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could anyone advise on the solution to this problem:

For a group $G$ and $g_{i}\in G$, $[g_{1}, g_{2}]= g^{-1}_{1}g^{-1}_{2}g_{1}g_{2}$ is the commutator. Let $[G,G]=<g^{-1}_{i}g^{-1}_{j}g_{i}g_{j}; g_{i},g_{j}\in G>$. Define $G^{(1)}=[G,G], G^{(i+1)} = [G^{(i)},G^{(i)}]$ so we have the derived series $...G^{(3)}\subseteq G^{(2)} \subseteq G^{(1)}$

(a) Compute derived series of $S_{4}$ (b) Prove $[S_{5}, S_{5}]=A_{5}, [A_{5},A_{5}]=A_{5}$

For (a), is there a pattern to observe? It seems the elements of $S^{(i)}_{4}$ gets more and more complicated.

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It would make more sense to do (b) before (a), strange. –  anon Sep 16 '13 at 2:18
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1 Answer

up vote 1 down vote accepted

For (a), you need to know the following facts :

  1. For any normal subgroup $H < G$, $G/H$ is abelian iff $[G,G]\subset H$. Hence, $[S_4,S_4] \subset A_4$
  2. $(123) = (12)^{-1}(132)^{-1}(12)(132)$. Use the fact that all 3-cycles are conjugate in $S_4$ to conclude that $|[S_4,S_4]| > 8$ Hence, $$ [S_4,S_4] = A_4 $$
  3. Let $V = \{1, (12)(34), (14)(23), (13)(24)\}$. By 1, $[A_4,A_4] \subset V$
  4. Let $H = [A_4,A_4]$ and suppose $H\neq V$, then $H$ must have order 2 (since $A_4$ has no other subgroups. Hence, $H\subset Z(A_4)$. But then $Z(A_4) \subset V$, which means $V$ has a non-identity element from the centre. This is not true. Hence, $$ [A_4,A_4] = V $$
  5. $V$ is abelian, so $[V,V] = \{e\}$

For (b), you can follow a similar line of argument, except you will need the fact that $A_5$ is simple.

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