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 gaussrelatz July 14th, 2012 05:06 AM

Solvable Groups

Now i know the definition of solvable groups and derived series etc... however im having trouble seeing why $S_3$ and $S_4$ are solvable, yes in the examples it does somehow have the identity element as one of its subgroups hence sufficing to be solvable but i don't see why, $G=S_3 , G^'=A_3, G^{''}=A^{'#39;}_3=1$ the $G^'$=derived series. i just dont see how the symmetric group 3 when derived with the commutator (according to the derivedseries definition) leads to the alternating group of $A_3$. SImilarly the same confusion lies for me in the fact that when $G=S_4$ then $G^'=A_4$, and somehow klein 4-group also appears from the derived series of $S_4$. I really dont get this, can anyone please give me a line to line explanation, i would really appreciate it

 mathbalarka July 14th, 2012 07:38 AM

Re: Solvable Groups

$\text{Start with the definition, } G' = < [x,y] = xyx^{-1}y^{-1} \: \mid \: x,y \in G >$

$\text{Since every commutator is an even permutation, A_n contains [S_n, S_n] (n \geq 3)}$

$\text{We can obviously obtain a 3-cycle as a commutator. [S_n, S_n] contains all the 3-cycles.$

$\text{Now, since 3-cycles generate A_n, [S_n, S_n] contains A_n implies A_n is equivalent to [S_n, S_n]$

$\text{QED}$

 gaussrelatz July 14th, 2012 08:29 AM

Re: Solvable Groups

Thank you for your reply but i don't completely understand your explanation. could you be alittle more clear, sorry im kind of new at this part of group theory.

 Deveno July 17th, 2012 01:35 AM

Re: Solvable Groups

note that:

(a c b) = (a b)(a c):

a-->c-->c
b-->b-->a
c-->a-->b

(here i have applied (a c) first, and (a b) second, as with composition of functions)

note as well that (a c b)^2 = (a b c):

a-->c-->b
b-->a-->c
c-->b-->a

which means that:

(a b c) = (a c b)^2 = (a b)(a c)(a b)(a c).

since (a b)^-1 = (a b), and (a c)^-1 = (a c), we have, letting x = (a b), y = (a c):

(a b c) = xyx^-1y^-1, so every 3-cycle (a b c) is a commutator.

this shows that for n ? 3, An is contained in [Sn,Sn], since the 3-cycles generate An, for n ? 3.

but clearly any commutator is of the form:

(odd)(odd)(odd)(odd), which means we have odd+odd+odd+odd = even number of transpositions, or
(odd)(even)(odd)(even), which means we have odd+even+odd+even = even number of transpositions, or
(even)(even)(even)(even), which means we have even+even=even=even = even number of transpositions, or
(even)(odd)(even)(odd), which means we have even+odd+even+odd = even number of transpostions

so in all possible cases, xyx^-1y^-1 is an even permutation, regardless of whether x,y are even or odd.

this means that [Sn,Sn] is contained in An, from which we conclude that An IS the commutator subgroup of Sn.

(If A is a subset of B, and B is a subset of A, then A = B).

now let's look specifically at [A4,A4], the commutator subgroup of A4. so we're looking at products of the form xyx^-1y^-1 for our basic commutators, where x and y are 3-cycles.

now if y is a power of x, then it commutes with x, so that xyx^-1y^-1 = yxx^-1y^-1 = yey^-1 = yy^-1 = e, which isn't terribly interesting. so we can assume that if x = (a b c), that y permutes a,b,d or a,c,d or b,c,d. thus it suffices to determine:

(a b c)(a b d)(a c b)(a d b)
(a b c)(a c d)(a c b)(a d c)
(a b c)(b c d)(a c b)(b d c)
(a b c)(a d b)(a c b)(a b d)
(a b c)(a d c)(a c b)(a c d)
(a b c)(b d c)(a c b)(b c d)

i'll do the first 3:

(a b c)(a b d)(a c b)(a d b):

a-->d-->d-->a-->b
b-->a-->c-->c-->a
c-->c-->b-->d-->d
d-->b-->a-->b-->c

so (a b c)(a b d)(a c b)(a d b) = (a b)(c d)

(a b c)(a c d)(a c b)(a d c):

a-->d-->d-->a-->b
b-->b-->a-->c-->a
c-->a-->c-->d-->d
d-->c-->b-->b-->c

so (a b c)(a c d)(a c b)(a d c) = (a b)(c d)

(a b c)(b c d)(a c b)(b d c):

a-->a-->c-->d-->d
b-->d-->d-->b-->c
c-->b-->a-->a-->b
d-->c-->b-->c-->a

so (a b c)(b c d)(a c b)(b d c) = (a d)(b c)

after verifying the other 3 products you see that we get one of (a b)(c d), (a c)(b d) or (a d)(b c) in every case.

that is, if y is not in <x>, then xyx^-1y^-1 is a double 2-cycle, and every such double 2-cycles can be realized as a commutator. thus:

[A4,A4] = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.

this is a non-cyclic group of order 4 (since all the non-identity elements are of order 2), and is isomorphic to the klein 4-group.

an alternative definiton of solvable is:

if there exist subgroups:G = H1,H2,...Hk = {e} such that H(j+1) is normal in Hj, and Hj/H(j+1) is abelian.

clearly A3 is normal in S3, and S3/A3 has order 2, and {e} is normal in A3, and A3/{e} = A3 is abelian (cyclic, even). so we have the solvable subgroup series S3,A3,{e}.

clearly A4 is normal in S4, and S4/A4 has order 2 (so is abelian). showing V = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is normal in A4 is left to you (hint: why does the fact that these are all conjugate in A4 make this easy?). note that A4/V has order 3, so is necessarily abelian.

finally, V itself is abelian, so we can take V/{e}, to get the composition series: S4,A4,V,{e}.

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