Results 1 to 7 of 7

Math Help - symmetric group

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    170

    symmetric group

    Why is the symmetric group  G of n letters not solvable for  n > 4 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by particlejohn View Post
    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?)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    170
    Quote Originally Posted by ThePerfectHacker View Post
    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 ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by particlejohn View Post
    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).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by ThePerfectHacker View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by NonCommAlg View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jun 2008
    Posts
    170
    yes
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Symmetric Group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 16th 2010, 01:09 AM
  2. symmetric group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 25th 2009, 08:04 AM
  3. Symmetric group
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 25th 2009, 01:05 PM
  4. Symmetric group of P
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 19th 2008, 06:40 AM
  5. Symmetric Group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 22nd 2008, 04:29 PM

Search Tags


/mathhelpforum @mathhelpforum