Math 545 1. Let G Be a Group and a ∈ G. (A) the Order of a (In G) Is The

Math 545

1. Let G be a group and a ∈ G. (a) The order of a (in G) is the least positive integer n, written |a| = n, such that an = e. Thus |a| = n =⇒ a1 6= e, a2 6= e, . . . , a(n−1) 6= e, and an = e. (b) |a| = ∞ ⇐⇒ am 6= e∀m ∈ Z+ ⇐⇒ a−m 6= e∀m ∈ Z+ (since am 6= e =⇒ (am)−1am 6= (am)−1e =⇒ e 6= (am)−1 =⇒ e 6= a−m) ⇐⇒ am 6= e∀m 6= 0 ∈ Z ⇐⇒ am = e (m ∈ Z) =⇒ m = 0 ⇐⇒ am = e (m ∈ Z) ⇐⇒ m = 0 (since m = 0 =⇒ am = e)

2. Find orders of all elements of Z4. |0| = 1. |1| = 4, since 1.1 = 1 6= 0, 2.1 = 2 6= 0, 3.1 = 3 6= 0, and 4.1 = 4 ≡4 0. |2| = 2, since 1.2 = 2 6= 0, 2.2 = 4 ≡4 0. |3| = 4, since 1.3 = 3 6= 0, 2.3 = 6 ≡4 2 6= 0, 3.3 = 9 ≡4 1 6= 0, and 4.3 = 12 ≡4 0. 3. Find orders of all elements of U(14). U(14) = {1, 3, 5, 9, 11, 13}. |1| = 1 (since 1.1 ≡12 1) 1 2 3 4 5 |5| = 6, since 5 = 5 6= 1, 5 = 25 ≡14 11 6= 1, 5 ≡14 13 6= 1, 5 ≡14 9 6= 1, 5 ≡14 3 6= 1 and 6 1 2 3 5 ≡14 15 ≡14 1. Similarly, |3| = 6. |9| = 3, since 9 = 9 6= 1, 9 = 81 ≡14 11, 9 ≡14 99 ≡14 1. 1 2 2 Similarly, |11| = 3. |13| = 2 (since (13) = 13 6= 1, (13) ≡14 (−1) = 1).

4. Find orders of all elements of Z. |0| = 1 ( 0 is the identity of the additive group Z ). Let a 6= 0 ∈ Z. Then na 6= 0∀n ∈ Z+. =⇒ |a| = ∞ So all nonzero elements of Z have inﬁnite order.

5. Find orders of all elements in R?, the group of nonzero real numbers under multiplication. |1| = 1 ( 1 is the identity of the multiplicative group R? ) and | − 1| = 2. Let a ∈ R? such that a∈ / {−1, 1}. Then an 6= 1∀n ∈ Z+. =⇒ |a| = ∞. So all elements of R?, other than 1 and -1, have inﬁnite order. 6. Prove that in any group, an element and its inverse have the same order. Case 1. |a| 6= ∞ =⇒ ∃m ∈ Z+ such that am = e ⇐⇒ ∃m ∈ Z+ such that (a−1)m = e ⇐⇒ |a−1|= 6 ∞(since am = e ⇐⇒ (am)−1am = (am)−1e ⇐⇒ e = (am)−1 ⇐⇒ (a−1)m = e). −1 m1 Case 2. Assume |a| = m1 and |a | = m2. Prove m1 = m2. a = e(since |a| = m1) ⇐⇒ m1 −1 −1 −1 −1 −1 (a ) = e ⇐⇒ ⇐⇒ (a )m1 = e (notation and e = e). Now (a )m1 = e and −1 |a | = m2. Then m2 ≤ m1. −1 m2 −1 m2 −1 m2 m2 (a ) = e (since |a | = m2)(a ) = e (notation) ⇐⇒ a = e. Now a = e and |a| = m1 −1 m2 −1 −1 −1 m =⇒ m1 ≤ m2 ((a ) ) (notation) = e (since |a | = m2) = e. Now a2 = e and |a| = m1 −1 m1 m1 −1 −1 −1 =⇒ m1 ≤ m2 (def. of order). Also, (a ) = (a ) (notation) = e = e and |a | = m2 =⇒ m2 ≤ m1.

7. Prove that an abelian group with two elements of order 2 must have a subgroup of order 4. Let G be an abelian group and a, b ∈ G such that |a| = 2 and |b| = 2. Then H = {e, a, b, ab} is closed and therefore, by ﬁnite subgroup test, a subgroup of G.

8. Find the order of each element of the group Z10, the group of integers under addition modulo 10. |1| = |3| = |7| = |9| = 10, |2| = |4| = |6| = |8| = 5, |5| = 2, |0| = 1.

9. (a) Find the order of each element of the group U(30), the group of positive integers less than 30 and relatively prime to 30 under multiplication modulo 30. U(30) = {1, 7, 11, 13, 17, 19, 23, 29}: |1| = 1, |7| = 4, |11| = 2, |13| = 4, |17| = 4, |19| = 2, |23| = 4, and |29| = 2.

1 (b) Give a multiplication table for U(30). U(30) = {1, 7, 11, 13, 17, 19, 23, 29}: |1| = 1, |7| = 4, |11| = 2, |13| = 4, |17| = 4, |19| = 2, |23| = 4, and |29| = 2. 1 7 11 13 17 19 23 29 1 1 7 11 13 17 19 23 29 7 7 19 17 1 29 13 11 23 11 11 17 1 23 7 29 13 19 13 13 1 23 19 11 7 29 17 17 17 29 7 11 19 23 1 13 19 19 13 29 7 23 1 17 11 23 23 11 13 29 1 17 19 7 29 29 23 19 17 13 11 7 1

(c) Find all cyclic subgroups U(30) of order 4. < 7 >= {1, 7, 19, 13} =< 13 >, < 17 >= {1, 17, 19, 23 >=< 23 > (d) Find all noncyclic subgroups of U(30) of order 4. {1, 11, 19, 29}. 7. Let G be a group and x ∈ G. If |x| = 20, then ﬁnd |x2|, |x5|, |x7|, |x11| and |x15|. k |x| Use the formula |x | = gcd(|x|,K) . 2 20 5 20 7 20 |x | = gcd(20,2) = 10, |x | = gcd(20,5) = 4, |x | = gcd(20,7) = 20, 11 20 15 20 |x | = gcd(20,11) = 20, |x | = gcd(20,15) = 4. 8. Find a multiplication table for the group G that contains elements a and b such that |a| = 7, |b| = 2 and |ab| = 2. (ab)2 = e =⇒ abab = e =⇒ ba = a−1b−1 = a6b (since |a| = 6 and |b| = 2 =⇒ a7 = e and b2 = e =⇒ a6 = a−1 and b = b−1). Thus the given group is {e, a, a2, a3, a4, a5, a6, b, ab, a2b, a3b, a4b, a5b, a6b}. A multiplication table is given below: e a a2 a3 a4 a5 a6 b ab a2b a3b a4b a5b a6b e e a a2 a3 a4 a5 a6 b ab a2b a3b a4b a5b a6b a a a2 a3 a4 a5 a6 e ab a2b a3b a4b a5b a6b b a2 a2 a3 a4 a5 a6 e a a2b a3b a4b a5b a6b b ab a3 a3 a4 a5 a6 e a a2 a3b a4b a5b a6b b ab a2b a4 a4 a5 a6 e a a2 a3 a4b a5b a6b b ab a2b a3b a5 a5 a6 e a a2 a3 a4 a5b a6b b ab a2b a3b a4b a6 a6 e a a2 a3 a4 a5 a6b b ab a2b a3b a4b a5b b b a6b a5b a4b a3b a2b ab e a6 a5 a4 a3 a2 a ab ab b a6b a5b a4b a3b a2b a e a6 a5 a4 a3 a2 a2b a2b ab b a6b a5b a4b a3b a2 a e a6 a5 a4 a3 a3b a3b a2b ab b a6b a5b a4b a3 a2 a e a6 a5 a4 a4b a4b a3b a2b ab b a6b a5b a4 a3 a2 a e a6 a5 a5b a5b a4b a3b a2b ab b a6b a5 a4 a3 a2 a e ab a6b a6b a5b a4b a3b a2b ab b a6 a5 a4 a3 a2 a e

10. If H and K are subgroups of G, show that H ∩ K is a subgroup of G. Apply the 1-step subgroup test: (1): e ∈ H and e ∈ K (since H and K are subgroups Of G). Thus e ∈ H ∩ K. (2) Let a, b ∈ H ∩ K. Then a, b ∈ H and a, b ∈ K. So a, b−1 ∈ H and a, b−1 ∈ K (since H and K are subgroups Of G). So ab−1 ∈ H and ab−1 ∈ K (since H and K are subgroups Of G). Hence ab−1 ∈ H ∩ K.

11. Let G be a ﬁnite abelian group and let a, b ∈ G. Prove that the set < a, b >= {aibj|i, j ∈ Z} is a subgroup of G. Apply the One-Step subgroup Test to H =< a, b >= {aibj|i, j ∈ Z} . First, show that H 6= ∅. The identity e = a0b0 of G is in H. Next, let aibj, akbl ∈ H. Prove (aibj)(akbl)−1 ∈ H. (aibj)(akbl)−1 = (aibj)(bl)−1(ak)−1(since (ab)−1 = b−1a−1 ∀a, b ∈ G) = (aibj)b−la−k (notation) = ai(bjb−l)a−k (associative law in the group G) = aia−k(bjb−l) (since G is abelian and bjb−l, a−k ∈ G, (bjb−l)a−k = a−k(bjb−l)) = ai−kbj−l (laws of exponent). So (aibj)(akbl)−1 = ai−kbj−l ∈ H (since i − k, j − l ∈ Z).