FIRST MIDTERM MATH 18.703, MIT, SPRING 13 You Have 80 Minutes

FIRST MIDTERM MATH 18.703, MIT, SPRING 13

You have 80 minutes. This test is closed book, closed notes, no calculators. There are 7 problems, and the total number of points is 100. Show all your work. Please make your work as clear and easy to follow as possible. Problem Points Score Points will be awarded on the basis of neatness, the use of complete sentences and the correct pre- 1 15 sentation of a logical argument. 2 15 Name: 3 15 Signature: 4 10 Student ID #: 5 20 6 10 7 10 8 10 Presentation 5 Total 100

1 1. (15pts) Give the deﬁnition of a group.

Solution: A group G is a set together with a rule of multiplication m: G × G −→ G, such that (1) multiplication is associative, that is, (gh)k = g(hk), for every g, h and k ∈ G. (2) there is an identity element e ∈ G, so that eg = ge = g for every g ∈ G. (3) every element g ∈ G has an inverse g−1, gg−1 = g−1g = e.

(ii) Let G be a group and let S be a subset of G. Give the deﬁnition of the subgroup generated by S.

Solution: The smallest subgroup which contains S.

(iii) Let G be a group and H a subgroup. Give the deﬁnition of a left coset.

Solution: Let g ∈ G. The left coset generated by g is gH = { gh | h ∈ H }.

1 2. (15pts) (i) Find the parity of the following element of S9, 123456789 . 435712694

Solution: This is equal to (1, 4, 7, 6, 2, 3, 5)(8, 9), a product of 6+1 transpositions. This permutation is odd.

(ii) Compute the conjugate of σ by τ, where σ = (1, 5, 6)(4, 3, 7, 2) and τ = (1, 5)(6, 3, 2).

Solution: (5, 1, 3)(4, 2, 7, 6).

(iii) Is it possible to conjugate σ to σ′, where σ = (1, 5, 4)(2, 3) and σ′ = (1, 5)(3, 4, 2)? If so, ﬁnd an element τ so that σ′ is the conjugate of σ by τ. Otherwise explain why it is impossible.

Solution: Yes, it is possible. 12345 τ = . 31524

2 3. (15pts) (i) Give a description of the group D3 of symmetries of a triangle.

o Solution: Let I be the identity, R rotation through 120 and let F1, 2 F2, F3 be the three ﬂips. Then G = { I,R,R ,F1,F2,F3 }.

(ii) List all subgroups of D3.

2 Solution: {e}, {I,Fi}, i = 1, 2, 3, {I,R,R }, and ﬁnally the whole of G.

(iii) Find the left cosets, for one subgroup of order two and one subgroup of order three.

Solution: Take H = {I,F1}. Then there are three left cosets,

[I]= H = {I,F1} =[F1]

[F2]= F2H = {F2,R} =[R] 2 2 [F3]= F3H = {F3,R } =[R ]

Take H = {I,R,R2}. Then there are two left cosets, [I]= H = {I,R,R2} =[R]=[R2]

[F1]= F2H = {F1,F2,F3} =[F2]=[F3].

3 4. (10pts) (i) Let S be a set and let A(S) be the set of permutations of S. Show that A(S) is a group where the multiplication law is composition of functions.

Solution: The composition of permutations is a permutation, so we do get a well-deﬁned multiplication. Composition of functions is associative, so multiplication is associative. The identity function IS : S −→ S plays the role of the identity, f ◦ IS = IS ◦ f. If f is a permutation then the inverse function f is a permutation and this plays the role of the inverse.

(ii) Let Aut(G) be the set of automorphisms of a group G. Show that Aut(G) is a subgroup of A(G), the permutations of the elements of G.

Solution: The identity is an automorphism so Aut(G) is not empty and we just have to show that the composition of two automorphisms is an automorphism and that the inverse of an automorphism is an automorphism. Suppose that σ and τ are two automorphisms of G. Let ρ = τσ = τ ◦ σ. If g and h ∈ G then ρ(gh)= τ(σ(gh)) = τ(σ(g)σ(h)) = τ(σ(g))τ(σ(h)) = ρ(g)ρ(h). Thus ρ is a homomorphism. We already know that that ρ is a bijection, so that ρ is an automorphism and so Aut(G) is closed under products. Let k = gh. As σ is a permutation, we may ﬁnd g′, h′ and k′ such that σ(g′)= g, σ(h′)= h and σ(k)= k′. As σ′ is a homomorphism, σ(g′h′)= σ(g′)σ(h′) = gh. Applying σ−1 to both sides we get g′h′ = k′, that is, σ−1(gh)= σ−1(g)σ−1(h), so that σ−1 is an automorphism and so Aut(G) is a subgroup.

4 5. (20pts) (i) Let a ∈ G. Prove that the map −1 σ = σa : G −→ G given as σ(g)= aga , is an automorphism of G. Solution: Let g, h ∈ G. Then σ(gh)= a(gh)a−1 =(aga−1)(aha−1) = σ(g)σ(h). Thus σ is a group homomorphism. It is easy to see that the inverse τ of σ is given by conjugation with a−1, τ(g) = a−1ga. Thus σ is an automorphism.

(ii) Let φ: G −→ A(G) be the map which sends a to φ(a)= σa. Show that φ is a group homomorphism. Solution: Let a, b ∈ G and c = ab. If g ∈ G then −1 σc(g)=(ab)g(ab) = a(bgb−1)a−1

= σa(σb(g)).

Thus σc = σaσb and so φ(ab)= φ(a)φ(b) and φ is a homomorphism. (iii) Show that the image H = φ(G) is isomorphic to G/Z, where Z is the centre of G. Solution: Suppose that a ∈ G is in the kernel of φ. Then aga−1 = a for all g ∈ G, so that ag = ga and so a ∈ Z. Conversely if a ∈ Z then ag = ga for all g ∈ G so that aga−1 = g and so a belongs to the kernel. Thus the kernel of φ is Z and H ≃ G/Z by the ﬁrst isomorphism theorem. (iv) Show that H is normal in Aut(G). Solution: Let a ∈ G and ψ ∈ Aut(G). Set b = ψ(a). If g ∈ G then g = τ(h), some h ∈ G and −1 (τσaτ )(g)= τ(σa(h)) = τ(aha−1) = τ(a)τ(h)τ(a−1) −1 = bgb = σb(g). −1 Thus τσaτ = σb ∈ H and so H is normal in Aut(G).

5 6. (10pts) State and prove Fermat’s Theorem (in the proof of Fermat’s Theorem, you may assume Lagrange’s Theorem).

Solution: Fermat’s Theorem states that if p is a prime number and a is any integer, then ap = a mod p. First note that if a is not coprime to p, then p divides a. In this case a =0 mod p and the equation above is trivially satisﬁed, as it reads zero is equal to zero. So now suppose that p is coprime to a. We are going to prove more, p−1 that a =1 mod p. Let G = Up be the group of units mod p. Then G has p − 1 elements. Now g =[a] deﬁnes an element of G and if k is the order of g, then by Lagrange, k divides the order of G, that is, k divides p − 1. But the order of an element of a group is equal to the smallest k such that gk = e. As k divides p − 1, there is an m such that km = p − 1 and it follows that gp−1 = e. But then ap−1 =1 mod p.

6 7. (10pts) Let G be a group and let Z be its centre. Prove that if G/Z is cyclic, then G is abelian.

Solution: Suppose that G/Z is generated by aZ. Pick x and y in G. Then xZ = aiZ and yZ = ajZ, for some i and j and there are z and z′ in Z such that x = aiz and y = ajz′. We have xy = aizajz′ = aiajzz′ = ai+jzz′. Similarly yx = ai+jzz′. Thus xy = yx and G is abelian.

7 Bonus Challenge Problem 8. (10pts) Give an example of a countable group that is not ﬁnitely generated (that is, a group which is not generated by any ﬁnite subset).

Solution: There are two natural examples. The ﬁrst is to look at the rational numbers under addition. Q is cer- tainly countable. Suppose that g1,g2,...,gk were a ﬁnite set of generators. Each gi is a rational number, say of the form a i . bi

Now let b be the least common multiple of the b1,b2,...,bk. Then any rational number which is a finite sum or difference of the g1,g2,...,gk will be of the form a , b for some integer a. But most rationals are not of this form. Thus Q is not finitely generated. The second is to look at the group A(N) of permutations of the natural numbers. Now this is not countable, but consider the subgroup G consisting of all permutations that fix all but finitely many natural numbers. Note that A(N) contains a nested sequence of copies of Sn, for all n, in an obvious way and that G is in fact the union of these finite subgroups. In particular G is countable, as it is the countable union of countable sets. Now suppose that g1,g2,...,gk were a finite set of generators. Then in fact there is some n such that gi ∈ Sn, for all i. As Sn is a subgroup of G, it follows that

hg1,g2,...,gki⊂ Sn =6 G, a contradiction. Put differently, no finite subset generates G, since any finite subset will only permute finitely many natural numbers.