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 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 $S_3$ 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
 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 $\text{Start with the definition, } G' = < [x,y] = xyx^{-1}y^{-1} \: \mid \: x,y \in G >$ $\text{Since every commutator is an even permutation, A_n contains [S_n, S_n] (n \geq 3)}$ $\text{We can obviously obtain a 3-cycle as a commutator. [S_n, S_n] contains all the 3-cycles.$ $\text{Now, since 3-cycles generate A_n, [S_n, S_n] contains A_n implies A_n is equivalent to [S_n, S_n]$ $\text{QED}$
 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 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 , 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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# commutator subgroup of a4 klein

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