MA2202 Algebra I AY 2007/2008 Sem 2 NATIONAL UNIVERSITY OF SINGAPORE MATHEMATICS SOCIETY PAST YEAR PAPER SOLUTIONS with credits to He Jinxin MA2202 Algebra I AY 2007/2008 Sem 2 Question 1 (a) Firstly, note that since Z/(192 ) × Z/(172 ) is cyclic, it has unique subgroups of each order dividing 172 × 192 . Let H be a subgroup of Z/(172 ) × Z/(192 ), Hence |H| = 1 or 17 or 19 or 172 or 192 or 17 × 19 or 17 × 192 or 172 × 19 or 172 × 192 . So the subgroups of Z/(172 ) × Z/(192 ) are: {0} × {0}, H17 × {0}, {0} × H19 , H17 × H19 , Z/(172 ) × {0}, {0} × Z/(192 ), Z/(172 ) × H19 , H17 × Z/(192 ) and Z/(172 ) × Z/(192 ). (b) Let a, b ∈ Z/(3) such that Z/(3) × Z/(3) = hai × hbi. Then the subgroups of Z/(3) × Z/(3) are {0} × {0}, hai × {0}, {0} × hbi, h(a, b)i,h(2a, b)i, hai × hbi. Question 2 (i) Let the order of g 14 be n, then (g 14 )n = 1. Hence 14n ≡ 0 mod 30, which means 14n = 30k for some integer k. So 7n = 15k, since 3 | 15k and 5 | 15k but 3 - 7 and 5 - 7, we have 15 | n. Then since (g 14 )15 = g 210 = (g 30 )7 = 1, we have n | 15. Hence we have n = 15. i.e. o(g 14 ) = 15. (ii) ∀ m such that o(g m ) = o(g 14 ) = 15, then (g m )15 = 1 and (g m )k 6= 1 for k = 1, 2, ...14. So we have 15m ≡ 0 mod 30, then m ≡ 0 mod 2, we set m = 2l. And km 6≡ 0 mod 30, then 2lk 6≡ 0 mod 30, lk 6≡ 0 mod 15 for all k = 1, 2, ..., 14. Therefore, l is not divisible by 3 and 5. Hence l = 1 or 2 or 4 or 6 or 7 or 8 or 11 or 13 or 14. Hence the set of elements of G whose order is equal to o(g 14 ) is {2, 4, 8, 14, 16, 22, 26, 28}. Alternative solution: Since hgi is cyclic it has exactly 1 subgroup of order 15. Hence all elements of order 15 in hgi lie in this subgroup and are exactly the generator of this subgroup. Hence there are φ(15) = 8 such element. Let h = g 14 then hhi is the unique subgroup of order 15 and the generators of this subgroup are precisely elements of the form hm such that m ∈ {0, ..., 14} and gcd(m, 15) = 1. Hence set of elements of order 15 is {(g 14 )m | gcd(m, 15) = 1 and m ∈ {0, 1, ..., 14}}. NUS Math LaTeXify Proj Team Page: 1 of 6 NUS Mathematics Society MA2202 Algebra I AY 2007/2008 Sem 2 Question 3 (a) Since h(12)(34), (13)(24)i = {1, (12)(34), (13)(24), (14)(23)} is the subgroup of all 2, 2 cycles and conjugation preserves cycle structure. h(12)(34), (13)(24)i C S4 . (b) No.(15)(12)(34)(15) = (15)(34) ∈ / h(12)(34), (13)(24)i. (c) Claim: For any σ ∈ S3 V there exist unique p ∈ S3 ,q ∈ V such that pq = σ. Proof: Existence: Since S3 ⊆ S4 , for every α ∈ V we have αS3 = S3 α. Therefore we have S3 V = V S3 as groups. Hence for all σ ∈ S3 V there exist a finite sequence of p1 p2 , p3 , ..., pn ∈ S3 and q1 , q2 , q3 , ..., qn ∈ V such that p1 q1 p2 q2 ...pn qn = σ. Since S3 V = V S3 for each pi , qj such that i, j ∈ {1, 2, ..., n}. There exists p0i ∈ S3 and qj0 ∈ V such that pi qj = qj0 p0i . Therefore we may rearrange and rewrite p1 q1 p2 q2 ...pn qn to p01 p02 ...p0n q10 q20 ...qn0 such that p0i ∈ S3 and qj0 ∈ V with i, j ∈ {1, 2, ..., n} and σ = p1 q1 p2 q2 ...pn qn = (p01 p02 ...p0n )(q10 q20 ...qn0 ). Uniqueness: Suppose p, p0 ∈ S3 and q, q 0 ∈ V such that pq = p0 q 0 . Hence we have p0−1 p = q 0 q −1 . Hence q 0 q −1 ∈ S3 ∩ V . Hence we have q 0 q −1 (4) = 4 and q 0 q −1 = (1). Therefore q = q 0 and p = p0 . Hence we may define the map f : S3 V → S3 pq 7→ p where p ∈ S3 , q ∈ V such that pq = σ is the unique decomposition of σ.This is a well defined function by the previous claim. Claim:This map is a well-defined homomorphism. Proof: For any p1 q1 , p2 q2 ∈ S3 V , we have ¡ ¢ f (p1 q1 p2 q2 ) = f p1 p2 (p2−1 q1 p2 )q2 = p1 p2 since V C S3 V = = f (p1 q1 )f (p2 q2 ) Also, since f is surjective and ker(f ) = V . By First Isomorphism Theorem, we have S3 ∼ = S3 V /V . (d) Since S3 ∼ = S3 V /V , we have |S3 ||V | = |S3 V |. Hence |S3 V | = 24. Since S3 V ⊆ S4 and |S3 V | = |S4 |, we have S3 V = S4 . NUS Math LaTeXify Proj Team Page: 2 of 6 NUS Mathematics Society MA2202 Algebra I AY 2007/2008 Sem 2 Question 4 µ d e f g (a) For all A ∈ CG (P ), let A = . If AP = P A, then µ d e f g ¶ ¶µ ² 0 0 δ ¶ µ = ² 0 0 δ ¶µ d e f g ¶ Hence we have, eδ = ²e f ² = δf ² 6= δ which derives e = f = 0. ½µ Then CG (P ) = D2 (R) = d 0 0 g ¶ ¾ |d, g ∈ R . (b) Obviously Z(G) ⊃ CG (P ). That is the center Z(G) has the form µ ¶ d 0 0 g . If d 6= g, then µ 1 3 4 2 ¶µ d 0 0 g ¶ µ 6= d 0 0 g ¶µ 1 3 4 2 ¶ . If d = g,we have µ µ x y for all w z Therefore d 0 0 d ¶µ x y w z ¶ µ = x y w z ¶µ d 0 0 d ¶ µ = dx dy dw dz ¶ ¶ ∈ G. ½µ Z(G) = r 0 0 r ¶ ¾ |r 6= 0 ∈ R . Question 5 (a) (i) Because G2 is abelian, then for all g, g 0 ∈ G1 f ((g −1 g 0−1 gg 0 )) = (f (g −1 )f (g 0−1 )f (g)f (g 0 )) = (f (g −1 )f (g 0 )f (g 0−1 )f (g)) = (f (g −1 g)f (g 0−1 g 0 )) = 1 . Therefore [g, g 0 ] ∈ ker(f ) for all g, g 0 ∈ G1 . Since [G1 , G1 ] is generated by {[g, g 0 ]|g, g 0 ∈ G1 }, hence we have [G1 , G1 ] ⊂ ker(f ). (ii) If [Gi , Gi 0 ] ⊂ ker(f ), since f is surjective, for any b1 , b2 ∈ G2 , there exists g1 , g2 such that NUS Math LaTeXify Proj Team Page: 3 of 6 NUS Mathematics Society MA2202 Algebra I AY 2007/2008 Sem 2 f (g1 ) = b1 , f (g2 ) = b2 . Then 1 = f (g1−1 g1 g2−1 g2 ) = f (g1 −1 )f (g1 )f (g2−1 )f (g2 ) = (f (g2 )f (g1 ))−1 f (g1 )f (g2 ) . Hence f (g1 )f (g2 ) = f (g2 )f (g1 ). Then b1 b2 = b2 b1 , i.e. G2 is abelian. (b) Let φN1 : G → G/N1 and φN2 : G → G/N2 be the canonical quotient maps,hence ker(φN1 ) = N1 and ker(φN2 ) = N2 . They are both surjective and since both G/N1 and G/N2 are abelian, by 5(i) we have [G, G] ≤ N1 and [G, G] ≤ N2 . Hence we have [G, G] ≤ N1 ∩ N2 . Since N1 C G and N2 C G, we have N1 ∩ N2 C G. Let φN1 ∩N2 : G → G/N1 ∩ N2 be the canonical quotient map, hence ker(φN1 ∩N2 ) = N1 ∩ N2 . Since [G, G] ≤ N1 ∩ N2 , by 5(ii) we have G/N1 ∩ N2 is abelian. Question 6 (a) Since |G| = 6, let g ∈ G\{1}. Then order of g divides 6. Hence o(g) = 2,3 or 6. Also, note that any group of order 3 must be cyclic since 3 is prime. If G only has elements of order dividing 2 then for all g ∈ G, g = g −1 . Hence g1 g2 = (g1 g2 )−1 = g2−1 g1−1 = g2 g1 . Then G is abelian and any subgroup is normal. Hence |G/hgi| = 3 for any g ∈ G\{1}. Hence G/hgi is generated by hhgi for some h ∈ / hgi. Hence we have h3 ∈ hgi but h3 = h ∈ hgi which is a contradiction !! Hence G must contain elements of order 3 or 6. If o(g) = 6, then G ∼ = Z6 . Hence {0, 2, 4} would be a cyclic normal subgroup of order 3 since Z6 is abelian. If o(g) = 3, then hgi is a cyclic normal subgroup of order 3 since it is of index 2. (b) < h > CG, then g < h > g −1 =< h > for any g ∈ G. Hence < ghg −1 >=< H >. (c) Since G contains a normal subgroup H of order 3. Hence |G/H| = 2. Therefore G/H = haHi for some a ∈ / H and a2 ∈ H. Let H = hbi. Hence a2 = bk where k ∈ {0, 1, 2}. Case 1) k = 1 or 2. Then a2 6= 1 and o(a2 ) = 3 since a2 ∈ hbi. Hence a6 = 1 and a2 6= 1. If a3 = 1 then we have a ∈ H since abk = 1 which is a contradiction!! Since o(a)|6 and a2 6= 1 and a3 6= 1,we have o(a) = 6. Hence G = hai ∼ = Z/(6). Case 2) k = 0. Then a2 = 1 and b3 = 1 and there exist a m ∈ {0, 1, 2} such that aba−1 = bm a2 ba−2 = abm a−1 b = abm a−1 b = (aba−1 )m NUS Math LaTeXify Proj Team Page: 4 of 6 NUS Mathematics Society MA2202 Algebra I AY 2007/2008 Sem 2 Hence we have (bm )m = b hence 3|m2 − 1. Hence m = 1or 2. If m = 1 then ab = ba. Then o(ab) = lcm(o(a), o(b)) = 6 and G = habi ∼ = Z/(6). If m = 2 then we have aba−1 = b−1 and a2 = 1 and b3 = 1 which are the defining conditions for D6 ∼ = S3 . Hence G ∼ = S3 . Question 7 (a) For any r ∈ H since r 6= 0 1r is well-defined, r · 1r = 1, r · 1 = 1. For any r1 , r2 ∈ H, r1 · r2 ∈ H since the product of positive real numbers is positive. Therefore H ≤ G. (b) Consider the homomorphism f : G → {1, −1} with ( 1 if r > 0 f (r) = −1 if r < 0 . Then we conclude that ker(f ) = H, so G/H ' {1, −1}. So [G : H] = 2. (c) ϕ is surjective. Since ∀y ∈ H,from Real Analysis we know that there exist a positive real solution x to the equation xm − y = 0. Hence ϕ(x) = xm = y. (d) Since G is abelian, any subgroup of normal. Hence let K be a subgroup of finite index in G. Then H ∩ K is also of finite index in G since [G : H ∩ K] = [G : H][H : H ∩ K] = [G : H][G : K]. In particular we have [H : H ∩ K] = m < ∞ where m ∈ N. Hence consider the homomorphism φ : H/H ∩ K → H/H ∩ K hH ∩ K 7→ (hH ∩ K)m = hm H ∩ K Note that since m = |H/H ∩ K|. This is the trivial map(i.e Im(φ) = {1}. But from part (c), this map is surjective. Hence |H/H ∩ K| = 1. Therefore we have H ∩ K = H. Hence H = K or there exist a ∈ K such that a ∈ / H. Hence G = aH ∪ H ⊂ K. Therefore either H = K or G = K. Question 8 (a) True. (g1 ∗ g2 ) ∗ g3 = 1, then g1 ∗ (g2 ∗ g3 ) = 1 ⇒ (g2 ∗ g3 )−1 = g1 ⇒ (g2 ∗ g3 ) ∗ g1 = 1. (b) False. (1) is not in T . (c) False. Z/(2) × Z/(4) is not cyclic even though Z/(2) and Z/(4) are both cyclic. (d) False. Consider (12) and (123). Then o((12)) = 2 and o((123)) = 3 and o((12)(123)) = o((23)) = 2 6= lcm(2, 3). NUS Math LaTeXify Proj Team Page: 5 of 6 NUS Mathematics Society MA2202 Algebra I AY 2007/2008 Sem 2 (e) True. We only need to show that for any h ∈ H ,h−1 ∈ H. For h ∈ H, ∃i such that h1 i = 1. Otherwise |H| = ∞. Since hi−1 = h−1 since hi−1 h = 1, we have h−1 = hi−1 ∈ H. (f) True. ker(f ) = {1}, so f is injective hence |G1 | ≤ Im(f ) and since |G1 | = |G2 | < ∞, Im(f ) ≥ G2 . Hence f is surjective and hence an isomorphism. (g) False. o(A5 ) = 60 but A5 doesn’t have subgroup of order 30 since A5 is simple. Or G = Z/(2) × Z/(2), so |G| = 4 but G has no elements of order 4 since G is not cyclic. (h) False. N C G ⇒ ∀g ∈ G, gN=Ng, but not necessarily gn = ng ∀n ∈ N . An example would be G = S3 and N = A3 . Hence (12)(123) = (32) 6= (13) = (123)(12) (i) True. We have H1 ∩ H2 ≤ H1 and H1 ∩ H2 ≤ H2 . Hence |H1 ∩ H2 | divides gcd(|H1 |, |H2 |) = 1. Hence |H1 ||H2 | |H1 ∩ H2 | = 1. Therefore |H1 H2 | = |H = |H1 | × |H2 |. 1 ∩H2 | (j) True. Since N1 , N2 C G we have N1 N2 ≤ G. Hence for any g ∈ G. we have gN1 N2 = (gN1 )N2 = (N1 g)(N2 ) = N1 (gN2 ) = N1 (N2 g) = N1 N2 g. Since gN1 N2 = N1 N2 g we have N1 N2 C G. NUS Math LaTeXify Proj Team Page: 6 of 6 NUS Mathematics Society