MATH 3005 ABSTRACT ALGEBRA I FINAL SOLUTION Write Your Answer Neatly and Show Steps. Any Electronic Devices Including Calculator

MATH 3005 ABSTRACT ALGEBRA I FINAL SOLUTION

SPRING 2014 - MOON

Write your answer neatly and show steps. Any electronic devices including calculators, cell phones are not allowed. (1) Write the deﬁnition. (a) (3 pts) Cyclic group. A group G is cyclic if it is generated by an element a ∈ G, i.e., G = hai. Equivalently, G is cyclic if there is a ∈ G such that G = {an | n ∈ Z}.

(b) (3 pts) The external direct product of two groups. The external direct product G⊕H of two groups G and H is a group deﬁned as G ⊕ H = {(a, b) | a ∈ G, b ∈ H}.

The binary operation is deﬁned by (a1, b1) ◦ (a2, b2) = (a1a2, b1b2).

(c) (3 pts) For a group G and a normal subgroup N, the quotient group G/N. (Describe it as a set and provide the deﬁnition of its binary operation.) As a set, the quotient group is G/N = {aN | a ∈ G}, the set of all distinct cosets. The binary operation is deﬁned by aN ◦ bN = (ab)N.

(d) (3 pts) The kernel of a group homomorphism. For a group homomorphism φ : G → G, ker φ = {g ∈ G | φ(g) =e ¯} where e¯ is the identity of G.

(e) (3 pts) For a permutation group G on a set S and a ∈ S, the stabilizer of a. The stabilizer is stab(a) = {g ∈ G | g(a) = a}.

Date: May 7, 2014. 1 MATH 3005 Final Spring 2014 - Moon (2) (6 pts) For a group G and its subgroup H, the normalizer of H is deﬁned as N(H) = {g ∈ G | ghg−1 ∈ H for all h ∈ H}. Show that N(H) is a subgroup of G. For the identity e ∈ G, ehe−1 = h ∈ H for all h ∈ H. So e ∈ N(H) and N(H) 6= ∅. −1 0 If g1, g2 ∈ N(H), then for any h ∈ H, g1hg1 = h ∈ H. Then −1 −1 −1 0 −1 0 g1g2h(g1g2) = g1g2hg2 g1 = g1h g1 . Because h ∈ H and g1 ∈ N(H), 0 −1 g1h g1 ∈ H. Therefore g1g2 ∈ N(H). Finally, if g ∈ N(H) and h ∈ H, then ghg−1 = h0 for some h0 ∈ H, so gh = h0g. Then g−1h(g−1)−1 = g−1hg = g−1gh0 = h0 ∈ H. Therefore g−1 ∈ N(H) and N(H) ≤ G. • Showing the nonemptyness: 1 pt. • Proving the closedness: 3 pts. • Showing the existence of inverse: 2 pts.

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(3) (a) (5 pts) In S7, what is the maximum order of an element?

If σ = σ1σ2 ··· σk ∈ Sn is a product of disjoint cycles σ1, σ2, ··· , σk, then |σ| = lcm(|σ1|, |σ2|, ··· , |σk|). For σ ∈ S7, if we describe σ as a product of disjoint cycles, then the length distributions of them are one of 7, 6 + 1, 5 + 2, 5 + 1 + 1, 4 + 3, 4 + 2 + 1, 4 + 1 + 1 + 1, 3 + 3 + 1, 3 + 2 + 2, 3 + 2 + 1 + 1, 3 + 1 + 1 + 1 + 1, 2 + 2 + 2 + 1, 2 + 2 + 1 + 1 + 1, 2+1+1+1+1+1, 1+1+1+1+1+1+1. In each case the least common multiple is 7, 6, 10, 5, 12, 4, 4, 3, 6, 6, 3, 2, 2, 2, 1. Therefore the maximum order is 12. • Indicating the fact that the order of σ is the least common multiple of the length of disjoint cycles: 2 pts. • Getting the answer 12: +3 pts. • If there is no explanation of the answer, -2 pts.

(b) (4 pts) Find an element σ ∈ S7 with the maximum order. Write σ as a product of disjoint cycles and a product of 2-cycles.

By (a), when σ ∈ S7 is a product of two disjoint cycles σ1 and σ2 of length 4, 3 respectively, the length is 12. One example is σ = (1234)(567). If we write it as a product of 2-cycles, then σ = (14)(13)(12)(57)(56). • Finding an element of order 12 and writing it as a product of disjoint cycles: 2 pts. • Describing it as a product of 2-cycles: 2 pts.

σ−1 = (1432)(576)

(d) (4 pts) In A7, what is the maximum order of an element? A cycle of odd (resp. even) length is even (resp. odd). So a permutation of type 4 + 3, 5 + 2 is odd, but length 7 cycle is even. It has the maximum order 7.

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(4) (6 pts) Let G be a cyclic group of order 36. Draw the subgroup lattice of G. Let G = hai. Then any subgroup of G is cyclic and generated by ak for some positive divisor k of 36. Therefore the subgroups of G are G = hai, ha2i, ha3i, ha4i, ha6i, ha9i, ha12i, ha18i, {e} = ha36i. Note that hami ≤ hani if and only if n|m. Therefore the subgroup lattice of G is G

ha2i ha3i

ha4i ha6i ha9i

ha12i ha18i

{e}.

• Listing all subgroups: 2 pts. • Drawing incorrect paths or missing paths/nodes: -1 pt per 2 paths. • Using incorrect names for nodes: -2 pts. • Flipping the diagram vertically: -1 pt.

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(5) (7 pts) Prove that S4 6≈ S3 ⊕ Z4.

In S4, if one write σ ∈ S4 as a product of disjoint cycles, then its length can be decomposed 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1. So the maximum order of an element is 4. On the other hand, in S3 there is an element τ of order 3. In Z4, |1| = 4. Then |(τ, 1)| = lcm(|τ|, |1|) = 12. Therefore S2 6≈ S3 ⊕ Z4. • Explaining an idea to proof: 1 pt. • Finding the maximum order of one group: 3 pts each. • Alternatively, counting the number of elements of ﬁxed order in a group: 3 pts each.

(6) (7 pts) Show that a group of prime order is cyclic. Let G be a group of prime order p. If a ∈ G is nonidentity element, |a|||G| = p. Because a 6= e, |a|= 6 1 and |hai| = |a| = p. Therefore G = hai and G is cyclic. • Stating the deﬁnition of cyclic group: 1 pt. • Stating Lagrange’s theorem correctly: 2 pts. • Proving the result correctly: 7 pts.

(7) (9 pts) Let G be a group of order 55. Show that G has a normal subgroup of order 11. Because 11 is a prime divisor of |G| = 55, there is an element a ∈ G with |a| = 11. Then hai is a subgroup of order 11. Let H and K be two subgroups of order 11. Then H ∩ K is a subgroup of H, so |H ∩ K|||H| = 11. Therefore |H ∩ K| is 1 or 11. But |H||K| 121 55 = |G| ≥ |HK| = = . |H ∩ K| |H ∩ K| Thus |H ∩ K| = 1 is impossible. So |H ∩ K| = 11 and H = K. In other words, there is a unique subgroup of order 11 in G. Let H be the subgroup. For any element g ∈ G, gHg−1 is a subgroup of G and |gHg−1| = |H| = 11. From above observation, gHg−1 = H. This implies that H/G. • Showing the existence of order 11 subgroup: 3 pts. • Proving that an order 11 subgroup is unique: 3 pts. • Showing that the unique order 11 subgroup is normal: 3 pts.

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(8) Let p be a prime and let G = Zp2 ⊕ Zp2 . (a) (6 pts) Determine the number of elements of order p in G. 2 Note that a ∈ Zp2 has order 1, p, or p . |a| = p if and only if gcd(a, p2) = p2/p = p. Therefore there are p − 1 such elements, p, 2p, 3p, ··· , (p − 1)p. Let H be the set of such elements. For any (a, b) ∈ Zp2 ⊕ Zp2 , |(a, b)| = lcm(|a|, |b|). Therefore there are three possibilities of |(a, b)| = p. Case 1. |a| = p, |b| = 1. Then b = 0. So there are p − 1 such elements. {(a, 0) | a ∈ H} is the set of such elements. Case 2. |a| = 1, |b| = p. Similarly, there are p − 1 such elements. {(0, b) | b ∈ H} is the set of such elements. Case 3. |a| = |b| = p. In this case, {(a, b) | a, b ∈ H} is the set of such elements. So the number of elements is (p − 1)2. In summary, there are (p − 1)2 + 2(p − 1) = p2 − 1 elements of order p in G. • Listing all possibilities to get an order p elements: 3 pts. • Counting the number of elements of oder p in G: 3 pts.

(b) (6 pts) Find the number of subgroups of order p in G. Note that for any order p element a ∈ G, hai is a subgroup of order p. Therefore every order p element is in an order p subgroup. Con- versely, any nonidentity element in an order p subgroup has order p. So an order p subgroup has exactly p − 1 order p elements. Moreover, if H and K are two distinct order p subgroups, then |H ∩ K|||H| = p and |H ∩K|= 6 p. Therefore |H ∩K| = 1 and H ∩K = {e}. Therefore for each order p element, there is a unique subgroup of order p containing it. In summary, the number of subgroups of order p is (p2 −1)/(p−1) = p + 1.

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(9) (a) (6 pts) Show that Z5 ⊕ Z5/h(1, 1)i ≈ Z5. Because |(1, 1)| = lcm(|1|, |1|) = 5, |h(1, 1)i|. Therefore | ⊕ | 25 | ⊕ /h(1, 1)i| = Z5 Z5 = = 5. Z5 Z5 |h(1, 1)i| 5 A group of prime order is cyclic. Therefore it is a cyclic group of order 5, which is isomorphic to Z5.

• Computing the order of Z5 ⊕ Z5/h(1, 1)i: 3 pts. • Concluding that it is isomorphic to Z5: 3 pts.

(b) (6 pts) By using the first isomorphism theorem, show that ∗ GL(2, R)/SL(2, R) ≈ R . Define a map φ : GL(2, R) → R∗ as φ(A) = det A. Then φ(AB) = det(AB) = det A det B = φ(A)φ(B), so it is a homomorphism. For any c ∈ R∗, c 0 φ( ) = c. 0 1 Therefore φ is onto and φ(GL(2, R)) = R∗. Furthermore, A ∈ ker φ ⇔ det A = 1 ⇔ A ∈ SL(2, R). So ker φ = SL(2, R). By the first fundamental theorem, ∗ GL(2, R)/SL(2, R) = GL(2, R)/ ker φ ≈ φ(GL(2, R)) = R .

• Stating the ﬁrst isomorphism theorem: 1 pt. • Giving the homomorphism φ: 2 pts. • Showing that φ is a homomorphism: 1 pt. • Proving that φ is onto: 1 pt. • Indicating that ker φ = SL(2, R): 1 pt.

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(10) (10 pts) Determine the number of ways in which the vertices of a square can be colored with four colors. Two colorings are equivalent if one is obtained from another by applying a permutation induced by the sym- metry group D4 of the square. 0 ◦ Let D4 = {e, R90,R180,R270,H,V,D,D } where Rθ is the rotation by θ and H,V are horizontal/vertical flips, D,D0 are two diagonal flips. Let S be the set of all colorings. Then |S| = 44 = 256. For e, |fix(e)| = |S| = 256. For R90, if a coloring is in fix(R90), then all vertices must be the same color. So |fix(R90)| = 4. By the same idea, |fix(R270)| = 4, too. For R180, a coloring is in fix(R180) exactly when the opposite vertices 2 have the same color. Therefore |fix(R180)| = 4 = 16. A coloring is fixed by H when two adjacent vertices connected by a vertical edge have the same color. So |fix(H)| = 42 = 16. By the same token, we have |fix(V )| = 16. Finally, for two diagonal flips D and D0, there are two fixed vertices on the flipping axis and a pair of vertices which share the same color. Therefore |fix(D)| = |fix(D0)| = 43 = 64. By Burnside’s theorem, the number of orbits is 1 X 1 440 |fix(g)| = (256 + 4 + 16 + 4 + 16 + 16 + 64 + 64) = = 55. |D4| 8 8 g∈D4

• Stating Burnside’s theorem: 2 pts. • Getting the number of all possible colorings before identiﬁcation: 2 pts. • Counting the number of ﬁxed colorings for each element of D4: 1 pt each.