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Homework Two: Group Actions and Cosets

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Homework Two: Group Actions and Cosets Homework Two: Group actions and cosets Though we're still in the midst of exploring what in the world groups and their homomorphisms can be, we've learned a fairly sophisticated no- tion: group actions. We need to get used to these (hence the homework!). For the record, we recall four definitions. Definition 0.1 (Group actions). Let G be a group and X some set. A left group action of G on X is a function µ : G × X ! X satisfying: (1) For every x 2 X, µ(e; x) = x. (2) For every g; h 2 G and x 2 X, µ(gh; x) = µ(g; µ(h; x)). Writing lazily µ(g; x) =: gx, we can re-write these properties as: (1) For every x 2 X, ex = x. (2) For every g; h 2 G and x 2 X,(gh)x = g(hx). Likewise, a right group action is a function µ : X × G ! X such that (1) xe = x and (2)( xg)h = x(gh). Example 0.2. Any group G acts on itself by taking µ = m. Condition (1) above is implied by having an identity, and (2) above is implied by associativity. In fact, G acts on itself both on the right, and on the left, in either case taking µ = m. A much more interesting example, as it turns out, is given by fixing a subgroup H ⊂ G. Then the multiplication map H × G ! G; (h; g) 7! hg is a left action of H on G. Likewise, G × H ! G; (g; h) 7! gh is a right action of H on G. 1 Fall 2017 Math 122 Homework Example 0.3. A special example is given by the subgroup S1 ⊂ C×, where C× is the group of non-zero complex numbers under multiplication. Then the picture of the action 1 × × iθ iθ S × C ! C ; (e ; z) 7! e z is suggestive of the terminology \orbit." Definition 0.4 (Orbits). Given a left action, G × X ! X, the orbit of an element x 2 X is the collection Ox := fy 2 X such that y = gx for some g 2 G g: Likewise, given a right action X × G ! X, the orbit of an element x 2 X is the collection Ox := fy 2 X such that y = xg for some g 2 G g: Example 0.5. In the example of the circle acting on C×, an orbit of a complex number z 2 C× is the collection of all complex numbers with norm jjzjj. For no good reason whatsoever (except history), we have a different term for an orbit when we have an action of a subgroup H ⊂ G on G. Definition 0.6 (Cosets). Fix a subgroup H ⊂ G. Note there is a right action of H on G. Then the orbit of g 2 G is (confusingly) called the left coset of H with respect to g. You can just call it \the coset of g" for short. This orbit is denoted gH: To be explicit, gH is the set of all elements g0 of G such that g0 = gh for some h 2 H. Example 0.7. The coset eH is equal to the set H. Note also that if h 2 H (and hence h 2 G), we have that hH also equals H. This helps illuminate a frequent notational confusion: even if g 6= g0, it is possible that gH = g0H; this happens if and only if g and g0 are in the same orbit. Definition 0.8 (G=H). Let H ⊂ G be a subgroup. Then G=H is the set of cosets of the right action of H on G. (I.e., the set of orbits of the right action.) Remark 0.9. When G is abelian, there is no difference between right and left actions of a subgroup H on G. 2 Fall 2017 Math 122 Homework 1. Subgroups of the integers, and their cosets Fix an integer n. Let nZ be the set of all integers that are divisible by n. (I.e., the set of all integers of the form an, where a is an integer|positive or negative or zero.) Note that nZ = (−n)Z, and that 0Z is a set with one element: zero. (a) Show that for any integer n, nZ is a subgroup of Z (under addition). (b) Conversely, show that any subgroup of Z is of the form nZ. (Hint: If a subgroup H is not the zero subgroup, what's the smallest positive number in H?) (c) Let H = nZ be a subgroup of Z. Write out all the elements of Z=H. (Recall each element of Z=H is an orbit; write out each one. Make sure you delineate between the n = 0 and n 6= 0 cases.) Is this related to the \Z=nZ" we wrote down in class? 2. Orbits and stabilizers Let G be a group, X some set, and fix a left group action G × X ! X. (Throughout, remember that inverses are your friend; as are the two defining properties of a group action.) (a) Fix x 2 X. Let Gx ⊂ G be the subset of those g such that gx = x. Show that Gx is a subgroup of G. (b) Suppose x 2 X and y 2 X are in the same orbit. Exhibit a group isomorphism between Gx and Gy. (c) For any x 2 X, exhibit a bijection between G=Gx and the orbit Ox of x. 3. Sizes of cosets Fix a group G and a subgroup H. (a) Let g1H and g2H be two arbitrary cosets. Exhibit a bijection between them. (Recall that g1H is the set of all elements of G that equal g1h for some h 2 H.) (Remark: Note this exercise means that any two 3 Fall 2017 Math 122 Homework orbits of the (right) action of H on G have are in bijection; but for more general group actions, orbits may have different sizes!) (b) If G is a finite group, show that jHj divides jGj. (Hint, H acts on G, and any set with a group action is a union of its orbits.) This statement is called Lagrange's Theorem, and you have proven it. 4. (Optional) Two proper subgroups cannot span This problem proves a (perhaps?) surprising fact about any group G: If H and K are two proper subgroups, their union H [ K cannot equal G. (a) Let X and Y be two sets, each with a G action. Fix two elements x0 2 X and y0 2 Y , and suppose that for every g 2 G, either x0 or y0 is fixed by g. Prove that either x0 or y0 is actually fixed by every element of g 2 G. (b) Let H ⊂ G and K ⊂ G be two proper subgroups of G. (That is, H 6= G and K 6= G.) Show that H [ K 6= G. (Note: However, elements of H and K may generate G. Generation is a notion we haven't defined yet, but you might be able to guess what it means.) (c) Exhibit a group G such that, for three proper subgroups H; K; L, we have that H [ K [ L = G. 5. (Optional) SU(2) and unit quaternions This problem, strictly speaking, has nothing to do with group actions or cosets. It's just a famous and surprising group isomorphism|the sphere in R4 admits a group structure, and it's isomorphic to SU(2). The group SU(2) comes up in physics|certain homomorphisms out of SU(2) are called spin representations; SU(2) is also sometimes called Spin(3) in the literature. (a) Consider the set S3 of vectors in R4 that have norm one. For conve- 4 nience, we will write an element of R by the couple (a0;~a) where a0 is a real number and ~a is a vector in R3. For two such elements, we define ~ ~ ~ ~ (a0;~a)(b0; b) := (a0b0 − ~a · b ; a0b + b0~a + ~a × b): 4 Fall 2017 Math 122 Homework Here, the · is the usual dot product, and × is the cross product of 3 ~ vectors in R . Show that if both (a0;~a) and (b0; b) have unit norm, so does their product. (b) Show that (1;~0) is the unit for this operation. (c) Show that (a0; −~a) is the inverse to (a0;~a). Don't bother showing associativity yet. (d) On the other hand, let SU(2) denote the set of 2x2 matrices A with complex number entries, such that A∗A = 1 and det A = 1. Here, A∗ is the matrix obtained by conjugating the entries of A and then transposing. Noting that (AB)∗ = B∗A∗, show SU(2) is a group under matrix multiplication. (e) Show that any element of SU(2) is of the form x −y y x where x and y are complex numbers, and jxj2 + jyj2 = 1. (f) Letting ~i;~j;~k be the usual basis vectors for R3, define the map φ : S3 ! SU(2) ~ by sending a unit vector (a0; a1~i + a2~j + a3k) to the matrix 1 0 i 0 0 1 0 i a + a + a + a : 0 0 1 1 0 −i 2 −1 0 3 i 0 Show that φ is a bijection. (g) Show that φ satisfies φ(gh) = φ(g)φ(h) for any g; h 2 S3. (h) Conclude that the multiplication we defined on S3 is associative. (Note that S3 is hence a group; working backwards, the two previous parts show that φ is a group isomorphism.) 5.
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