July 14th, 2012, 05:06 AM  #1 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  Solvable Groups
Now i know the definition of solvable groups and derived series etc... however im having trouble seeing why and 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, the =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 . SImilarly the same confusion lies for me in the fact that when then , and somehow klein 4group also appears from the derived series of . I really dont get this, can anyone please give me a line to line explanation, i would really appreciate it

July 14th, 2012, 07:38 AM  #2 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Solvable Groups 
July 14th, 2012, 08:29 AM  #3 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  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.

July 17th, 2012, 01:35 AM  #4 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  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 3cycle (a b c) is a commutator. this shows that for n ? 3, An is contained in [Sn,Sn], since the 3cycles 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 3cycles. 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 2cycle, and every such double 2cycles 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 noncyclic group of order 4 (since all the nonidentity elements are of order 2), and is isomorphic to the klein 4group. 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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