1. ## symmetric group

Why is the symmetric group $G$ of $n$ letters not solvable for $n > 4$?

2. Originally Posted by particlejohn
Why is the symmetric group $G$ of $n$ letters not solvable for $n > 4$?
The group $S_n$ is not solvable for $n\geq 5$. This is an interesting result which is not so easy to prove.

We will restate the conditions of a group being solvable with a criterion. Let $G$ be a finite group. The group $G$ is solvable if and only if $G^{(k)} = \{\text{id}\}$ for some positive integer $k$. The notational meaning of $G^{(k)}$ is the $k$-th derived subgroup. It is defined as follows. An element of $G$ is a commutator iff it has the form $aba^{-1}b^{-1}$. The group generated by the set of all commutators is the commutator subgroup. We denote it by $G'$. We define $G^{(1)} = G', G^{(2)} = (G')', G^{(3)} = ((G')')', ...$. Thus, the equivalent definition of a finite solvable group is a group such that taking the commutator subgroup enough times will produce the trivial subgroup. The other important property of commutator subgroups is that if $N$ is a normal subgroup and $G/N$ is abelian then $G' \subseteq N$.

We can now prove that $S_n$ is not solvable for $n\geq 5$. Our first step will be to show $(A_n)' = A_n$ where $A_n$ is the alternating group. Let $(a,b,c)$ be a $3$-cycle in $A_n$. Then $(a,b,c)=(a,b,d)(a,c,e)(a,d,b)(a,e,c) = (a,b,d)(a,c,e)(a,b,d)^{-1}(a,e,c)^{-1}$. Thus, any $3$-cycle is a commutator. But any element of $A_n$ can be expressed as a product of $3$-cycles (we will prove this in the end). Thus, this shows $(A_n)' = A_n$. Since $A_n$ is a normal subgroup with $S_n/A_n \simeq \mathbb{Z}_2$ which is abelian it means (by above) $(S_n)'\subseteq A_n$. But since $A_n\subseteq S_n \implies A_n = (A_n)'\subseteq (S_n)'$. Thus, $(S_n)' = A_n$. This means $(S_n)^{(k)} = A_n$ for all $k\geq 1$. Thus, $S_n$ is not solvable.

It just remains to show any element in $A_n$ $(n\geq 5)$ is a product of $3$-cycles. Remember if $\sigma \in A_n$ then $\sigma$ is a product of an even # of transpositions i.e. $2$-cycles (by definition). We can therefore pair each transposition with another transposition. The product of two transpositions must take one of three forms: (i) $(a,b)(a,b)$ (ii) $(a,b)(b,c)$ (iii) $(a,b)(c,d)$. In (i) we can write $(a,b,c)(a,b,c)(a,b,c)$. In (ii) we can write $(a,b,c)$. In (iii) we can write $(a,b,c)(b,c,d)$. Thus, we have expressed everything in terms of $3$-cycles.

There is a more elegant proof that $S_n$ is not solvable but it uses a fact which is also takes work proving. That is $A_n$ is a simple group for $n\geq 5$. This time we prove non-solvability directly by definition. Write $\{ \text{id} \} \subset A_n \subset S_n$. This is a composition series because the factors $A_n / \{ \text{id} \} \simeq A_n$ and $S_n/A_n \simeq \mathbb{Z}_2$ are simple. It follows by the Jordan Holder theorem (one of my favorite theorems) that this composition series is unique. Since $A_n$ is not an abelian group it means it is impossible to get a composition series with abelian groups. Thus, $S_n$ is not solvable.

(Do you see where we use the fact $n\geq 5$?)

3. Originally Posted by ThePerfectHacker
The group $S_n$ is not solvable for $n\geq 5$. This is an interesting result which is not so easy to prove.

We will restate the conditions of a group being solvable with a criterion. Let $G$ be a finite group. The group $G$ is solvable if and only if $G^{(k)} = \{\text{id}\}$ for some positive integer $k$. The notational meaning of $G^{(k)}$ is the $k$-th derived subgroup. It is defined as follows. An element of $G$ is a commutator iff it has the form $aba^{-1}b^{-1}$. The group generated by the set of all commutators is the commutator subgroup. We denote it by $G'$. We define $G^{(1)} = G', G^{(2)} = (G')', G^{(3)} = ((G')')', ...$. Thus, the equivalent definition of a finite solvable group is a group such that taking the commutator subgroup enough times will produce the trivial subgroup. The other important property of commutator subgroups is that if $N$ is a normal subgroup and $G/N$ is abelian then $G' \subseteq N$.

We can now prove that $S_n$ is not solvable for $n\geq 5$. Our first step will be to show $(A_n)' = A_n$ where $A_n$ is the alternating group. Let $(a,b,c)$ be a $3$-cycle in $A_n$. Then $(a,b,c)=(a,b,d)(a,c,e)(a,d,b)(a,e,c) = (a,b,d)(a,c,e)(a,b,d)^{-1}(a,e,c)^{-1}$. Thus, any $3$-cycle is a commutator. But any element of $A_n$ can be expressed as a product of $3$-cycles (we will prove this in the end). Thus, this shows $(A_n)' = A_n$. Since $A_n$ is a normal subgroup with $S_n/A_n \simeq \mathbb{Z}_2$ which is abelian it means (by above) $(S_n)'\subseteq A_n$. But since $A_n\subseteq S_n \implies A_n = (A_n)'\subseteq (S_n)'$. Thus, $(S_n)' = A_n$. This means $(S_n)^{(k)} = A_n$ for all $k\geq 1$. Thus, $S_n$ is not solvable.

It just remains to show any element in $A_n$ $(n\geq 5)$ is a product of $3$-cycles. Remember if $\sigma \in A_n$ then $\sigma$ is a product of an even # of transpositions i.e. $2$-cycles (by definition). We can therefore pair each transposition with another transposition. The product of two transpositions must take one of three forms: (i) $(a,b)(a,b)$ (ii) $(a,b)(b,c)$ (iii) $(a,b)(c,d)$. In (i) we can write $(a,b,c)(a,b,c)(a,b,c)$. In (ii) we can write $(a,b,c)$. In (iii) we can write $(a,b,c)(b,c,d)$. Thus, we have expressed everything in terms of $3$-cycles.

There is a more elegant proof that $S_n$ is not solvable but it uses a fact which is also takes work proving. That is $A_n$ is a simple group for $n\geq 5$. This time we prove non-solvability directly by definition. Write $\{ \text{id} \} \subset A_n \subset S_n$. This is a composition series because the factors $A_n / \{ \text{id} \} \simeq A_n$ and $S_n/A_n \simeq \mathbb{Z}_2$ are simple. It follows by the Jordan Holder theorem (one of my favorite theorems) that this composition series is unique. Since $A_n$ is not an abelian group it means it is impossible to get a composition series with abelian groups. Thus, $S_n$ is not solvable.

(Do you see where we use the fact $n\geq 5$?)

But its not solvable by radicals for $n >4$?

4. Originally Posted by particlejohn
And its not solvable by radicals for $n >4$?
A polynomial in Q[x] is solvable by radicals if and only if its Galois group is a solvable group.

If the degree of f(x) is at most four then its Galois group is embedded in $S_4$ which is solvable. However if f(x) has degree 5 then for it to be unsolvable we required that its Galois group is $S_5$. We need to actually construct such a polynomial (because just because it has degree 5 does not mean it automatically is unsolvable).

5. Originally Posted by ThePerfectHacker
The group $S_n$ is not solvable for $n\geq 5$.
There is a more elegant proof that $S_n$ is not solvable but it uses a fact which is also takes work proving. That is $A_n$ is a simple group for $n\geq 5$. This time we prove non-solvability directly by definition. Write $\{ \text{id} \} \subset A_n \subset S_n$. This is a composition series because the factors $A_n / \{ \text{id} \} \simeq A_n$ and $S_n/A_n \simeq \mathbb{Z}_2$ are simple. It follows by the Jordan Holder theorem (one of my favorite theorems) that this composition series is unique. Since $A_n$ is not an abelian group it means it is impossible to get a composition series with abelian groups. Thus, $S_n$ is not solvable.
you don't need Jordan-Holder theorem: since $A'_n$ is a normal subgroup of $A_n,$ and $A_n$ is a simple non-abelian group for n > 4,

we get: $A'_n=A_n.$ since $S_n/A_n,$ is abelian, we have $S_n ' \subseteq A_n,$ which again gives us $S_n '=A_n.$ hence $S_n^{(k)}=A_n \neq \{1\}, \ \forall k,$

i.e. $S_n$ is not solvable.

6. Originally Posted by NonCommAlg
you don't need Jordan-Holder theorem: since $A'_n$ is a normal subgroup of $A_n,$ and $A_n$ is a simple non-abelian group for n > 4,

we get: $A'_n=A_n.$ since $S_n/A_n,$ is abelian, we have $S_n ' \subseteq A_n,$ which again gives us $S_n '=A_n.$ hence $S_n^{(k)}=A_n \neq \{1\}, \ \forall k,$

i.e. $S_n$ is not solvable.
I know, I was just presenting two different approaches.

7. yes