Download ABSTRACT ALGEBRA 1 COURSE NOTES, LECTURE 10: GROUPS

ABSTRACT ALGEBRA 1 COURSE NOTES, LECTURE 10: GROUPS ACTING
ON SETS.
ANDREW SALCH
Definition 0.1. Let G be a group, S a set. By an action of G on S we mean a function
ρ:GˆS ÑS
satisfying the two axioms:
‚ (Associativity:) ρpgh, sq “ ρpg, ρph, sqq for all g, h P G and all s P S .
‚ (Unitality:): ρp1, sq “ s for all s P S .
If S is a set equipped with an action by G, we sometimes say that S is a G-set.
When g P G and s P S , for convenience we usually write gs instead of ρpg, sq.
The idea behind a G-set is that sometimes a set arises in mathematics (or elsewhere)
which has natural symmetries given by the group G. Below are some examples:
Example 0.2.
‚ Let G be the group C2 , the group with two elements. Then a C2 -set
(as you have already seen!) is a set S equipped with a “bilateral symmetry,” i.e., a
bijection σ : S Ñ S such that σ ˝ σ “ id.
‚ For example, let k be the real numbers, or the complex numbers (in fact, k could
be any field, a notion we haven’t talked about yet), and consider the zeroes px, yq P
kˆk of some polynomial f px, y2 q. (Writing f px, y2 q means that we’re talking about
a polynomial in the variables x and y, but whenever y appears in the polynomial, it
is taken to an even power; for example, y2 ´ x3 ´ x, or y4 ´ xy2 ´ x3 , etc.) Every
time px, yq is a zero of f px, y2 q, so is px, ´yq, since y2 “ p´yq2 . Consequently the
set of zeroes of f px, y2 q is a C2 -set: the nonidentity element σ P C2 acts on the set
of zeroes by sending px, yq to px, ´yq.
‚ Suppose instead that we have a polynomial in n variables over the real numbers: f px1 , . . . , xn q P Rrx1 , . . . , xn s. Let Z Ď Cn denote the set of zeroes of
f px1 , . . . , xn q in the complex numbers. Then Z has a C2 -action given by coordinatewise complex conjugation: given a point px1 , . . . , xn q P Z, we let σpx1 , . . . , xn q “
px1 , . . . , xn q, the same point but with each coordinate complex-conjugated. (You
can easily check that px1 , . . . , xn q is in Z if px1 , . . . , xn q is.)
Notice that the set of points z P Z such that σpzq “ z, i.e., the “fixed points” of
the C2 -action on Z, is exactly the set of zeroes px1 , . . . , xn q of f px1 , . . . , xn q all of
whose coordinates are real numbers, i.e., the set of points z P Z such that σpzq “ z
is exactly the set of zeroes of f px1 , . . . , xn q in the real numbers! We will return to
this idea when we get around to doing some Galois theory.
‚ Let S 1 be the unit circle, centered at the origin, in the complex plane C. Then S 1
is actually a subgroup of Cˆ , the group of units in the field C (this is easy to prove:
just show that if you multiply together two complex numbers of norm one, you get
a complex number of norm one.) You can think of the group S 1 as the group of
Date: October 2016.
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ANDREW SALCH
rotations: the point p1, 0q “ 1 ` 0i is the identity element and represents rotation
by zero radians, while the point pcosθ, sin θq “ cos θ ` i sin θ represents rotation
by θ radians.
Then saying that a set has an S 1 -action is a precise and rigorous way of saying that that set has “rotational symmetry,” the way that (for example) cones and
cylinders do.
‚ There are many, many more examples of G-sets—in fact, the study of sets (or
more structured objects, like topological spaces, manifolds, etc.) equipped with
actions of a group G is one of the main motivations for all of group theory!
Definition 0.3. Let G be a group, let S be a G-set, and let x P S . By the orbit of x, written
Gx, we mean the subset
Gx “ tgx : g P Gu
of S .
By the stabilizer of x, written G x , we mean the subgroup
G x “ tg P G : gx “ xu
of G.
Finally, we write S G for the fixed points of G acting on S :
S G “ ts P S : gs “ s @ g P Gu Ď S .
Proposition 0.4. Let G be a group, let S be a G-set, and let x P S . Then there is a bijection
between the orbit Gx and the number of cosets of G x in G, that is
#pGxq “ rG : G x s.
Proof. Let C be the set of cosets of G x in G, and let f : C Ñ Gx be the function sending a
coset rgG x s to gx. We need to check that f is well-defined, so suppose that rgG x s “ rhG x s.
Then h´1 g P G x , so h´1 gx “ x, so hx “ gx. So f prgG x sq “ f prhG x sq. So f is indeed
well-defined.
Suppose now that f prgG x sq “ f prhG x sq. Then hx “ gx, so h´1 gx “ x, so h´1 g P G x ,
so rgG x s “ rhG x s. So f is one-to-one. Meanwhile, any element gx in Gx is equal to
f prgG x sq, so f is surjective. So f is a bijection.
Theorem 0.5. Let G be a finite group, and let S be a finite G-set. Then we have an equality:
ÿ
(0.0.1)
#pS q “ #pS G q ` rG : G x s,
Gx
where the sum is taken over all the orbits Gx Ď S containing more than one element.
Proof. This is basically the same argument as what we used to prove the class equation:
each element of S is in exactly one orbit of the G-action on S . The elements whose orbits
contains only a single element are exactly the elements of S fixed by the action of G.
Meanwhile, the size of each of the remaining orbits Gx is the number of cosets rG : Gxs,
by Proposition 0.4.
Corollary 0.6. Let G be a finite p-group acting on a finite set S . Then we have the congruence
(0.0.2)
#pS q ” #pS G q mod p.
ABSTRACT ALGEBRA 1 COURSE NOTES, LECTURE 10: GROUPS ACTING ON SETS.
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Proof. Since G is a finite p-group, #pGq is a power of p. For any subgroup H of G, we
have
#pGq “ rG : 1s “ rG : HsrH : 1s “ rG : Hs#pHq,
so rG : Hs must be a power of p (since the only divisors of a power of p are other powers
of p) for every subgroup H of G. If H is a subgroup of G and H ‰ G, then rG : Hs must
furthermore be a power of p greater than 1 (i.e., not p0 “ 1). So each of the terms rG : Gxs
in equation 0.0.1 is divisible by p. reduce equation 0.0.1 modulo p to get congruence 0.0.2.
Corollary 0.7. Let G be a finite p-group. Then the number of elements x of G such that
x p “ 1 is divisible by p.
Proof. Let S be the set of all p-tuples px1 , . . . , x p q P G ˆ ¨ ¨ ¨ ˆ G whose product is equal to
the identity element: x1 . . . x p “ 1. Then #pS q “ #pGq p´1 , since x1 , . . . , x p´1 can be any
elements of G whatsoever, but x p is necessarily the inverse x p “ px1 . . . x p´1 q´1 . Since
#pGq is a p-group, #pS q is a power of p.
Meanwhile, we let C p act on S by cyclic permutation of the coordinates, that is, choose
a generator σ for C p , and let σpx1 , . . . , x p q “ px2 , . . . , x p , x1 q. Then the fixed points S C p
are exactly the set of p-tuples px, x, x, . . . , xq such that x p “ 1. By Corollary 0.6 and by the
fact that #pS q is divisible by p, we have that #pS C p q is divisible by p.
Corollary 0.8. Every finite p-group contains at least p elements x such that x p “ 1.
Exercise 0.9. (Fun, definitely requires some thought.) Let p be an odd prime number,
and let Z Ď Z{pZ ˆ Z{pZ be the set of solutions px, yq to x2 ` y2 “ 1. Suppose that
Z{pZ contains a square root of 21 , i.e., suppose that there exists an integer n such that 2n2
is congruent to 1 modulo p. Prove that #pZq is even.