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1 GROUP ACTIONS KEITH CONRAD 1. Introduction The symmetric groups S n, alternating groups A n, and (for n 3) dihedral groups D n behave, by their very definition, as permutations on certain sets. The groups S n and A n both permute the set {1, 2,..., n} and D n can be considered as a group of permutations of a regular n-gon, or even just of its n vertices, since rigid motions of the vertices determine where the rest of the n-gon goes. If we label the vertices of the n-gon in a definite manner by the numbers from 1 to n then we can view D n as a subgroup of S n. For instance, the labeling lets us regard the 90 degree counterclockwise rotation r in D 4 as (1234) and the reflection s across the x-axis as (12)(34). The rest of the elements of D 4, as permutations of the vertices, are in the table below. 1 r r 2 r 3 s sr sr 2 sr 3 (1) (1234) (13)(24) (1432) (12)(34) (24) (14)(23) (13) If we label the vertices in a different way (e.g., swap the labels 1 and 2), we may get a different subgroup of S 4. More abstractly, if we are given any set X (not necessarily the set of vertices of a square), then the set Sym(X) of all permutations of X is a group under composition, and the subgroup Alt(X) of even permutations of X is a group under composition. If we list the elements of X in a definite order, say as X = {x 1,..., x n }, then we can think about Sym(X) as S n and Alt(X) as A n, but a listing in a different order leads to different identifications of Sym(X) with S n and Alt(X) with A n. The abstract symmetric groups Sym(X) really do arise naturally: Theorem 1.1 (Cayley). Every finite group G can be embedded in a symmetric group. Proof. To each g G, define the left multiplication function l g : G G, where l g (x) = gx for x G. Each l g is a permutation of G as a set, with inverse l g 1. So l g belongs to Sym(G). Since l g1 l g2 = l g1 g 2 (that is, g 1 (g 2 x) = (g 1 g 2 )x for all x G), associating g to l g gives a homomorphism of groups, G Sym(G). This homomorphism is one-to-one since l g determines g (after all, l g (e) = g). Therefore the correspondence g l g is an embedding of G as a subgroup of Sym(G). Allowing an abstract group to behave like a group of permutations, as happened in the proof of Cayley s theorem, is a useful tool. 1

2 2 KEITH CONRAD Definition 1.2. An action of a group G on a set X is the choice, for each g G, of a permutation π g : X X such that the following two conditions hold: π e is the identity: π e (x) = x for each x X, for every g 1 and g 2 in G, π g1 π g2 = π g1 g 2. Example 1.3. Let S n act on X = {1, 2,..., n} in the usual way. Here π σ (i) = σ(i) in the usual notation. Example 1.4. Any group G acts on itself (X = G) by left multiplication functions. That is, we set π g : G G by π g (h) = gh for all g G and h G. Then the conditions for being a group action are eh = h for all h G and g 1 (g 2 h) = (g 1 g 2 )h for all g 1, g 2, h G, which are both true since e is an identity and multiplication in G is associative. In practice, one dispenses with the notation π g and writes π g (x) simply as g x or gx. This is not meant to be an actual multiplication of elements from two possibly different sets G and X. It is just the notation for the effect of g (really, the permutation associated to g) on the element x. In this notation, the axioms for a group action take the following form: For each x X, e x = x. For every g 1, g 2 G and x X, g 1 (g 2 x) = (g 1 g 2 ) x. The basic idea in any group action is that the elements of a group are viewed as permutations of a set in such a way that composition of the corresponding permutations matches multiplication in the original group. To get used to the notation, let s prove a simple result. Theorem 1.5. Let G act on X. If x X, g G, and y = g x, then x = g 1 y. If x x then g x g x. Proof. From y = g x we get g 1 y = g 1 (g x) = (g 1 g) x = e x = x. To show x x = gx gx, we show the contrapositive: if g x = g x then applying g 1 to both sides gives g 1 (g x) = g 1 (g x ), so (g 1 g) x = (g 1 g) x, so x = x. Another way to think about an action of a group on a set is that it is a certain homomorphism. Here are the details. Theorem 1.6. Actions of the group G on the set X are the same as group homomorphisms from G to Sym(X), the group of permutations of X. Proof. Suppose we have an action of G on X. We view g x as a function of x (with g fixed). That is, for each g G we have a function π g : X X by π g (x) = g x. The axiom e x = x says π e is the identity function on X. The axiom g 1 (g 2 x) = (g 1 g 2 ) x says π g1 π g2 = π g1 g 2, so composition of functions on X corresponds to multiplication in G. Moreover, π g is an invertible function since π g 1 is an inverse: the composite of π g and π g 1 is π e, which is the identity function on X. Therefore π g Sym(X) and g π g is a homomorphism G Sym(X). Conversely, suppose we have a homomorphism f : G Sym(X). For each g G, we have a permutation f(g) on X, and f(g 1 g 2 ) = f(g 1 ) f(g 2 ). Setting g x = f(g)(x) defines a group action of G on X, since the homomorphism properties of f yield the defining properties of a group action.

3 GROUP ACTIONS 3 From this viewpoint, the set of g G that act trivially (g x = x for all x X) is simply the kernel of the homomorphism G Sym(X) associated to the action. Therefore those g that act trivially on X are said to lie in the kernel of the action. We will not often use the interpretation of Theorem 1.6 before Section 6. Until then we take the more concrete viewpoint of a group action as a kind of product g x of G with X, taking values in X subject to the properties e x = x and g 1 (g 2 x) = (g 1 g 2 ) x. Here is an outline of later sections. Section 2 describes several concrete examples of group actions and also some general actions available from any group. Section 3 describes the important orbit-stabilizer formula. The short Section 4 isolates an important fixed-point congruence for actions of p-groups. Sections 5 and 6 give applications of group actions to group theory. In Appendix A, group actions are used to derive three classical congruences from number theory. 2. Examples Example 2.1. We can make R n act on itself by translations: for v R n, let T v : R n R n by T v (w) = w + v. The axioms for a group action are: T 0 (w) = w and T v1 (T v2 (w)) = T v1 +v 2 (w). These are true from properties of vector addition. Example 2.2. Let G be the group of Rubik s cube: all sequences of motions on the cube (keeping center facelets in fixed locations). This group acts on two different sets: the 12 edge cubelets and the 8 corner cubelets. Or we could let G act on the set of all 20 non-centerface cubelets together. Example 2.3. For n 3, each element of the group D n acts as a rigid motion of a regular n-gon in the plane, either a rotation or a reflection. We can also view D n as acting just on the n vertices of a regular n-gon. Knowing where the vertices go determines the rest of the rigid motion, so the effect of D n on the vertices tells us all we need to know to determine the rigid motion on the n-gon. Restricting the action of D n from an n-gon to its vertices, and labeling the vertices as 1, 2,..., n in some manner, makes D n act on {1, 2,..., n}. Example 2.4. The group GL n (R) acts on vectors in R n in the usual way that a matrix can be multiplied with a (column) vector: A v = Av. In this action, the origin 0 is fixed by every A while other vectors get moved around (as A varies). The axioms of a group action are properties of matrix-vector multiplication: I n v = v and A(Bv) = (AB)v. Example 2.5. The group S n acts on polynomials f(t 1,..., T n ), by permuting the variables: (2.1) (σ f)(t 1,..., T n ) = f(t σ(1),..., T σ(n) ). This is a change of variables T i T σ(i) in f(t 1,..., T n ). For example, (12)(23) = (123) in S 3 and (12) ((23) (T 2 + T 2 3 )) = (12) (T 3 + T 2 2 ) = T 3 + T 2 1 and (123) (T 2 + T 2 3 ) = T 3 + T 2 1, giving the same result both ways. It s obvious that (1) f = f. To check that σ (σ f) = (σσ ) f for all σ and σ in S n, so (2.1) is a group action of S n on polynomials in n variables, let s compute Since (σ (σ f))(t 1,..., T n ) = (σ f)(t σ(1),..., T σ(n) ). (2.2) (σ f)(t 1,..., T n ) = f(t σ (1),..., T σ (n)),

4 4 KEITH CONRAD making the substitution T i T σ(i) on both sides of (2.2), so T σ (i) T σ(σ (i)), shows Thus (σ f)(t σ(1),..., T σ(n) ) = f(t σ(σ (1)),..., T σ(σ (n))). (σ (σ f))(t 1,..., T n ) = (σ f)(t σ(1),..., T σ(n) ) = f(t σ(σ (1)),..., T σ(σ (n))) = f(t (σσ )(1),..., T (σσ )(n)) = ((σσ ) f)(t 1,..., T n ). Lagrange s study of this group action (ca. 1770) marked the first systematic use of symmetric groups in algebra. Lagrange wanted to understand why nobody had found an analogue of the quadratic formula for roots of a polynomial of degree greater than four. While Lagrange was not completely successful, he found in this group action that there are some different features in the cases n 4 and n = 5. Since f and σ f have the same degree, and if f is homogeneous then σ f is homogeneous, this action of S n can be restricted to the set of polynomials in n variables with a fixed degree or the set of homogeneous polynomials in n variables with a fixed degree. An example is the set of homogeneous linear polynomials {a 1 T a n T n }, where (2.3) σ (c 1 T c n T n ) = c 1 T σ(1) + + c n T σ(n) = c σ 1 (1)T c σ 1 (n)t n. Example 2.6. Here is a tricky example, so pay attention. Let S n act on R n by permuting the coordinates: for σ S n and v = (c 1,..., c n ) R n, set π σ (v) = (c σ(1),..., c σ(n) ). For example, let n = 3, σ = (12), σ = (23), and v = (5, 7, 9). Then and π σ (π σ (v)) = π (12) (π (23) (5, 7, 9)) = π (12) (5, 9, 7) = (9, 5, 7) π σσ (5, 7, 9) = π (123) (5, 7, 9) = (7, 9, 5), which do not agree, so sending v to π σ (v) in R n is not a group action of S n on R n. A peculiar thing happens if we calculate π σ (π σ (v)) and π σσ (v) in general to see what is going wrong, since it is easy to convince ourselves that we do have a group action: π σ (π σ (c 1,..., c n )) = π σ (c σ (1),..., c σ (n)) = (c σ(σ (1)),..., c σ(σ (n))) = (c (σσ )(1),..., c (σσ (n)) = π σσ (c 1,..., c n ), which suggests π σ π σ = π σσ, and that is not what we saw in the numerical example above. What happened?!? The mistake really is in the general calculation, not the example. Try to find the error before reading further. The mistake was in the second line, when we computed π σ (c σ (1),..., c σ (n)) by applying σ to the indices σ (i). A vector does not remember the indices of its coordinates after they are permuted: to compute π (12) (π (23) (5, 7, 9)) = π (12) (5, 9, 7), the next step treats (5, 9, 7) as a new vector with coordinates indexed 1, 2, 3, in that order even though that is not their order in the original vector (5, 7, 9). The computation of π σ (v) always requires indexing the coordinates of v from 1 to n in that order. Thus when we compute π (12) (π (23) (c 1, c 2, c 3 )) = π (12) (c 1, c 3, c 2 ), for the next step let s write (c 1, c 3, c 2 ) = (d 1, d 2, d 3 ). Then π (12) (π (23) (c 1, c 2, c 3 )) = π (12) (c 1, c 3, c 2 ) = π (12) (d 1, d 2, d 3 ) = (d 2, d 1, d 3 ) = (c 3, c 1, c 2 ),

5 which does not agree with GROUP ACTIONS 5 π (12)(23) (c 1, c 2, c 3 ) = π (123) (c 1, c 2, c 3 ) = (c 2, c 3, c 1 ). In general, for σ and σ in S n, and v = (c 1,..., c n ) in R n, π σ (π σ (v)) = π σ (c σ (1),..., c σ (n)) = π σ (d 1,..., d n ) where d i = c σ (i) = (d σ(1),..., d σ(n) ) = (c σ (σ(1)),..., c σ (σ(n))) = (c (σ σ)(1),..., c (σ σ)(n)) = π σ σ(v), so in general π σ π σ = π σ σ rather than π σσ. To make things turn out correctly, we should redefine the effect of S n on R n to use the inverse: set σ v = (c σ 1 (1),..., c σ 1 (n)). Then σ (σ v) = (σσ ) v and we have a group action of S n on R n, which in fact is essentially the action of S n from the previous example on homogeneous linear polynomials (see (2.3)). The lesson from these last two examples is that when S n permutes variables in a polynomial then it acts directly, but when S n permutes coordinates in a vector then it has to act using inverses. It s easy to remember that when S n acts on variables or coordinates that it acts without inverses in one case and with inverses in the other case, but it s easy to forget which case is which. At least remember that you need to be careful. Example 2.7. Let G be a group acting on the set X, and S be any set. Write Map(X, S) for the set of all functions f : X S. It is natural to try defining an action of G on the set Map(X, S) by the rule (2.4) (π g f)(x) = f(gx), where gx is the action of g G on x X. While π g f is a function X S, sending each f to π g f is usually not an action of G on Map(X, S) even though it is easy to confuse yourself into thinking it is: for g and h in G, and x X, π g ((π h f)(x)) = π g (f(hx)) = f(g(hx)) = f((gh)x) = (π gh f)(x). This holds for all x X, so π g (π h f) = π gh f, right? No. The calculation is bogus since the first expression π g ((π h f)(x)) is nonsense: (π h f)(x) = f(hx) belongs to S and no action of G has been defined on S. Even if it had been, π g is applied to functions X S, not to elements of S. The mistake was confusing (π g f)(x) in (2.4) with π g (f(x)), which is meaningless. A correct calculation is (π g (π h f))(x) = (π h f)(gx) = f(h(gx)) = f((hg)x) = (π hg f)(x). Therefore π g (π h f) = π hg f, so we find the same problem as with the false action in Example 2.6. To get a group action of G on Map(X, S) when G already acts on X (from the left), we need to replace g with g 1 in (2.4) and set Now (g f)(x) = f(g 1 x). (g (h f))(x) = (h f)(g 1 x) = f(h 1 (g 1 x)) = f((gh) 1 x) = ((gh) f)(x),

7 GROUP ACTIONS 7 Example When A is a subset of G, and g G, the subset ga = {ga : a A} has the same size as A. Therefore G acts by left multiplication on the set of subsets of G, or even on the subsets with a fixed size. Example 2.8 is the special case of one-element subsets of G. Notice that, when H G is a subgroup, gh is usually not a subgroup of G, so the left multiplication action of G on its subsets does not convert subgroups into other subgroups. Example As a special case of Example 2.12, let S 4 act on the set of pairs from {1, 2, 3, 4} by the rule σ {a, b} = {σ(a), σ(b)}. There are 6 pairs: x 1 = {1, 2}, x 2 = {1, 3}, x 3 = {1, 4}, x 4 = {2, 3}, x 5 = {2, 4}, x 6 = {3, 4}. The effect of (12) on these pairs is (12)x 1 = x 1, (12)x 2 = x 4, (12)x 3 = x 5, (12)x 4 = x 2, (12)x 5 = x 3, (12)x 6 = x 6. Thus, as a permutation of the set {x 1,..., x 6 }, (12) acts like (x 2 x 4 )(x 3 x 5 ). That is interesting: we have made a transposition in S 4 look like a product of two 2-cycles in S 6. In particular, we have made an odd permutation of {1, 2, 3, 4} look like an even permutation (on a new set). Example Let G be a group. When A G, gag 1 is a subset with the same size as A. Moreover, unlike the left multiplication action of G on its subsets, the conjugation action of G on its subsets transforms subgroups into subgroups: when H G is a subgroup, ghg 1 is also a subgroup. For instance, three subgroups of S 4 with size 4 are {(1), (1234), (13)(24), (1432)}, {(1), (2134), (23)(14), (2431)}, {(1), (12)(34), (13)(24), (14)(23)}. Under conjugation by S 4, the first two subgroups can be transformed into each other, but neither of these subgroups can be conjugated to the third subgroup: the first and second subgroups have an element with order 4 while the third one does not. While the left multiplication action of G on itself (Example 2.8) associates different permutations to different group elements, the conjugation action of G on itself (Example 2.9) can make different group elements act in the same way: if g 1 = g 2 z, where z is in the center of G, then g 1 and g 2 have the same conjugation action on G. Group actions where different elements of the group act differently have a special name: Definition A group action of G on X is called faithful (or effective) if different elements of G act on X in different ways: when g 1 g 2 in G, there is an x X such that g 1 x g 2 x. Note that when we say g 1 and g 2 act differently, we mean they act differently somewhere, not everywhere. This is consistent with what it means to say two functions are not equal: they take different values somewhere, not everywhere. Example The action of G on itself by left multiplication is faithful: different elements send e to different places. Example The action of G on itself by conjugation is faithful if and only if G has a trivial center, because g 1 gg1 1 = g 2 gg2 1 for all g G if and only if g2 1 g 1 is in the center of G. When D 4 acts on itself by conjugation, the action is not faithful since r 2 acts trivially (it is in the center), so 1 and r 2 act in the same way.

8 8 KEITH CONRAD Example When H is a subgroup of G and G acts on G/H by left multiplication, two elements g 1 and g 2 act in the same way on G/H precisely when g 1 gh = g 2 gh for all g G, which means g2 1 g 1 g G ghg 1. So the left multiplication action of G on G/H is faithful if and only if the subgroups ghg 1 (as g varies) have trivial intersection. Example The action of GL 2 (R) on R 2 is faithful, since we can recover the columns of a matrix by acting it on ( 1 0) and ( 0 1). Non-faithful actions are as important as non-injective group homomorphisms (in fact, that is precisely what a non-faithful action is from the viewpoint of Theorem 1.6). Remark What we have been calling a group action could be called a left group action, while a right group action, denoted xg, has the properties xe = x and (xg 1 )g 2 = x(g 1 g 2 ). The exponential notation x g in place of xg works well here, especially by writing the identity in the group as 1: x 1 = x and (x g 1 ) g 2 = x g 1g 2. The distinction between left and right actions is how a product gg acts: in a left action g acts first and g acts second, while in a right action g acts first and g acts second. Right multiplication of G on itself (or more generally right multiplication of G on the space of right cosets of a subgroup H) is an example of a right action. To take a more concrete example, the action of GL n (R) on row vectors of length n is most naturally a right action since the product va (not Av) makes sense when v is a row vector and A GL n (R). The wrong definitions of actions π g in Examples 2.6 and 2.7, which were wrong because formulas came out backwards (π g π h = π hg ) are legitimate right actions of G. Many group theorists (unlike most other mathematicians) like to define the conjugate of h by g as g 1 hg instead of as ghg 1, and this convention fits well with the right (but not left) conjugation action: setting h g = g 1 hg we have h 1 = h and (h g 1 ) g 2 = h g 1g 2. The difference between left and right actions of a group is largely illusory, since replacing g with g 1 in the group turns left actions into right actions and conversely because inversion reverses the order of multiplication in G. We saw this idea at work in Example 2.8 and in Example 2.6. We will not use right actions (except in Example 3.23), so for us group action means left group action. 3. Orbits and Stabilizers The information encoded in a group action has two basic parts: one part tells us where points go and the other part tells us how points stay put. The following terminology refers to these ideas. Definition 3.1. Let G act on X. For each x X, its orbit is Orb x = {g x : g G} X and its stabilizer is Stab x = {g G : g x = x} G. (The stabilizer of x is often denoted G x in the literature, where G is the group.) We call x a fixed point for the action when g x = x for every g G, that is, when Orb x = {x} (or equivalently, when Stab x = G). Writing the definition of orbits and stabilizers in words, the orbit of a point is a geometric concept: it is the set of places where the point can be moved by the group action. On the other hand, the stabilizer of a point is an algebraic concept: it is the set of group elements that fix the point.

9 GROUP ACTIONS 9 We will often refer to the elements of X as points and we will refer to the size of an orbit as its length. If X = G, as in Examples 2.8 and 2.9, then we think about elements of G as permutations when they act on G and as points when they are acted upon. Example 3.2. When GL 2 (R) acts in the usual way on R 2, the orbit of 0 is {0} since A 0 = 0 for every A in GL 2 (R). The stabilizer of 0 is GL 2 (R). The orbit of ( ) 1 0 is R 2 {0}, in other words every non-zero vector can be obtained from ( ) 1 0 by applying a suitable invertible matrix to it. Indeed, if ( ( a b) 0, then we have a ) b = ( a 1 b 0 )( ) 1 0 and ( ) a b = ( a 0 b 1 )( ) 1 0. One of the matrices ( a 1 b 0 ) or ( a b 0 1 ) is invertible (since a or b is non-zero), so ( ) a b is in the GL2 (R)-orbit of ( ( 1 0). The stabilizer of 1 ) 0 is {( 1 x 0 y ) : y 0} GL 2 (R). Example 3.3. When the group GL 2 (Z) acts in the usual way on Z 2, the orbit of 0 is {0} with stabilizer GL 2 (Z). But in contrast to Example 3.2, the orbit of ( 1 0) under GL2 (Z) is not Z 2 {0}. Indeed, a matrix ( a c d b ) in GL 2(Z) sends ( ( 1 0) to a c), which is a vector with relatively prime coordinates since ad bc = ±1. (For instance, GL 2 (Z) can t send ( 1 0) to ( 2 ) ( 0.) Conversely, any vector m ) n in Z 2 with relatively prime coordinates is in the GL 2 (Z)- orbit of ( ) 1 0 : we can solve mx + ny = 1 for some integers x and y, so ( m y n x ) is in GL 2(Z) (its determinant is 1) and ( m n y x )( ( 1 0) = m n). Check as an exercise that the orbits in Z 2 under the action of GL 2 (Z) are the vectors whose coordinates have a fixed greatest common divisor. Each orbit contains one vector of the form ( ( d 0) for d 0, and the stabilizer of d ) 0 for d > 0 is {( 1 x 0 y ) : y = ±1} GL 2 (Z). Example 3.4. Identifying Z/(2) with the subgroup {±I n } of GL n (R) gives an action of Z/(2) on R n, where 0 acts as the identity and 1 acts by negation on R n. We can restrict this action of Z/(2) to the unit sphere of R n, and then it is called the antipodal action since its orbits are pairs of opposite points (which are called antipodal points) on the sphere. Example 3.5. When the Rubik s cube group acts on the non-centerface cubelets of Rubik s cube, there are two orbits: the corner cubelets and the edge cubelets. Example 3.6. For n 2, consider S n in its natural action on {1, 2,..., n}. What is the stabilizer of an integer k {1, 2,..., n}? It is the set of permutations of {1, 2,..., n} fixing k, which can be identified with the set of permutations of {1, 2,..., n} {k}. This is just S n 1 in disguise (once we identify {1, 2,..., n} {k} in a definite manner with the numbers from 1 to n 1). The stabilizer of any number in {1, 2,..., n} for the natural action of S n on {1, 2,..., n} is isomorphic to S n 1. Example 3.7. For n 2, the even permutations of {1, 2,..., n} that fix a number k can be identified with the even permutations of {1, 2,..., n} {k}, so the stabilizer of any point in the natural action of A n is essentially A n 1 up to relabelling. Remark 3.8. When trying to think about a set as a geometric object, it is helpful to refer to its elements as points, no matter what they might really be. For example, when we think about G/H as a set on which G acts (by left multiplication), it is useful to think about the cosets of H, which are the elements of G/H, as the points in G/H. At the same time, though, a coset is a subset of G. There is a tension between these two interpretations: is a left coset of H a point in G/H or a subset of G? It is both, and it is important to be able to think about a coset in both ways.

10 10 KEITH CONRAD All of our applications of group actions to group theory will flow from the relations between orbits, stabilizers, and fixed points, which we now make explicit in our three basic examples of group actions. Example 3.9. When G acts on itself by left multiplication, there is one orbit (g = ge Orb e ), Stab x = {g : gx = x} = {e} is trivial, there are no fixed points (if G > 1). Example When G acts on itself by conjugation, the orbit of a is Orb a = {gag 1 : g G}, which is the conjugacy class of a. Stab a = {g : gag 1 = a} = {g : ga = ag} is the centralizer of a, denoted Z(a). a is a fixed point when it commutes with all elements of G, and thus the fixed points of conjugation form the center of G. Example When G acts on G/H by left multiplication, there is one orbit (gh = g H Orb {H} ), Stab ah = {g : gah = ah} = {g : a 1 ga H} = aha 1, there are no fixed points (if H G). These examples illustrate several facts: an action need not have any fixed points (Example 3.9 with non-trivial G), different orbits can have different lengths (Example 3.10 with G = S 3 ), and the points in a common orbit don t have to share the same stabilizer (Example 3.11, if H is a non-normal subgroup). Example When G acts on its subgroups by conjugation, Stab H = {g : ghg 1 = H} is the normalizer N(H) and the fixed points are the normal subgroups of G. When G acts on X, any subgroup of G also acts on X. Let s look at two examples. Example Any subgroup K G acts on G/H by left multiplication. Then there could be more than one orbit (not all cosets gh can be reached from H by left multiplication by elements of K unless KH = G), Stab ah = aha 1 K, there are no fixed points (if H G). Example Any subgroup H G acts on G by right-inverse multiplication (Example 2.8). Then the orbits are the left H-cosets ({gh 1 : h H} = gh) Stab a is trivial, there are no fixed points (if H > 1). Here is the fundamental theorem about group actions. Theorem Let G act on X. a) Different orbits of the action are disjoint. b) For each x X, Stab x is a subgroup of G and Stab gx = g Stab x g 1 for all g G. c) For x X, gx = g x if and only if g and g lie in the same left coset of Stab x. In particular, if x and y are in the same orbit then {g G : gx = y} is a left coset of Stab x and different left cosets of Stab x correspond to different points in Orb x, so Orb x = [G : Stab x ].

11 GROUP ACTIONS 11 Parts b and c show the role of conjugate subgroups and cosets of a subgroup when working with group actions. The formula in part c that relates the length of an orbit to the index in G of a stabilizer for a point in the orbit, is called the orbit-stabilizer formula. Proof. a) We prove different orbits in a group action are disjoint by proving that two orbits which overlap must coincide. 1 Suppose Orb x and Orb y have a common element z: z = g 1 x, z = g 2 y. We want to show Orb x = Orb y. It suffices to show Orb x Orb y, since then we can switch the roles of x and y to get the reverse inclusion. For any point u Orb x, write u = gx for some g G. Since x = g 1 1 z, u = g(g 1 1 z) = (gg 1 1 )z = (gg 1 1 )(g 2y) = (gg1 1 g 2)y, which shows us that u Orb y. Therefore Orb x Orb y. b) To see that Stab x is a subgroup of G, we have e Stab x since ex = x, and if g 1, g 2 Stab x, then (g 1 g 2 )x = g 1 (g 2 x) = g 1 x = x, so g 1 g 2 Stab x. Thus Stab x is closed under multiplication. Lastly, gx = x = g 1 (gx) = g 1 x = x = g 1 x, so Stab x is closed under inversion. To show Stab gx = g Stab x g 1, for any x X and g G, observe that h Stab gx h (gx) = gx (hg)x = gx g 1 ((hg)x) = g 1 x (g 1 hg)x = x g 1 hg Stab x h g Stab x g 1, so Stab gx = g Stab x g 1. c) The condition gx = g x is equivalent to x = (g 1 g )x, which means g 1 g Stab x, or g g Stab x. Therefore g and g have the same effect on x if and only if g and g lie in the same left coset of Stab x. (Remember that for any subgroup H, g gh if and only if g H = gh.) Since Orb x consists of the points gx for varying g, and elements of G have the same effect on x if and only if they lie in the same left coset of Stab x, we get a bijection between the left cosets of Stab x and the points in the orbit of x: associate to any left coset g Stab x the point gx. (Think carefully about why this makes sense and is a bijection, which requires unwinding definitions.) Therefore the length of the orbit of x, which means Orb x, is the number of left cosets of Stab x in G. Remark That different orbits of a group action are disjoint includes as special cases two basic disjointness results in group theory: disjointness of conjugacy classes and disjointness of left cosets of a subgroup. The first result uses the action of a group on 1 The argument will be similar to the proof that different left cosets of a subgroup are disjoint: if the cosets overlap they coincide.

12 12 KEITH CONRAD itself by conjugation, having conjugacy classes as its orbits. The second result uses the right-inverse multiplication action of the subgroup on the group (Example 3.14). Corollary Let a finite group G act on a set. a) The length of every orbit divides the size of G. b) Points in a common orbit have conjugate stabilizers, and in particular the size of the stabilizer is the same for all points in an orbit. Proof. a) The length of an orbit is an index of a subgroup, so it divides G. b) If x and y are in the same orbit, write y = gx. Then Stab y = Stab gx = g Stab x g 1, so the stabilizers of x and y are conjugate subgroups. Corollary Let G act on X, where G and X are finite. Let the different orbits of X be represented by x 1,..., x t. Then (3.1) X = t Orb xi = i=1 t [G : Stab xi ]. Proof. Different orbits are disjoint and the orbit-stabilizer formula tells us how large each orbit is. Example For any finite group G, each conjugacy class in G has size dividing the size of G, since a conjugacy class in G is an orbit in the conjugation action of G on itself, so Corollary 3.17a applies. Moreover, for the conjugation action (3.1) is the class equation. Example Let S 3 act on itself by conjugation. Its orbits are its conjugacy classes, which are i=1 {(1)}, {(12), (13), (23)}, {(123), (132)}. The conjugacy class of (12) has size 3 and the stabilizer of (12) is its centralizer {(1), (12)}, which has index 3 in S 3. Example In D 6 which elements commute with s? This is asking for {g D 6 : gs = sg}. Three such elements are 1, s, and r 3 (since r n/2 Z(D n ) for even n). Let s interpret the condition gs = sg as gsg 1 = s: the task is now computing the stabilizer of s when D 6 acts on itself by conjugation. To compute the stabilizer, let s first compute the orbit: how many different values of gsg 1 are there as g runs over D 6? Elements of D 6 are r k (rotations) and r k s (reflections, so equal to their inverses). From r k sr k = r 2k s, (r k s)s(r k s) 1 = r k ssr k s = r k r k s = r 2k s the different gsg 1 as g varies in D 6 is {r even s} = {s, r 2 s, r 4 s}. Since the D 6 -orbit of s has size 3, the stabilizer of s has index 3 in D 6 and thus size D 6 /3 = 12/3 = 4. We already know 1, s, and r 3 are in the stabilizer, so being a group means r 3 s is in the stabilizer too. That is a fourth element, and there are only 4 elements, so {g D 6 : gs = sg} = {1, s, r 3, r 3 s}. Example We examine now a geometric example. Consider the figure F below, a hexagon with an X drawn inside of it. Which elements of D 6 preserve this figure?

13 GROUP ACTIONS 13 F For g D 6, g(f ) = F means g Stab F. To compute Stab F we first compute the orbit of F : it s easier to figure out all the ways F can change than to figure out all the ways F can stay the same, and these are related by the orbit-stabilizer formula. By rotating and reflecting it is clear that g(f ), as g runs over D 6, has only the 3 values below. F F F Since F has an orbit of size 3, its stabilizer in D 6 has index 3, so Stab F = D 6 /3 = 12/3 = 4. From the 180-degree rotational symmetry of F, r 3 Stab F. From the symmetry of F by a reflection across the x-axis, s Stab F. Since Stab F is a subgroup of D 6, Stab F contains r 3 s. Thus {1, r 3, s, r 3 s} Stab F, and we are done since we know Stab F = 4. While F looks like F, it is not equal to F. What are Stab F and {g D 6 : g(f ) = F }? We can compute both as soon as we know just one g sending F to F. Since F = r(f ) we can use g = r: Theorem 3.15b says Stab F = Stab r(f ) = r Stab F r 1 = r{1, r 3, s, r 3 s}r 1 = {1, r 3, r 2 s, r 5 s} and Theorem 3.15c says {g D 6 : g(f ) = F } = r Stab F = r{1, r 3, s, r 3 s} = {r, r 4, rs, r 4 s}. Example The 2 2 matrices ( a c d b ) GL 2(R) whose columns add up to 1 are a group. This can be checked by a tedious calculation. But this can be seen more simply by observing that the column sums are the entries in the vector-matrix product (1 1)( a c d b ), so the matrices of interest are those satisfying (1 1)( a c d b ) = (1 1). This is the stabilizer of (1 1) in the (right!) action of GL 2 (R) on R 2 viewed as row vectors by v A = va, so it is a subgroup of GL 2 (R) since the stabilizers of a point are always a subgroup. (Theorem 3.15 for right group actions should be formulated and checked by the reader.) Moreover, because (0 1)( ) = (1 1), Stab (1 1) and Stab (0 1) are conjugate subgroups in GL 2 (R). Since Stab (0 1) = {( a 0 1 b ) GL 2(R)} = Aff(R), a model for our column-sum-1 group is its conjugate subgroup Aff(R). Explicitly, ( Stab (1 1) = Stab (0 1)( ) = ) 1 ( 0 1 Aff(R) 1 1 Example As a cute application of the orbit-stabilizer formula we explain why HK = H K / H K for subgroups H and K of a finite group G. Here HK = {hk : h H, k K} is the set of products, which usually is just a subset (not a subgroup) of G. To count the size of HK, let the direct product group H K act on the set HK like this: (h, k) x = hxk 1. ).

14 14 KEITH CONRAD Check this gives a group action (the group is H K and the set is HK). There is only one orbit since e = ee HK and hk = (h, k 1 ) e. Therefore the orbit-stabilizer formula tells us H K HK = Stab e = H K {(h, k) : (h, k) e = e}. The condition (h, k) e = e means hk 1 = e, so Stab e = {(h, h) : h H K}. Therefore Stab e = H K and HK = H K / H K. Example We now discuss the original version of Lagrange s theorem in group theory. Here is what he proved: for any polynomial f(t 1,..., T n ) in n variables, the number of different polynomials we can obtain from f(t 1,..., T n ) through all permutations of its variables is a factor of n!. For instance, taking n = 3, consider the polynomial T 1. If we run through all six permutations of the set {T 1, T 2, T 3 }, and apply each to T 1, we get 3 different results: T 1, T 2, and T 3. The polynomial (T 1 T 2 )T 3 + (T 2 T 3 )T 1 + (T 3 T 1 )T 2 has only 2 possibilities under any change of variables: itself or its negative (check this for yourself). The polynomial T 1 + T2 2 + T 3 3 has 6 different possibilities. The number of different polynomials we find in each case is a factor of 3!. To explain Lagrange s general observation, we apply the orbit-stabilizer formula to the group action in Example 2.5. That was the action of S n on n-variable polynomials by permutations of the variables. For an n-variable polynomial f(t 1,..., T n ), the different polynomials we obtain by permuting its variables are exactly the polynomials in its S n - orbit. Therefore, by the orbit-stabilizer formula, the number of polynomials we get from f(t 1,..., T n ) by permuting its variables is [S n : H f ], where H f = {σ S n : σ f = f}. This index is a divisor of n!. In a group action, the length of an orbit divides G, but the number of orbits usually does not divide G. For example, S 4 has 5 conjugacy classes, and 5 does not divide 24. But there is an interesting relation between the number of orbits and the group action. Theorem Let a finite group G act on a finite set X with r orbits. Then r is the average number of fixed points of the elements of the group: r = 1 Fix g (X), G where Fix g (X) = {x X : gx = x} is the set of elements of X fixed by g. g G Don t confuse the set Fix g (X) with the fixed points for the action: Fix g (X) is only the points fixed by the element g. The set of fixed points for the action of G is the intersection of the sets Fix g (X) as g runs over the group. Proof. We will count {(g, x) G X : gx = x} in two ways. By counting over g s first we have to add up the number of x s with gx = x, so {(g, x) G X : gx = x} = g G Fix g (X). Next we count over the x s and have to add up the number of g s with gx = x, i.e., with g Stab x : {(g, x) G X : gx = x} = Stab x. x X

15 Equating these two counts and dividing by G gives Fix g (X) = Stab x. g G x X By the orbit-stabilizer formula, G / Stab x = Orb x, so Fix g (X) = G Orb x. g G x X GROUP ACTIONS 15 Divide by G : 1 Fix g (X) = 1 G Orb x. g G x X Let s consider the contribution to the right side from points in a single orbit. If an orbit has n points in it, then the sum over the points in that orbit is a sum of 1/n for n terms, and that is equal to 1. Thus the part of the sum over points in an orbit is 1, which makes the sum on the right side equal to the number of orbits, which is r. Theorem 3.26 is often called Burnside s lemma, but it is not due to him [3]. He included it in his widely read book on group theory. Example We will use a special case of Theorem 3.26 to prove for all a Z and m Z + that m (3.2) a (k,m) 0 mod m. k=1 When m = p is a prime number, the left side is (p 1)a + a p = (a p a) + pa, so (3.2) becomes a p a mod p, which is Fermat s little theorem. Thus (3.2) can be thought of as a generalization of Fermat s little theorem to all moduli that is essentially different from the generalization called Euler s theorem, which says a ϕ(m) 1 mod m if (a, m) = 1: (3.2) is true for all a Z. Our setup leading to (3.2) starts with any finite group G and comes from [2]. For a positive integer a, G acts on the set of functions Map(G, {1, 2,..., a}) by (g f)(h) = f(g 1 h). This is a special case of the group action at the end of Example 2.7, where G acts on itself by left multiplication. We want to apply Theorem 3.26 to this action, so we need to understand the fixed points (really, fixed functions) of each g G. We have g f = f if and only if f(g 1 h) = f(h) for all h G, which is the same as saying f is constant on every left coset g h in G. The number of left cosets of g in G is [G : g ] = m/ord(g), where m = G and ord(g) is the order of g, so the number of functions fixed by g is a m/ord(g), since the value of the function on each coset can be chosen arbitrarily in {1,..., a}. Therefore Theorem 3.26 implies (1/m) g G am/ord(g) is a positive integer, so (3.3) a m/ord(g) 0 mod m. g G Since (3.3) depends on a only through the value of a mod m, it holds for all a Z, not just a > 0. Taking G = Z/(m), any k G has additive order m/(k, m), so (3.3) becomes m a (k,m) 0 mod m. k=1

16 16 KEITH CONRAD Next we turn to the idea of two different actions of a group being essentially the same. Definition Two actions of a group G on sets X and Y are called equivalent if there is a bijection f : X Y such that f(gx) = g(f(x)) for all g G and x X. Actions of G on two sets are equivalent when G permutes elements in the same way on the two sets after matching up the sets appropriately. When f : X Y is an equivalence of group actions on X and Y, gx = x if and only if g(f(x)) = f(x), so the stabilizer subgroups of x X and f(x) Y are the same. Example Let R act on a linear subspace Rv 0 R n by scaling. This is equivalent to the natural action of R on R by scaling: let f : R Rv 0 by f(a) = av 0. Then f is a bijection and f(ca) = (ca)v 0 = c(av 0 ) = cf(a) for all c in R and a R. Example Let S 3 act on the conjugacy class {(12), (13), (23)} by conjugation. This action on a 3-element set, described in Table 1 below, looks like the usual action of S 3 on {1, 2, 3} if we identify (12) with 3, (13) with 2, and (23) with 1. Therefore this action of S 3 on {(12), (13), (23)} by conjugation is equivalent to the natural action of S 3 on {1, 2, 3}. π π(12)π 1 π(13)π 1 π(23)π 1 (1) (12) (13) (23) (12) (12) (23) (13) (13) (23) (13) (12) (23) (13) (12) (23) (123) (23) (12) (13) (132) (13) (23) (12) Table 1. Example Let GL 2 (R) act on the set B of ordered bases (e 1, e 2 ) of R 2 in the natural way: if A GL 2 (R) then A(e 1, e 2 ) := (Ae 1, Ae 2 ) is another ordered basis of R 2. This action of GL 2 (R) on B is equivalent to the action of GL 2 (R) on itself by left multiplication. The reason is that the two columns of a matrix in GL 2 (R) are a basis of R 2 (with the first and second columns being an ordering of basis vectors: the first column is the first basis vector and the second column is the second basis vector) and two square matrices multiply through multiplication on the columns: A( a c d b ) = (A( ( a c) A b d) ). Letting f : B GL2 (R) by f( ( ( a c), b ) d ) = ( a b c d ) gives a bijection and f(a(e 1, e 2 )) = A f(e 1, e 2 ) for all A GL 2 (R) and (e 1, e 2 ) B. Example Let H and K be subgroups of G. The group G acts by left multiplication on G/H and G/K. If H and K are conjugate subgroups then these actions are equivalent: write K = g 0 Hg0 1 and let f : G/H G/K by f(gh) = gg0 1 K. This is well-defined since, for h H, ghg0 1 K = ghg 1 0 g 0Hg0 1 = ghg0 1 = gg0 1 K. The reader can check f(g(g H)) = gf(g H) for g G and g H G/H, and f is a bijection. If H and K are non-conjugate then the actions of G on G/H and G/K are not equivalent: corresponding points in equivalent actions have the same stabilizer subgroup, but the stabilizer subgroups of left cosets in G/H are conjugate to H and those in G/K are conjugate to K, and none of the former and latter are equal.

17 GROUP ACTIONS 17 The left multiplication action of G on a left coset space G/H has one orbit. It turns out all actions with one orbit are essentially of this form: Theorem An action of G with one orbit is equivalent to the left multiplication action of G on a left coset space. Proof. Suppose that G acts on X with one orbit. Fix x 0 X and let H = Stab x0. Every x X has the form gx 0 for some g G, and all elements in a left coset gh have the same value at x 0 : for all h H, (gh)(x 0 ) = g(hx 0 ) = g(x 0 ). Let f : G/H X by f(gh) = gx 0. This is well-defined, as we just saw. Moreover, f(g g H) = gf(g H) since both sides equal gg (x 0 ). We will show f is a bijection, so the action of G on X is equivalent to the left multiplication action of G on G/H. Since X has one orbit, X = {gx 0 : g G} = {f(gh) : g G}, so f is onto. If f(g 1 ) = f(g 2 ) then g 1 x 0 = g 2 x 0, so g2 1 g 1x 0 = x 0. Since x 0 has stabilizer H, g2 1 g 1 H, so g 1 H = g 2 H. Thus g is one-to-one. A particular case of Theorem 3.33 says that an action of G is equivalent to the left multiplication action of G on itself if and only if the action has one orbit and the stabilizer subgroups are trivial. Definition The action of G on X is called free when every point has a trivial stabilizer. Example The left multiplication action of a group on itself is free with one orbit. Example The antipodal action of Z/(2) on a sphere (where the nontrivial element acts by negation) is a free action. There are uncountably many orbits. Free actions show up quite often in topology, and Example 3.36 is a typical illustration of that. Example For an integer n 2, let X n be the set of roots of unity of exact order n in C, so X n = ϕ(n). (For instance, X 4 = {i, i}.) The group (Z/(n)) acts on X n by a ζ = ζ a. Since every element of X n is a power of every other element of X n using exponents relatively prime to n, this action of (Z/(n)) has a single orbit. Since ζ a = ζ only if a 1 mod n (ζ has order n), stabilizers are trivial. Thus (Z/(n)) acting on X n is equivalent to the multiplication action of (Z/(n)) on itself, except there is no naturally distinguished element of X n while 1 is a distinguished element of (Z/(n)). It is worth comparing faithful and free actions. An action is faithful (Definition 2.15) when g 1 g 2 g 1 x g 2 x for some x X (different elements of G act differently at some point) while an action is free when g 1 g 2 g 1 x g 2 x for all x X (different elements of G act differently at every point). Since g 1 x = g 2 x if and only if g2 1 g 1x = x, we can describe faithful and free actions in terms of fixed points: an action is faithful when each g e has Fix g (X) X while an action is free when each g e has Fix g (X) =. 4. Actions of p-groups The action of a group of prime power size has special features. When G = p k for a prime p, we call G a p-group. For example, D 4 is a 2-group. Because all subgroups of a p-group have p-power index, the length of an orbit under an action by a p-group is a multiple of p unless the point is a fixed point, when its orbit has length 1. This leads to an important congruence modulo p when a p-group is acting.

18 18 KEITH CONRAD Theorem 4.1 (Fixed Point Congruence). Let G be a finite p-group acting on a finite set X. Then X {fixed points} mod p. Proof. Let the different orbits in X be represented by x 1,..., x t, so Corollary 3.18 leads to (4.1) X = t Orb xi. i=1 Since Orb xi = [G : Stab xi ] and G is a power of p, Orb xi 0 mod p unless Stab xi = G, in which case Orb xi has length 1, i.e., x i is a fixed point. Thus when we reduce both sides of (4.1) modulo p, all terms on the right side vanish except for a contribution of 1 for each fixed point. That implies X {fixed points} mod p. Keep in mind that the congruence in Theorem 4.1 holds only for actions by groups with prime-power size. When a group of size 9 acts we get a congruence mod 3, but when a group of size 6 acts we do not get a congruence mod 2 or 3. Corollary 4.2. Let G be a finite p-group acting on a finite set X. If X is not divisible by p, then there is at least one fixed point in X. If X is divisible by p, then the number of fixed points is a multiple of p (possibly 0). Proof. When X is not divisible by p, neither is the number of fixed points (by the fixed point congruence), so the number of fixed points can t equal 0 (after all, p 0) and thus is 1. On the other hand, when X is divisible by p, then the fixed point congruence shows the number of fixed points is 0 mod p, so this number is a multiple of p. Example 4.3. Let G be a p-subgroup of GL n (Z/(p)), where n 1. Then there is a nonzero v (Z/(p)) n such that gv = v for all g G. Indeed, because G is a group of matrices it naturally acts on the set V = (Z/(p)) n. (The identity matrix is the identity function and g 1 (g 2 v) = (g 1 g 2 )v by the rules of matrix-vector multiplication.) Since the set V has size p n 0 mod p, the number of fixed points is divisible by p. The number of fixed points is at least 1, since the zero vector is a fixed point, so the number of fixed points is at least p. A non-zero fixed point for a group of matrices can be interpreted as a simultaneous eigenvector with eigenvalue 1. These are the only possible simultaneous eigenvectors for G in (Z/(p)) n since every element of G has p-power order and the only element of p-power order in (Z/(p)) is 1 (so a simulatenous eigenvector for G in (Z/(p)) n must have eigenvalue 1 for each element of the group). Theorem 4.1 can be used to prove existence theorems involving finite groups (nonconstructively) if we can interpret a problem in terms of fixed points. For example, an element of a group is in the center precisely when it is a fixed point for the conjugation action of the group on itself. Thus, if we want to show some class of groups has non-trivial centers then we can try to show there are fixed points other than the identity element for the conjugation action.

ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

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