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 ∈ 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 homomorphism with kernel An. Applying FIT, we get ∼ Sn/An = Z2, hence |Sn/An| = [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 {e, r }. (4) Let G be a group acting on a set X. If O(x) = O(y) for all x, y ∈ X, then G is cyclic. Solution. False- Consider the natural action of Sn on {1 ··· n}. 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 {2a | a ∈ Z, a 6= 0} ⊂ Z. But it is not a subgroup. (7) Let K be a subgroup of G such that ab = ba for all a, b ∈ 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 of the group Q under addition is of the form x 7→ qx for some q ∈ 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 ∈ 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 ∈ 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 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 ∈ 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 ∈ 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] ∈ 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 ∈ g(K ∩ N)g−1, for g ∈ K. Since h ∈ N, and N is normal in G, and g ∈ 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) ∈ ∼ 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 . 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. note that 0 + x = x, and for n, m ∈ Z, n + (m + x) = (n + m) + x. This is an action! (34) If G is a cyclic group and H is a subgroup of G, then H is normal in G. Solution. True- Any cyclic group is abelian and any subgroup of any abelian group is normal. (35) If His a normal subgroup of a group G, and N is a subgroup of a group M such that N ∼= H, then N is a normal subgroup of M. Solution. False-Take H = N = hxi M = D4 and G the Klein four subgroup in D4. (36) Let K be a subgroup in G. If a ≡ b( mod K) and c ≡ d( mod K); then ac ≡ bd( mod K): 5

Solution. False-Take a non normal subgroup to find a counter example. (37) An is an abelian subgroup of Sn. Solution. False- You can find non commuting elements in A4. (A3 is abelian) (38) Let X be any non-empty set. Let G by any group. Then G acts on X. Solution. True- We can always consider the trivial action that sends every x ∈ X to itself. We can check that it is an action by checking the two axioms. (39) Let G; Hbe finite cyclic groups where the order of H divides the order of G. Then there exists a surjective homomorphism from G → H. Solution. True- Let G =< g >, and let H =< h > (true since they are both cyclic). Let. Define a map as follows: φ : G → H, gn → hn. We can show that this is a group homomorphism (check!) and surjective (check!) (40) Every group has at least 2 distinct normal subgroups. Solution. FALSE- This is tricky. The only exception to this statement is the trivial group: the group of order 1. (41) There exists a finite simple group G that has a subgroup H with |H| = |G|/2 , and |G| ≥ 4. Solution. FALSE-Since H has index two, it must be normal. |G| ≥ 4 implies that H is non-trivial. Hence G having a non- trivial normal subgroup is not simple. (42) Every infinite abelian group has at least one element of infinite order. Solution. False- As we show in the homework every element of Q/Z has finite order. (43) There exists a non-cyclic group of order 17. Solution. False, as all groups with prime order is cyclic. (44) Z4 × Z4 isomorphic to Z2 × Z8. Solution. False, as Z2 × Z8 has an element (0, 1) of order 8 but there is no element of order 8 in Z4 × Z4. (45) Every ring is an abelian group under multiplication. Solution. False- Every ring is an abelian group under addition (theorem 7.1), however, a ring is never a group under multi- plication: R has additive identity 0, but 0 does not have an inverse, so R cannot be a group under multiplication. (46) Suppose group G contains a subgroup K, and suppose that for some a, b in G, aK 6= bK. Then there is at least one element c in G such that c ∈ aK and c ∈ bK. 6

Solution. False. We have shown that cosets aK, bK are either disjoint or equal. Thus if aK 6= bK, their intersection is empty and such a c could not exist. (47) Suppose G is a group and a ∈ G, and suppose a has infinite order. Then there exist integers k, n st. ak = an. Solution. False. Suppose ak = an and suppose k > n. Then multiplying both sides by a( − j)a(i − j) = a(j − j) = a0 = e. Since i − j > 0, this proves a has finite order, contradicting our assumption that it has infinite order. (48) Suppose G is a group and |G| = 6. Then G must be isomorphic to Z6. Solution. False- by counterexample: S3 has order 6 but does not contain any element of order 6, so cannot be isomorphic to Z6. (49) Every subgroup of a cyclic group G is normal. Solution. True. We know that every subgroup of an abelian group is normal. Every cyclic group is abelian, so every sub- group of a cyclic group is normal. (50) Consider D37 and let R be the subgroup of rotations in D37. ∼ Then D37/R = {±1}. Solution. True- because in PS7 we showed that the map D37 → {±1}, which sends rotations to 1 and reflections to −1, is a group homomorphism. Obviously it is also surjective. The kernel of this map is the subgroup R of rotation. By FIT, the statement is true. (51) Consider Dn and let G =< y > be the cyclic subgroup of Dn ∼ generated by a reflection over the y-axis. Then Dn/G = An. Solution. False- Since G is not a normal subgroup, Dn/G is not a valid quotient group, so the statement is false. (52) Let G be a group such that |G| = 17. G has an element of order 17. Solution. This statement is true because any group of prime order is cyclic. (53) Z41 has a subgroup of order 4. Solution. This statement is false because we proved that a finite cyclic group is simple iff n is prime. Since 41 is prime, Z41 is simple, so it only has subgroups of order 41 and 1. (54) Let H 6= {e} be a subgroup of Z7. Then the smallest Sn (mean- ing the smallest n) that H embeds in is S7. Solution. True- This question uses the theorem stating that a finite cyclic group of order n is simple iff n is prime. Since 7 is prime, Z7 is simple. Since H is nontrivial, H must be 7

Z7. Since it has an element of order 7, the smallest subgroup of permutations that it can be isomorphic to is a subgroup generated by a 7-cycle. (55) There is an isomorphism between U26/h5i and Z3. Solution. True- U26 = {1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25}, so U26 has 12 elements. h5i = {5, 25, 21, 1}. So |U26/h5i| = |U26|/|h5i| = 12/4 = 3. Recall that upto isomorphism there is only one group of order 3: Z3. (56) If every proper subgroup of a group G is cyclic, then G is also cyclic. Solution. False. Counterexample: G = Z2 ×Z2. The elements are {(0, 0), (0, 1), (1, 0), (1, 1)}. All the proper subgroups of this group are cyclic, but since there is no element has order 4, the group G cannot be cyclic. (57) M and Nare two subgroups of group G. Let MN = {ab | a ∈ M, b ∈ N}, then MN is a subgroup of G. Solution. False. Counterexample: Let B be the S4, and M = {(1), (12)}, N = {(1), (13)}.Then MN = {(1)(1), (1)(13), (12)(1), (12)(13)} = {(1), (13), (12), (132)}. We can tell that this set is not closed under multiplication since the product of (13) and (12) which is the permutation (123) is not in the set, so this set MN is not a subgroup of G, and it gives a contradiction. (58) If f : G → H is a surjective group homomorphism, and G is abelian, then H is also abelian. Solution. True. Consider any two elements of a, b in H. Then, there exists x, y in G such that f(x) = a, f(y) = b. We have, ab = f(x) ∗ f(y) = f(xy) = f(yx) = f(y) ∗ f(x) = ba, which means that for any two arbitrary elements a, b in H, we have ab = ba. Hence, H is abelian. (59) Let R be a commutative ring and I a proper ideal of R. Define φ : R → R/I by φ(r) = r + I, then φ is surjective. Solution. True. Let r + I ∈ R/I be an arbitrary element in codomain of φ. Then φ(r) = r + I. (60) Every group G of order 2017 has exactly one distinct right coset. Solution. False. 2017 is prime, the only divisors are 1 and itself so by Lagrange the only subgroups are {e} and G itself. The trivial subgroups partitions G into 2017 cosets and the G itself partitions G into one coset. (61) Every group is isomorphic to a subgroup of a permutation group. Solution. True. Cayley’s theorem. 8

(62) An is a normal subgroup of Sn. Solution. True. Consider the sign homomorphism from Sn. Then An is the kernel of this homomorphism so it is normal by the first isomorphism theorem. (63) Sn is generated by two elements. Solution. True. Take (12) and (12 ::: n). (you need to do some work to see this: the key is to notice that (1, 2, 3, . . . , n)i(1, 2)(1, 2, 3, . . . , n)−i = (i + 1, i + 2)). (64) Suppose F is a field and H is a subring of F . Then, H is also a field. Solution. False- Take Z inside Q. (65) Suppose D is an integral domain and H is a subring of D. Then, H is also an integral domain. Solution. True- if x is a zero devisor in H, then it is also a zero divisor in D. (66) Let D be the ideal of a ring R. Then, D forms a group under the multiplication operation. Solution. False-Consider the ideal (2) in Z. The element 2 has no multiplicative inverse in (2). (67) If a is congruent to 1 mod n, then (a, n) = 1. Solution. True- a ≡ 1 mod n means n|a−1, that is nq = a−1 hence nq − a = 1. (68) All prime ideals are maximal. Solution. False- Consider the ideal (2, x) in Z[x]. (69) All subgroups of S3 are normal. Solution. False- Consider the subgroup generated by (12). (70) All subgroups of S5 are cyclic. Solution. False- Consider the S5 it self. (71) S7 is isomorphic to some Zp where p is prime. Solution. False- The order if on group is prime when the other one is not. (72) If N is a normal subgroup of G then for any n in N and a in G, an = na. Solution. False- Group of rotations in D4 is normal but you can find elements in D4 that don’t commute with r3 (find them) (73) Every group of order n is isomorphic to a group Sn. Solution. False-The order of Sn is n! so it is not isomorphic to a group of order n unless n = 2. (74) Every group has a cyclic subgroup. Solution. True-Take the group generated by any element. 9

(75) All groups of order two are isomorphic. Solution. True- Up to isomorphism there is only one group of any prime order, including 2. (76) Let G be a group with order n. Then if d|n there is a subgroup H in G with order d. Solution. False- (77) The centre of a group is abelian. Solution. True- Use the definition of centre to see this. (78) Let a be an element of a group such that a2 = a then a is the identity. Solution. True- Multiply both side the inverse of a. (79) The group of units of Zn is cyclic. × Solution. False- Consider Z8 . (80) Every cyclic group is of a prime order. Solution. False- Consider Z8. (81) All normal subgroups of a group G are subgroups of the centre of G. Solution. False- Consider the group of rotations in D4. (82) An is normal in Sn for all natural number n. Solution. True- Because it has index two. (83) A3 is Abelian. Solution. True- Can be checked by hand. (84) If G is simple and Abelian, then G is cyclic. Solution. True- Consider the subgroup generated by a non- trivial element, it is a normal subgroup of G since G is abelian, and since it is simple it has to be entire group. So G is cyclic. (85) Let G be a group of prime order, then all the non-trivial group homomorphisms from G to an arbitrary group H are injective. Solution. True- The kernel of a non-trivial homomorphism is a proper subgroup whose order should divide p. Hence the order has to be one, and so the homomorphism is injective. (86) C[x]/(x2 + 2x + 10) is a field. Solution. x2 + 2x + 10 factors to (x + 1 − 3i)(x + 1 + 3i) in C, thus it is not irreducible in C and is not a field. (87) Suppose G is a cyclic group. Then its generating element is unique. Solution. False. Z3 is generated by [1] and [2] (88) There exists a group G such that |G| = 6 and is abelian and not cyclic Solution. False. Any group of order 6 is isomorphic to either Z6 (abelian and cyclic) or S3 (nonabelian, noncyclic). Thus there cannot exist a group that is both abelian and noncyclic) (89) Z8 is isomporphic to Z4 × Z2. 10

Solution. False, since Z4 × Z2 has no elements of order 8.. (90) If G is a group with order 37 then there is an element of G with order 37. Solution. True, since 37 is prime so G is cyclic so the gener- ating element of G has order 37. (91) Let G, H be groups. Let f be a group homomorphism from G to H. Then the image of f is a normal subgroup of H. Solution. False- Consider the group embedding from the sub- group generated by a reflection in D4 to D4. (92) Let f(x) be a polynomial in R[x]. If a + bi is root of f(x), then a − bx is also a root of f(x). Solution. True- f(¯z) = f(¯z). (93) Let K be a subgroup of D4 generated by r1, the rotation by 90 degrees. Then Kx = Ky where x is the reflection across the x axis and y is the reflection across the y axis. Solution. True- The rotation subgroup partition d4 into two cosets rotations and reflections. 2 (94) Consider the natural action of D4 on R . Then the |O(1, 0)| = 8. Solution. False- This orbit has 4 elements. (95) Group G acting on itself via conjugation. If H is a normal subgroup of G, then for all h in H, O(h) = H. Solution. False- Consider h = e.