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  1. #1
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    symmetric group

    Why is the symmetric group $\displaystyle G $ of $\displaystyle n $ letters not solvable for $\displaystyle n > 4 $?
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    Quote Originally Posted by particlejohn View Post
    Why is the symmetric group $\displaystyle G $ of $\displaystyle n $ letters not solvable for $\displaystyle n > 4 $?
    The group $\displaystyle S_n$ is not solvable for $\displaystyle n\geq 5$. 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 $\displaystyle G$ be a finite group. The group $\displaystyle G$ is solvable if and only if $\displaystyle G^{(k)} = \{\text{id}\}$ for some positive integer $\displaystyle k$. The notational meaning of $\displaystyle G^{(k)}$ is the $\displaystyle k$-th derived subgroup. It is defined as follows. An element of $\displaystyle G$ is a commutator iff it has the form $\displaystyle aba^{-1}b^{-1}$. The group generated by the set of all commutators is the commutator subgroup. We denote it by $\displaystyle G'$. We define $\displaystyle 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 $\displaystyle N$ is a normal subgroup and $\displaystyle G/N$ is abelian then $\displaystyle G' \subseteq N $.

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

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

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

    (Do you see where we use the fact $\displaystyle n\geq 5$?)
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    Quote Originally Posted by ThePerfectHacker View Post
    The group $\displaystyle S_n$ is not solvable for $\displaystyle n\geq 5$. 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 $\displaystyle G$ be a finite group. The group $\displaystyle G$ is solvable if and only if $\displaystyle G^{(k)} = \{\text{id}\}$ for some positive integer $\displaystyle k$. The notational meaning of $\displaystyle G^{(k)}$ is the $\displaystyle k$-th derived subgroup. It is defined as follows. An element of $\displaystyle G$ is a commutator iff it has the form $\displaystyle aba^{-1}b^{-1}$. The group generated by the set of all commutators is the commutator subgroup. We denote it by $\displaystyle G'$. We define $\displaystyle 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 $\displaystyle N$ is a normal subgroup and $\displaystyle G/N$ is abelian then $\displaystyle G' \subseteq N $.

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

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

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

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

    But its not solvable by radicals for $\displaystyle n >4 $?
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  4. #4
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    Quote Originally Posted by particlejohn View Post
    And its not solvable by radicals for $\displaystyle 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 $\displaystyle S_4$ which is solvable. However if f(x) has degree 5 then for it to be unsolvable we required that its Galois group is $\displaystyle S_5$. We need to actually construct such a polynomial (because just because it has degree 5 does not mean it automatically is unsolvable).
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    Quote Originally Posted by ThePerfectHacker View Post
    The group $\displaystyle S_n$ is not solvable for $\displaystyle n\geq 5$.
    There is a more elegant proof that $\displaystyle S_n$ is not solvable but it uses a fact which is also takes work proving. That is $\displaystyle A_n$ is a simple group for $\displaystyle n\geq 5$. This time we prove non-solvability directly by definition. Write $\displaystyle \{ \text{id} \} \subset A_n \subset S_n$. This is a composition series because the factors $\displaystyle A_n / \{ \text{id} \} \simeq A_n$ and $\displaystyle S_n/A_n \simeq \mathbb{Z}_2$ are simple. It follows by the Jordan Holder theorem (one of my favorite theorems) that this composition series is unique. Since $\displaystyle A_n$ is not an abelian group it means it is impossible to get a composition series with abelian groups. Thus, $\displaystyle S_n$ is not solvable.
    you don't need Jordan-Holder theorem: since $\displaystyle A'_n$ is a normal subgroup of $\displaystyle A_n,$ and $\displaystyle A_n$ is a simple non-abelian group for n > 4,

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

    i.e. $\displaystyle S_n$ is not solvable.
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  6. #6
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    Quote Originally Posted by NonCommAlg View Post
    you don't need Jordan-Holder theorem: since $\displaystyle A'_n$ is a normal subgroup of $\displaystyle A_n,$ and $\displaystyle A_n$ is a simple non-abelian group for n > 4,

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

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