for G such that 1. cq,0c2 ..... cx r e A, and 2. G c~l,0h ..... ct, is the pointwise stabilizer of A in G. Applying the base change algorithm if necessary, we may assume a strong generating set S of U relative to B is known. Let us return to the above example where G is the symmetries of the square acting on pairs of points and A is A 1 , the set of edges of the square. The points (x I =1 and cz2=3 form a base for G, so G1,3 = G1,3,4,6 = < identity > (and s=0). Hence, ~1 =1 and ~2=2 form a base for imr The stabiliser G 1 is generated by b• so {a,b,bxa} is a strong generating set of G relative to the chosen base. Furthermore, the stabiliser of 1 in imr is generated by bxa=(2,3). Hence, the set of images { -d, b, bxa } = { (1,3,4,2), (1,2)(3,4), (2,3) } is a strong generating set of im(p relative to the base [1,2]. The kernel of the homomorphism is the trivial subgroup, < identity > perform each of the basic tasks :
for G such that
1. cq,0c2 ..... cx r e A, and
2. G c~l,0h ..... ct, is the pointwise stabilizer of A in G.
Applying the base change
U relative to B is known.
Let us return to the above example where G is the symmetries of the square acting on pairs of
points and A is A 1 , the set of edges of the square. The points (x I =1 and cz2=3 form a base for
G, so G1,3 = G1,3,4,6 = < identity > (and s=0). Hence, ~1 =1 and ~2=2 form a base for imr
The stabiliser G 1 is generated by b• so {a,b,bxa} is a strong generating set of G
relative to the chosen base. Furthermore, the stabiliser of 1 in imr is generated by bxa=(2,3).
Hence, the set of images { -d, b, bxa } = { (1,3,4,2), (1,2)(3,4), (2,3) } is a strong generating
set of im(p relative to the base [1,2]. The kernel of the homomorphism is the trivial subgroup,
< identity >
perform each of the basic tasks :
Step by step
Solved in 2 steps
