symmetric group

**particlejohn** · July 12th 2008, 06:26 PM

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

**ThePerfectHacker** · July 12th 2008, 07:18 PM

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 $n\geq 5$ ?)

**particlejohn** · July 12th 2008, 08:00 PM

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 $n >4$ ?

**ThePerfectHacker** · July 12th 2008, 08:04 PM

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).

**NonCommAlg** · July 12th 2008, 10:48 PM

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.

**ThePerfectHacker** · July 13th 2008, 09:55 AM

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.

**particlejohn** · October 31st 2008, 10:34 PM

yes

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