Ludzie pragną czasami się rozstawać, żeby móc tęsknić, czekać i cieszyć się z powrotem.
But now an element ak ∈ hai with k = 0 , 1 , 2 , . . . , d − 1
®
has order d precisely if gcd( d, k) = 1 since this requires ak = hai and so for some u ∈ Z, uk ≡ 1
d
which happens precisely when gcd( d, k) = 1 as we know from Theorem 1.9. By the definition
of ϕ, there are ϕ( d) of such elements in hai, hence θ( d) = ϕ( d). Thus we have shown that in all cases θ( d) 6 ϕ( d).
Notice that if θ( d) < ϕ( d) for some d dividing |G|, this would give a strict inequality in place of Equation (2.2). Hence we must always have θ( d) = ϕ( d). In particular, there are ϕ( n) elements of order n, hence there must be an element of order n, so G is cyclic.
¤
Taking G = Up, the group of invertible elements of Z /p under multiplication, we obtain
Theorem 1.27.
7. Group actions
If X is a set and ( G, ∗) then a ( group) action of ( G, ∗) on X is a rule which assigns to each g ∈ G and x ∈ X and element gx ∈ X so that the following conditions are satisfied.
GpAc1 For all g 1 , g 2 ∈ G and x ∈ X, ( g 1 ∗ g 2) x = g 1( g 2 x).
GpAc2 For x ∈ X, ιx = x.
Thus each g ∈ G can be viewed as acting as a permutation of X.
Example 2.29. Let G 6 Sn and let X = n. For σ ∈ G and k ∈ n let σk = σ( k). This defines an action of ( G, ◦) on n.
Example 2.30. Let X ⊆ R n and let G 6 Sym( X) be a subgroup of the symmetry group of
X. For ϕ ∈ G and x ∈ X, let ϕx = ϕ( x). This defines an action of ( G, ◦) on X.
Suppose we have an action of a group ( G, ∗) on a set X. For x ∈ X, the stabilizer of x is Stab G( x) = {g ∈ G : gx = x} ⊆ G,
and the orbit of x is
Orb G( x) = {gx : g ∈ G} ⊆ X.
Notice that x = ιx, so x ∈ Orb G( x) and ι ∈ Stab G( x). Thus Stab G( x) 6= ∅ and Orb G( x) 6= ∅.
Theorem 2.31. For each x, y ∈ X,
a) Stab G( x) 6 G;
b) y ∈ Orb G( x) if and only if x ∈ Orb G( y) ;
c) y ∈ Orb G( x) if and only if Orb G( y) = Orb G( x) .
Proof.
a) If g 1 , g 2 ∈ Stab G( x) then by GpAct1,
( g 1 ∗ g 2) x = g 1( g 2 x) = g 1 x = x.
7. GROUP ACTIONS
39
By GpAct2, ιx = x, hence ι ∈ Stab G( x). Finally, if g ∈ Stab G( x) then by GpAct1 and GpAct2, g− 1 x = g− 1( gx) = ( g− 1 ∗ g) x = ιx = x,
hence g− 1 ∈ Stab G( x). So Stab G( x) 6 G.
b) If y ∈ Orb G( x), then y = gx for some g ∈ G. Hence x = ( g− 1 ∗ g) x = g− 1( gx) = g− 1 y and so x ∈ Orb G( y). The converse is similar.
c) If y ∈ Orb G( x) then by (b), x ∈ Orb G( y) and so x = ky for some k ∈ G. Hence if g ∈ G, gx = g( ky) = ( g ∗ k) y ∈ Orb G( y) and so Orb G( x) ⊆ Orb G( y). By (b), x ∈ Orb G( y) and so we also have Orb G( y) ⊆ Orb G( x). This gives Orb G( y) = Orb G( x).
Conversely, if Orb G( y) = Orb G( x) then y ∈ Orb G( y) = Orb G( x).
¤
Example 2.32. Let X = ¤ be the square with vertices A, B, C, D and let G = Sym(¤).
Determine Stab G( x) and Orb G( x) where
a) x is the vertex A;
b) x is the midpoint M of AB;
c) x is the point P on AB where AP : P B = 1 : 3.
Solution. Recall Example 2.16. We will write permutations of the vertices in cycle nota-
tion.
a) We have
Stab G( A) = {ι, ( B D) } .
Also, every vertex can be obtained from A by applying a suitable symmetry, hence
Orb G( x) = {A, B, C, D}.
b) A symmetry Ï• fixes the midpoint of AB if and only if it maps this edge to itself. The
symmetries doing this have one of the effects Ï•( A) = A, Ï•( B) = B or Ï•( A) = B, Ï•( B) = A.
Thus
Stab G( M) = {ι, ( A B)( C D) } .
Also, we can arrange to send A to any other vertex and B to either of the adjacent vertices of
the image of A, hence the orbit of M consists of the set of 4 midpoints of edges.