let G be Group; for N1, N2 being normal Subgroup of G holds [.N1,N2.] = [.N2,N1.]
let N1, N2 be normal Subgroup of G; [.N1,N2.] = [.N2,N1.]
( [.N1,N2.] is Subgroup of [.N2,N1.] & [.N2,N1.] is Subgroup of [.N1,N2.] )
by Lm2;
hence
[.N1,N2.] = [.N2,N1.]
by GROUP_2:55; verum