroups of a solvable group are solvable, $S_n$ cannot be solvable, either, for $n\ge 5$.
For an elementary proof of $[A_n,A_n]=A_n$ for $n\ge 5$ see here (using $3$-cycles):
How do I find the commutator subgroup of $A_n$ for $n \ge 5$?