Question

Prove that Sn is not solvable for n>4.

Answer 1

Solution : ASSUME $S_n$ is solvable for $n>4$. Then, we know the theorem,"(i) Every subgroup and every homomorphic image of a solvable group is solvable. (ii) If $N$ is a normal subgroup of a group $G$ such that $N$ and $G/N$ are solvable, then $G$ is solvable". Therefore ,subgroup $A_n$

is solvable. Since $A_n$ is nonabelian, then the commutator subgroup $A'_n \neq \{i\}$ (the

trivial group) because a group G is abelian if and only if $G'=\{e\}$ .

By Theorem "If $G$ is a group, then the commutator subgroup $G'$ is a normal

subgroup of $G$ and $G/G'$ is abelian. If $N$ is a normal subgroup of $G$ , then $G/N$ is

abelian if and only if N contains $G'$ ."

$\therefore A'_n$ is normal in $A_n$ .

By Theorem "The alternating group $A_n$ is simple if and only if $n\neq 4$ . "

$\therefore A_n$ is simple for $n>4$.

So, by the definition of simple group, it must be that $A'_n=A_n$ . But then the chain of derived subgroups $A^{(i)}$ consists only of copies of group $A_n$ and does not terminate at $\{i\}$ , implying that $A_n$ is not solvable, a CONTRADICTION. So the assumption that $S_n$ is solvable is false and, in fact, $S_n$ is not solvable for $n>4$ .