Symmetric Group on Greater than 4 Letters is Not Solvable

Theorem

Let $n \in \N$ such that $n > 4$.

Let $S_n$ denote the symmetric group on $n$ letters.

Then $S_n$ is not a solvable group.

Proof

As stated, let $n > 4$ in the below.

Recall the definition of solvable group:

A finite group $G$ is a solvable group if and only if it has a composition series in which each factor is a cyclic group.

A composition series for $G$ is a normal series for $G$ which has no proper refinement.

A normal series for $G$ is a sequence of (normal) subgroups of $G$:

$\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$

where $G_{i - 1} \lhd G_i$ denotes that $G_{i - 1}$ is a proper normal subgroup of $G_i$.

Consider the alternating group on $n$ letters $A_n$.

From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is a proper normal subgroup of $S_n$.

But from Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group, the only normal subgroup of $S_n$ is $A_n$

From Alternating Group is Simple except on 4 Letters, $A_n$ is simple.

That is, $A_n$ has no proper normal subgroup.

It follows that the only composition series for $S_n$ is:

$\set e \lhd A_n \lhd S_n$

From Alternating Group on More than 3 Letters is not Abelian, $A_n$ is not an abelian group.

From Cyclic Group is Abelian, $A_n$ is not cyclic.

Thus we have demonstrated that the only composition series for $S_n$ contains a factor which is not cyclic.

Hence the result.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 84$