Midterm 3 Practice Worksheet 1: Solutions#

(A) True. . (Every group with a prime number of elements is cyclic.)

(B) False. , but , so is not commutative, and a cyclic group is commutative. Also, the elements of have orders , , , never .

(C) True. , so is closed under addition modulo .

(D) True. The cycles and are disjoint, and disjoint cycles commute.

(E) False. By Lagrange’s theorem the number of elements of a subgroup divides , and does not.

(F) False. exactly when divides . In fact .

(G) True. The cycles are disjoint, so the order is .

(A) The multiples of modulo are , so . The multiples of are , so . The multiples of are , , , , so .

is cyclic. An element is a generator exactly when its multiples reach all elements before returning to , that is, when has order . This happens exactly when and have no common divisor greater than : the generators are . (For example, the multiples of are .)

(B) A finite subset is a subgroup exactly when it is closed under addition.

• : subgroup; , , .

• : subgroup; .

• : not a subgroup; is not in the set.

• : subgroup; , , .

• : not a subgroup; is not in the set. (Also, divides , so size alone does not rule it out; closure does.)

(C) Apply the right factor first. For : , , , so . For : , , , so . The two products are different, so is not commutative.

The elements of are (order ), the transpositions , , (order ), and the -cycles , (order ). No element has order , so is not cyclic. (The failure of commutativity gives a second reason: a cyclic group is commutative.)

(D) By Lagrange’s theorem the number of elements of a subgroup divides . So and are possible, and and are not. The order of an element is the number of elements of the subgroup it generates, so it also divides : no element has order . An element of order is possible; for example, has order in .

(E) exactly when divides . So , , and , but . Since , . Since , .

(F) For disjoint cycles, take the least common multiple of the lengths: has order ; has order ; has order .

The cycles of overlap, so compute the product first: , , , . So , which has order , not .

(A) Follow each element under , starting with the smallest element not yet used: , then , then . So The cycle type is and : the fixed point counts as a cycle.

(B) The order is . The inverse reads each cycle backwards: Check: .

(C) , where sends , , and fixes the rest: Following the cycles: , and and are fixed. So with cycle type .

(D) Now apply first: . , , , , , , and , . So . It has the same cycle type as , but it is a different permutation: sends , and sends .

(E) The written cycles move ; the fixed points in are . So , with cycle type , , and order .

(A) Follow the elements: , , . So with cycle type and (there are no fixed points).

(B) The order is . Reading each cycle backwards,

(C) , where sends , , : Following the cycle from : . So a single -cycle.

(A) The multiples of modulo are , , , so and has order . The multiples of are , , , , so and has order . Both orders divide , as Lagrange’s theorem requires.

(B) The multiples of modulo are all elements, so has order and is a generator. The multiples of are , only elements, so is not a generator. In general generates exactly when and have no common divisor greater than ; the generators are .

(C) The multiples of modulo are , , , so ; has order and is not a generator. The generators of are the elements with no common divisor with other than , that is, those not divisible by : .

(D) The multiples of modulo are , , , so and has order . The multiples of modulo are all elements, so has order and is a generator. (Equivalently, and have no common divisor greater than .)

(A) The rotation is , and moves each corner two steps: . The reflection fixes corner and exchanges the corners on the two sides of the axis: , so .

Right to left, : , , , , . So , the reflection across the axis through corner . Similarly : , the reflection across the axis through corner .

(B) : , , , , , so . Also , the same permutation. So . Since sends while sends , we have .

(C) The identity has order and . The four rotations are -cycles: order and . The five reflections each fix one corner and exchange the other four in two pairs: order and . No element has order , so is not cyclic; part (B) gives a second reason, since a cyclic group is commutative.