Math 412 Practice True/False Questions By the amazing Winter 2018 Sections 1 and 21 (1) Let N be a subgroup of G and let K be a normal subgroup of N. Then K is normal in G. Solution. False. Take N = hx; yi inside G = D4 and take K = hxi. (2) Let n 2 N. Then [Sn;An] = 2. Solution. True- Several ways to see this. Here is one: consider the map φ : Sn ! Z2, σ 7! sign(σ). which takes add permu- tations to 1 and even permutations to 0. This is a surjective group homomorphism with kernel An. Applying FIT, we get ∼ Sn=An = Z2, hence jSn=Anj = [Sn : An] = 2. (3) Suppose N is a normal subgroup of G. Then N ⊆ Z(G). Solution. False. Consider R = hr2i, the group of rotations in D4. This is a normal subgroup of D4 and is not contained in 2 the centre of D4, which is fe; r g. (4) Let G be a group acting on a set X. If O(x) = O(y) for all x; y 2 X, then G is cyclic. Solution. False- Consider the natural action of Sn on f1 ··· ng. This action has a single orbit but Sn is not cyclic for n > 2. (5) Let G be a group and let H be a nonempty finite subset of G that is closed under the operation of G. Then H is a subgroup. Solution. True- See Theorem 7.12. (6) Let G be a group and let H be a nonempty subset of G that is closed under the operation of G. Then H is a subgroup. Solution. False- Consider f2a j a 2 Z; a 6= 0g ⊂ Z. But it is not a subgroup. (7) Let K be a subgroup of G such that ab = ba for all a; b 2 K. Then K is a normal subgroup of G. Solution. False- Consider K = hxi in D4. K is abelian but not normal. (8) Any group automorphism of the group Q under addition is of the form x 7! qx for some q 2 Q. Solution. True- This is more of a short proof question. To prove this suppose φ is an arbitrary automorphism pf (Q; +). 1Alison Elgass, Charlemagne McHaffie, Michael McMain, Ian Cleary, Tiffany Liu, Yuhan Liang, Daoheng Niu, Thomas Eberle, Ryan Yao, Shihui Wang, Paulina Czarnecki, Kendra Robbins, Haidan Tang, Rithvik Pasumarty, Rupal Nigam, Sahil Farishta, Carolyn Evans, Kristin Dona, Tianyu (Justin) Yang, John Koenig, Julia Rakas, Bennet Sakelaris, Xiang Li (Olivia), Ally Young, Yujia Xie, Steven Schaefer, Jenna King 1 2 Then φ(1) = q for some q 2 Q. Hence, φ(−1) = −φ(1) = −q and consequently we can argue that φ(n) = n for all integer n. Observe that for every integer b, q = φ(1) = φ(b=b) = φ(b(1=b)) = bφ(1=b). So, φ(1=b) = q=b. Now let a=b 2 Q. Then φ(a=b) = φ(a(1=b)) = aφ(1=b) = aq=b = q(a=b). (9) All non-trivial proper subgroups of (R; +) are cyclic. Solution. False- There are many counter examples. Here is one: consider the addition group of rational numbers. It is not cyclic (why?) and it is a proper non-trivial subgroup of R. (10) Some abelian group of order 102 cannot have a subgroup of order 18. Solution. True- 18 doesn't divide 102. So Lagrange doesn't let us to have such subgroup. (11) A subgroup H of a group G is a normal subgroup if and only if the number of left cosets of H is equal to the number of right cosets of H. Solution. False- Number of right and left cosets are equal for any subgroup. Any non-normal subgroup serves as counter example. My favourite is hxi in D4. (12) Let R be a ring. Then (a + b)2 = a2 + 2ab + b2 for all a; b 2 R. Solution. False- Any two non-commuting elements in an non abelian ring serves as a counter example. Here is one: Take 1 0 0 1 and in M ( ). 0 0 0 0 2 R (13) There exists a subgroup G of S5 that is isomorphic to the group Z3 × Z3. (14) A group of order 8 must contain an element of order 2. Solution. True- If we use the structure theorem, we will see that a group of order 8 is one of the following: Z2 × Z2 × Z2, Z4 × Z2 and Z8. Each of these groups have an element of order 2: take (1; 0; 0), (0; 1) and 4 respectively. (15) If a non-trivial group has an action on itself by conjugation, then there exists at least one fixed point. Solution. True- The identity is fixed by conjugation. (16) The only generators of an additive cyclic group Z, are +1 and −1. Solution. True-because hni for any n 2 Z only contains mul- tiples of n, hence in order to generate the entire Z, n should be ±1. (17) If all subgroups of a group G are normal, the group is abelian. 3 Solution. False- One example I can think of is Q8, the group of quaternions, which we met in a problem set, we almost proved that all the subgroups are normal (one needs to show that there are only one subgroup of a given order ). (18) If G is a cyclic group, every element of G is a generator for G. Solution. False- Take [4] 2 Z8. (19) The order of the intersection of two subgroups is the least com- mon multiple of the orders of each of the subgroups. Solution. False- Take h(12)i and h(34)i in S4. They intersect only at identity, whereas the least common multiple of their order is 2. (20) The intersection of a subgroup K of G with a normal subgroup N is normal in K. Solution. True- Take ghg−1 2 g(K \ N)g−1, for g 2 K. Since h 2 N, and N is normal in G, and g 2 K ⊂ G, then ghg−1 is in N. On the other hand K is closed under multiplication so ghg−1 is in K. Hence it is in the K \ N. (21) Let G be a group. Let a and b be elements in G where a has order m and b has order n. The order of ab divides mn. Solution. (22) Z3 × Z5 is isomorphic to Z15. Solution. True- If we use the structure theorem, we can see that there is only one abelian group of order 3 × 5 = 15 up to isomorphic and that is Z3 × Z5. So Z15 must be isomorphic to Z3 × Z5. Without the structure theorem, consider (1; 1) 2 ∼ Z3 × Z5 (abusing notation), it has order 15 so Z15 = h(1; 1)i = Z3 × Z5. (23) Let N be a subgroup of Sn such that the order of N is odd. N is a subgroup of An.T (24) All ideals of Z are both prime and maximal. F (25) f : R ! R× which takes x 7! 2x by the function f(x) = 2x is a group isomorphism. F (not surjective) (p−1) (26) In Zp, let p be prime and a nonzero. Then a = 1. Solution. True- This is Fermat's little theorem. An easy proof is to note that the group of units of Zp is of order p − 1 and the order of every element in the group divides the order of the group by Lagrange. (27) For every n > 2, can find a group G having n elements not isomorphic to group Zn. Solution. True- If n is prime, then we know that there is only one group of prime order and that is isomorphic to Zn. But if n 4 is not prime that is not true. For example take n = 4. Z2 × Z2 is not isomorphic to Z4. × (28) Z37 is isomorphic to Z36. Solution. True- Both are cyclic groups with order 36. (29) A group with order 62 has an element with order above 32, then the group is a cyclic group. Solution. True- the order of an element in a group divides the order of the group; if a natural number is greater that 32 and divides 62, it is equal to 62. A group with an element of the order of the group is cyclic. (30) The largest order of the element in the group (Zn × Zm; +) is max(m; n). Solution. false- when m and n are coprime, this group is cyclic of order m × n. (31) Let set X has 20 elements and a group G of order 12 acts set X, it is possible for X to contain only two orbits. Solution. False- The orbits of an action partition X. If there are only twi orbits, their cardinality should add up to 20. On the other hand, cardinality of each orbit divides the order of the group by the orbit stabilizer, i.e, the possibilities are 1; 2; 3; 4; 6; 12. No two if these add up to 20. (32) Sm is a normal subgroup for any Sn where n ≥ m. Solution. false- Sm is a subgroup of Sn where n ≥ m. However it is usually not normal. Example: take S4, we know all its non- trivial normal subgroups are A4 and a group we met in one of the problem sets (isomorphic to Klein group). (33) It is possible for integers to act on the set of whole numbers Solution. True- Consider the map Z × R ! R, sending (n; x) to n + x.