Tuesday 22 January 2013

Commutator Subgroups

Definition The commutator $[g,h] = g^{-1} h^{-1} g h$ has the following properties $[h,g] = [g,h]^{-1}$, $[g,h] = 1$ iff $gh = hg$.

Proposition If $[x,y] \in G$ then $xG$ and $yG$ commute.
proof: $xGyG = xyG$ by the definition of coset groups. $yxG = yx(x^{-1}y^{-1}xy)G = xyG$ because $G = x^{-1}y^{-1}xyG$ since $x^{-1}y^{-1}xy \in G$.

Lemma If $[g,z] \in G$ then $z^{-1} g z G = g G = G$ then $g^z \in G$.

Definition If $H,K \le G$ their commutator $[H,K] = [K,H] = \langle [h,k] \in G|h \in H, g \in G \rangle$ (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 $[G,G] \text{char} G$
proof: Given $\alpha \in Aut(G)$, $[g,h]^\alpha = [g^\alpha,h^\alpha]$.

Proposition (The commutator subgroup is the smallest normal subgroup $N$ such that $G/N$ is abelian) Given $N \unlhd G$ then $G/N$ is abelian iff $[G,G] \le N$.
proof: (for all $g$,$h$ in $G$) $Ng Nh = Nh Ng$ iff $Ngh = Nhg$ iff $[g^{-1},h^{-1}] \in N$.


Theorem if $A \le B$ are both subgroups of $G$ then $[A,G] \le [B,G]$.
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 $G_i/G_{i-1}$ of a series for $G$ is central iff $[G,G_i] \le G_{i-1}$
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 $a \in H$ we $aH=H=Ha$ so the other coset must be all the other elements so for $b \in H$, $bH=Hb$.

Theorem Any index 2 subgroup contains the commutator subgroup.
proof: If $H$ has index 2 then $G/H \simeq C_2$ is abelian so $H$ contains $G'$.

Corollary The only index 2 subgroup of $S_n$ is $A_n$.

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