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Tuesday, 22 January 2013

Commutator Subgroups

Definition The commutator [g,h]=g1h1gh has the following properties [h,g]=[g,h]1, [g,h]=1 iff gh=hg.

Proposition If [x,y]G then xG and yG commute.
proof: xGyG=xyG by the definition of coset groups. yxG=yx(x1y1xy)G=xyG because G=x1y1xyG since x1y1xyG.

Lemma If [g,z]G then z1gzG=gG=G then gzG.

Definition If H,KG their commutator [H,K]=[K,H]=[h,k]G|hH,gG (Note! This is the group generated by commutators, not just the set of all commutators).

Definition The commutator subgroup or derived group of G is [G,G].

Proposition [G,G]charG
proof: Given αAut(G), [g,h]α=[gα,hα].

Proposition (The commutator subgroup is the smallest normal subgroup N such that G/N is abelian) Given NG then G/N is abelian iff [G,G]N.
proof: (for all g,h in G) NgNh=NhNg iff Ngh=Nhg iff [g1,h1]N.


Theorem if AB are both subgroups of G then [A,G][B,G].
proof: We show the stronger result that the generators of [A,G] are contained in the generators of [B,G] and this is just immediate from A being a subset of B.

Theorem A factor Gi/Gi1 of a series for G is central iff [G,Gi]Gi1
proof: todo

Theorem Any index 2 subgroup is normal.
proof: Let H have index 2 in G, then there are exactly 2 cosets of it. For aH we aH=H=Ha so the other coset must be all the other elements so for bH, bH=Hb.

Theorem Any index 2 subgroup contains the commutator subgroup.
proof: If H has index 2 then G/HC2 is abelian so H contains G.

Corollary The only index 2 subgroup of Sn is An.

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