Page 116 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
P. 116
3.5 Nilpotent groups and p -groups
gap> F:=FreeGroup( "x", "y", "z" );;
gap> AssignGeneratorVariables( F );;
gap> Comm( x * y, z ) = Comm( x, z )^y * Comm( y, z );
true
Let X , Y ⊂ G be non-empty sets. Define the commutator subgroup of X and Y by
[X , Y ] = 〈[x , y ] | x ∈ X , y ∈ Y 〉 and note that [X , Y ] = [Y, X ]. For any n 2 nonempty
subsets X1, X2, . . . , Xn of G denote
[X1, X2, . . . , Xn ] = [[X1, . . . , Xn−1], Xn ].
Note that [G ,G ] = G is just the derived subgroup of G . Define also X Y = 〈x y | x ∈ X , y ∈
Y 〉. If X is a subset and H G , then X ⊂ X H 〈X , H 〉. Thus, X H = X 〈X,H〉 is the normal
closure of X in 〈X , H 〉.
Here is an example:
gap> G := SmallGroup( 64, 52);;
gap> gen := GeneratorsOfGroup(G);;
gap> H := Subgroup(G, [gen[1]]);;
gap> K := Subgroup(G, [gen[2], gen[3]]);;
gap> C := CommutatorSubgroup(H,K);;
gap> Order(H);
2
gap> Order(K);
32
gap> Order(C);
16
Lemma 3.5.4 Let X ⊂ G and H G . Then
1. X K = 〈X , [X , K ]〉;
2. [X , K ]K = [X , K ];
3. if K = 〈Y 〉, then [X , K ] = [X , Y ]K .
PROOF. (1) Follows from x k = x [x , k ].
(2) For k , h ∈ K and x ∈ X we have [x , hk ] = [x , k ][x , h]k , so that [x , h]k ∈ [X , K ].
(3) It suffices to show that [x , k ] ∈ [X , Y ]K what we prove for k = y1±1y2±1 . . . yr±1 by
induction on r . For r = 1 we get [x , y1−1] = ([x , y1]y1−1 )−1 ∈ [X , Y ]K . For the inductive step
we write k = k yr±1. Then [x , k ] = [x , k yr±1] = [x , yr±1][x , k ]yr±1 ∈ [X , Y ]K by induction.
Corollary 3.5.5 If H = 〈X 〉 and K = 〈Y 〉, then [H , K ] = [X , Y ]HK .
PROOF. This follows from Lemma 3.5.4, (3).
gap> F:=FreeGroup( "x", "y", "z" );;
gap> AssignGeneratorVariables( F );;
gap> Comm( x * y, z ) = Comm( x, z )^y * Comm( y, z );
true
Let X , Y ⊂ G be non-empty sets. Define the commutator subgroup of X and Y by
[X , Y ] = 〈[x , y ] | x ∈ X , y ∈ Y 〉 and note that [X , Y ] = [Y, X ]. For any n 2 nonempty
subsets X1, X2, . . . , Xn of G denote
[X1, X2, . . . , Xn ] = [[X1, . . . , Xn−1], Xn ].
Note that [G ,G ] = G is just the derived subgroup of G . Define also X Y = 〈x y | x ∈ X , y ∈
Y 〉. If X is a subset and H G , then X ⊂ X H 〈X , H 〉. Thus, X H = X 〈X,H〉 is the normal
closure of X in 〈X , H 〉.
Here is an example:
gap> G := SmallGroup( 64, 52);;
gap> gen := GeneratorsOfGroup(G);;
gap> H := Subgroup(G, [gen[1]]);;
gap> K := Subgroup(G, [gen[2], gen[3]]);;
gap> C := CommutatorSubgroup(H,K);;
gap> Order(H);
2
gap> Order(K);
32
gap> Order(C);
16
Lemma 3.5.4 Let X ⊂ G and H G . Then
1. X K = 〈X , [X , K ]〉;
2. [X , K ]K = [X , K ];
3. if K = 〈Y 〉, then [X , K ] = [X , Y ]K .
PROOF. (1) Follows from x k = x [x , k ].
(2) For k , h ∈ K and x ∈ X we have [x , hk ] = [x , k ][x , h]k , so that [x , h]k ∈ [X , K ].
(3) It suffices to show that [x , k ] ∈ [X , Y ]K what we prove for k = y1±1y2±1 . . . yr±1 by
induction on r . For r = 1 we get [x , y1−1] = ([x , y1]y1−1 )−1 ∈ [X , Y ]K . For the inductive step
we write k = k yr±1. Then [x , k ] = [x , k yr±1] = [x , yr±1][x , k ]yr±1 ∈ [X , Y ]K by induction.
Corollary 3.5.5 If H = 〈X 〉 and K = 〈Y 〉, then [H , K ] = [X , Y ]HK .
PROOF. This follows from Lemma 3.5.4, (3).