SOLUTIONS Math A4900 Homework 5 10/4/2017 1. (DF 2.2.12(a)-(d)+) Symmetric polynomials. The group S n acts on the set tx 1, x 2,..., x n u by σ x i x σpiq. That action extends to a function S n ˆ A Ñ A, where A Zrx 1,..., x n s (polynomials in n variables with integer coefficients), defined by σ ppx 1,..., x n q ppx σp1q,..., x σpnq q. For example, p12q p2x 1 x 2 ` 5x 2 x 2 3 q 2x 2 x 1 ` 5x 1 x 2 3. (a) Let p 12x 5 1x 7 2x 4 18x 6 2x 3 ` 11x 1 x 2 x 3 x 4, σ p1234q, and τ p123q. Compute τ pσ pq, and pτσq p. Answer. We have σ p 12x 5 2x 7 3x 1 18x 6 3x 4 ` 11x 2 x 3 x 4 x 1 12x 1 x 5 2x 7 3 18x 6 3x 4 ` 11x 1 x 2 x 3 x 4 τ pσ pq 12x 2 x 5 3x 7 1 18x 6 1x 4 ` 11x 2 x 3 x 1 x 4 12x 7 1x 2 x 5 3 18x 6 1x 4 ` 11x 1 x 2 x 3 x 4 pτσq p. (b) Prove that this function is a group action of S n on Zrx 1,..., x n s. Proof. The left action of S n on rns induces an action on tx 1, x 2,..., x n u (via the isomorphism S n S tx1,x 2,...,x nu S tx1,x 2,...,x nuq. So for ppx 1, x 2,..., x n q P Zrx 1,..., x n s, 1 ppx 1, x 2,..., x n q ppx 1 1, x 1 2,..., x 1 n q ppx 1, x 2,..., x n q and g ph ppx 1, x 2,..., x n qq g ppx h 1, x h 2,..., x h n q g ppx g ph 1q, x g ph 2q,..., x g ph nq q ppx gh 1, x gh 2,..., x gh n q by the S n action on tx 1,..., x n u pghq ppx 1, x 2,..., x n q. 1
2 (c) Fix n 4. (i) Calculate the stabilizer of x 4 in S 4, and show that it is a subgroup isomorphic to S 3. Answer. If σ stabilizes x 4, then it must stabilize 4. So ps 4 q x4 t1, p12q, p13q, p23q, p123q, p132qu. This is exactly the set of permutations in S 4 that fix 4 and 5, which is naturally isomorphic to S 3 (via restriction of these bijections to bijections on r3s see 2(b)(i)). (ii) Show that the stabilizer of x 1 ` x 2 in S 4 is an abelian subgroup of order four. Answer. If σ stabilizes x 1 ` x 2, then it must setwise fix t1, 2u. So ps 4 q x1`x 2 t1, p12q, p34q, p12qp34qu xp12q, p34qy. Since the generators commute, this is an abelian group (of order 4). (iii) Give an example of a (non-constant) polynomial whose stabilizer is all of S 4. (Aside: We call such polynomials symmetric polynomials.) Answer. Examples: x 1 x 2 x 3 x 4, x 1 ` x 2 ` x 3 ` x 4, x 1 x 2 ` x 1 x 3 ` x 1 x 4 ` x 2 x 3 ` x 2 x 4 ` x 3 x 4.
2. (DF 1.7.6+) Groups actions and homomorphism. Recall that for a group G and a set A, is a group action if and only if G ˆ A Ñ A written pg, aq ÞÑ g a σ : G Ñ S A defined by σ g paq g a is a homomorphism (where σ g σpgq). We call σ the homomorphism corresponding to the group action G ü A. (a) Briefly show that G acts faithfully on A if and only if the kernel of the action is trivial, and that this happens if and only if the kernel of the corresponding homomorphism is trivial. Proof. A group G acts faithfully on a set A, by definition, if each g action induces a distinct permutation. Namely, every permutation is distinct from the action of 1. This is equivalent to the statement that for all g P G, there is some a P A such that g a 1 a a. In other words, the set tg P G g a a for all a P Au is just t1u, i.e. trivial. Conversely, if the kernel is trivial, and g a h a for all a P A, then ph 1 gq a h 1 pg aq h 1 ph aq ph 1 hq a a for all a P A. So h 1 g is in the kernel, which is trivial, so h g. In terms of the homomorphism, the kernel of the action is the same as the set tg P G g a a@a P Au tg P G σ g paq a@a P Au tg P G σ g 1u kerpσq. (b) Fix s P A. Recall that S A S A. (i) Prove that the set of permutations that fix s is a subgroup isomorphic to S A 1. [If this is not clear, then before you attempt this proof, consider the following examples: Let S 4 act on r4s t1, 2, 3, 4u, and calculate the set of permutations fixing 4, and then calculate the set of permutations fixing 2.] 3 Proof. If τ : A Ñ A is a bijection that fixes s P A, then the restriction of the map τ A s : A s Ñ A s is well-defined and bijective. So letting H tτ P S A τpsq su Ă S A, we have the map H Ñ S A s defined by τ ÞÑ τ A s is well-defined, bijective, and a homomorphism (which, incidentally, implies that H is a subgroup). So H S A s S A s S A 1.
(ii) Use (i) to redefine G s in terms of the corresponding homomorphism Explain why, in terms of the framework of homomorphisms (rather than of group actions as we did in class), that we expect the stabilizer G s to be a subgroup of G. Answer. We have G s tg P G g s su tg P G σ g psq su σ 1 phq, where H is the subgroup of S A from part (i). But the preimage of a subgroup is again a subgroup (the proof follows exactly analogously to the proof that the kernel is a subgroup), so G s ď G. 3. (DF 2.3,5) Cyclic groups (a) If x is an element of a finite group G and x G, prove that G xxy. Give an explicit example to show x G does not imply G xxy if G is an infinite group. Proof. Since x P G, we have xxy ď G. So if G is finite, G xxy G xxy G x 0. So G xxy H, implying G xxy. If G is not finite, G xxy is the difference of infinite orders, which is not well-defined. For example, in Z, 2 has infinite order, but x2y 2Z Z. (b) Prove that Z 2 ˆ Z 2 and Z 2 ˆ Z are not cyclic. Proof. Let x be a generator of Z 2. First, Z 2 ˆ Z 2 tp1, 1q, p1, xq, px, 1q, px, xqu where x 2 1. Since Z 2 ˆ Z 2 4, but all its non-identity elements have order 2, it must not be cyclic. For Z 2ˆZ, note that the binary operation in the first coordinate is and the binary operation in the second coordinate is `. We can show that this group is not cyclic in two ways: (1) Directly: Show that no one element generates all of Z 2 ˆ Z. Every element can be written as px, zq or p1, zq for some z P Z. If z 0, we have xpx, 0qy Z 2 ˆ t0u Z 2 ˆ Z. So assume z 0. Similarly, xp1, zqy ď t1u ˆ Z Z 2 ˆ Z. Finally, for z 0, xpx, zqy tpx a, azq a P Zu Z 2 ˆ Z because it does not contain px, 0q (if the second coordinate is 0, then a 0, so the first coordinate is x 0 1). So no one term generates all of Z 2 ˆ Z, implying that it is not cyclic. 4
(2) Using cyclic group theorems: Show Z 2 ˆ Z is infinite but not isomorphic to Z. Every element of Z either has order 1 (the identity) or has infinite order. On the other hand, px, 0q 2 px 2, 0 ` 0q p1, 0q 1, so px, 0q has order 2. Therefore there is no isomorphism Z Ñ Z 2 ˆ Z. (c) In class, we listed the cyclic subgroups of D 8. (i) Give an example of a proper subgroup of D 8 that is not cyclic. [Start by picking two elements and seeing what they generate; make sure they don t generate both r and s!] 5 Answer. The two subgroups of D 8 that are not cyclic are xs, r 2 y t1, s, sr 2, r 2 u and xsr, r 2 y t1, sr, r 2, sr 3 u. We know that neither is cyclic (1) because we re enumerated all cyclic groups, and (2) both have order 4, but neither has an order 4 element to generate the set. (ii) Draw the subgroup lattice of D 8. Answer. Continuing from part (i), we can calculate that these are the only two noncyclic subgroups as follows: If you add one more element to xry, it must be of the form sr k. But since r k P xry, we have s P xr, sr k y. So D 8 xs, ry ď xr, sr k y ď D 8, so ă r ą is maximal (is not properly contained in another proper subgroup). If you add one more element to xr 2 y, it s either of the form r 1, which will then generate xry; or sr k : xs, r 2 y t1, s, sr 2, r 2 u xsr 2, r 2 y, xsr, r 2 y t1, sr, sr 3, r 2 u xsr 3, r 2 y. If you add one more element to xsr k y t1, sr k u, it s either of the form r 1, which will then generate all of D 8, since it will contain s and r; similarly for adding sr k 1, since sr k sr k 1 s 2 r k`k 1 r; or r 2 or sr k, which will then generate xsr k, r 2 y. Finally, adding one more element to xs, r 2 y or xsr, r 2 y will result in generating the whole of D 8, as every choice falls into one of the cases above where the result contained both r and s.
6 So the full subgroup lattice is D 8 xs, r 2 y xry xsr, r 2 y xsy xsr 2 y xr 2 y xsry xsr 3 y 1 (d) List the subgroups of Z 45, giving a generator for each. Then draw the subgroup lattice of Z 45. Answer. The divisors of 45 3 2 5 are 1, 3, 5, 9, 15, and 45. For a 45, one generator of the unique cyclic subgroup of order a is 45{a. So the subgroups of Z 45 xxy are x1y, xx 15 y, xx 9 y, xx 5 y, xx 3 y, and xxy, where, since xx kd y ď xx d y for all k, d P Z, we have containment given by xxy Z 45 xx 3 y xx 5 y xx 9 y xx 15 y xx 45 1y 1
7 (e) Let Z 36 xxy. For which integers a does the map ψ a defined by ψ a : 1 Ñ x a extend to a well defined homomorphism from Z{48Z to Z 36? Can ψ a ever be a surjective homomorphism? [ Extend to means that since ψ a is a homomorphism, if you know the image of 1, then you also know the image of 1 ` 1 ` ` 1; and thus since 1 generates Z{48Z, you should know the image of every element of Z{48Z.] Answer. Note that the binary operation in the domain is addition, but the binary operation in the codomain is multiplication. So for ψ a to be well-defined, we need the relation ψ a p 1q 48 ψ a p 1 looooomooooon ` ` 1 q ψ a p0q 1. 48 terms The left-most term is pψ a p 1qq 48 px a q 48 x 48a (). So for ψ a to be well-defined, we need that x 48a 1. Since x 36, this means that 36 48a. Dividing both sides by the gcd p36, 48q 2 2 3 12, we get 36 48a ðñ 3 4a ðñ 3 a. So ψ a is well-defined if and only if a is a multiple of 3. However, if a is a multiple of 3, then xx a y ď xx 3 y Z 36, since 3 36. So no such homomorphism can be an isomorphism. (f) For a P Z, define σ a : Z n Ñ Z n by σ a pxq x a for all x P Z n. Show that σ a is an automorphism of Z n if and only if pa, nq 1. Proof. Since Z n xzy has presentation xz z n 1y, to check when σ a is a homomorphism, we only need to check when σ a p1q σ a pz n q. This is true for all a since σ a p1q 1 a 1 and σ a pz n q pz n q a 1 for all a. So what remains to be seen is when σ a is a bijection. Note that the image of σ a is σ a pxzyq σ a ptz l l P Zuq tpz l q a l P Zu tpz a q l l P Zu xz a y. So σ a is surjective (and therefore bijective since Z n is finite) exactly when z a is a generator of Z n (i.e. when xz a y Z n ). By Proposition 6 for 2.3, this happens exactly when pa, nq 1.
8 4. (DF 2.4.2,3) Generating groups (a) Show that if A Ď B, then xay ď xby. Proof. (Using the intersection of groups definition.) If A Ď B, then for all H containing B, H contains A as well. So xay č H Ď č H xby. AĎH BĎH HďG HďG (b) Give an example where A B, but xay xby. Answer. Since 1 P xay for any A Ď G, xay xt1u Y Ay. So, for example, A tru and B t1, ru in D 8. (c) Show that if H is an abelian subgroup of G, then xh, ZpGqy is abelian. Proof. (Using the closure definition.) By definition, anything in ZpGq commutes with anything in H. So, since H is abelian, the generators of xh, ZpGqy all pairwise commute. So words in those generators pairwise commute. Thus xh, ZpGqy is abelian. 5. (DF 2.4.14,15) Finitely generated groups A group is called finitely generated if there is a finite set A such that H xay. (a) Briefly show that every finite group and every cyclic group is finitely generated. Answer. If G is finite, the underlying set is finite and is a generating set for G. If G xxy is cyclic, it is generated by the finite set txu. (b) Prove that every finitely generated subgroup of Q is cyclic. [Show that if H ď Q is generated by the finite set A, then H ď x1{ky where k is the product of all the denominators that appear in A.] Answer. Let A Ă Q be finite, and let k be the product of all the denominators that appear in A. Then each a{b P A can be rewritten as a b ad k where d k{b (since k is the product of all the denominators, b k, so d is an integer). So a{b P xky, and thus A Ă x1{ky. Thus by problem 4(a), xay ď x1{ky. Since every subgroup of a cyclic group is cyclic, this implies that xay is cyclic.
9 (c) Show that Q is not cyclic, and conclude that Q is not finitely generated. Proof. Fix n{m P Q, n, mz, and consider xn{my tplnq{m l P ZZu. So a{b (a, b P Z in lowest terms) is in xn{my only if b m. So this set certainly does not contain 1{pn ` 1q P Q. Thus Q is not cyclic. (d) Give a proper subgroup of Q that is not cyclic. Answer. By part (b), any such subgroup cannot be finitely generated, so let s not even try. However, for a prime p, I claim that H ta{p l a P Z, l P Z ě0 u ň Q and H is not cyclic. To check, we see that 0 0{p 0 P H, so H H. Further, for a{p l, a{p l1 P H, a p l a apl1 a 1 p l P H. p l1 p l`l1 So H is a subgroup by the subgroup criterion. Further, using a similar reasoning as in the previous part, for any xn{my, there is some power L of p such that p L ą m (so that p L does not divide m), so that 1{p L R xn{my.