Download Introduction to Group Theory Solutions 5

Introduction to Group Theory
Solutions 5
Unit website: http://people.maths.bris.ac.uk/~majcr/IGT.html
Questions 1,2,3 to be handed in for tutorials.
1. Let U11 be the group of non-zero elements of Z/11Z under multiplication
(mod 11).
(a) For each element of U11 find its inverse.
(b) Prove that U11 is cyclic by finding a generator.
Solution:
(a) U11 = {[1], [2], [3], [4], [5], [6], [7], [8], [9], [10]}, since 11 is prime
and so hcf(a, 11) = 1 for all 0 < a < 11.
[1] is its own inverse.
[2][6] = [12] = [1], so [2] and [6] are inverses of one another.
[3][4] = [12] = [1], so [3] and [4] are inverses of one another.
[5][9] = [45] = [1], so [5] and [9] are inverses of one another.
[7][8] = [56] = [1], so [7] and [8] are inverses of one another.
[10][10] = [100] = [1] (or, simpler, [10] = [−1] and [−1][−1] = [1]),
so [10] is its own inverse.
(b) Calculating the powers of [2]: [2]2 = [4], [2]3 = [8], [2]4 =
[16] = [5], [2]5 = [2][5] = [10], [2]6 = [2][10] = [20] = [9], [2]7 =
[2][9] = [18] = [7], [2]8 = [2][7] = [14] = [3], [2]9 = [2][3] = [6], and
[2]10 = [2][6] = [12] = [1]. So ord([2]) = 10, and so U11 is cyclic, with
[2] as a generator. (We’ll soon see a short cut for answering questions
like this). [Other generators are [6], [7] and [8].]
2. Let Un be the multiplicative group of elements of Z/nZ with multiplicative inverses.
List the elements of Un and find the order of each element in the
following cases:
(1)
(2)
(3)
(4)
n = 9.
n = 12.
n = 14.
n = 15
In which of these cases is Un cyclic?
Solution:
(1) U9 = {[1], [2], [4], [5], [7], [8]}, since hcf(a, 9) = 1 unless 3 divides
a.
[1] is the identity and has order 1.
[2]2 = [4], [2]3 = [8], [2]4 = [16] = [7], [2]5 = [2][7] = [14] =
[5], [2]6 = [2][5] = [10] = [1], so ord([2]) = 6.
[4]2 = [16] = [7], [4]3 = [4][7] = [28] = [1], so ord([4]) = 3.
[5]2 = [25] = [7], [5]3 = [5][7] = [35] = [8], [5]4 = [5][8] =
[40] = [4], [5]5 = [5][4] = [20] = [2], [5]6 = [5][2] = [10] = [1], so
ord([5]) = 6.
[7]2 = [49] = [4], [7]3 = [7][4] = [28] = [1], so ord([7]) = 3.
[8]2 = [64] = [1], so ord([8]) = 2.
Since |U9 | = 6 and there are elements of order 6, U9 is cyclic.
(2) U12 = {[1], [5], [7], [11]} since hcf(a, 12) = 1 unless a is even or
divisible by 3.
[1] is the identity and has order 1.
[5]2 = [25] = [1], so ord([5]) = 2.
[7]2 = [49] = [1], so ord([7]) = 2.
[11]2 = [−1]2 = [1], so ord([11]) = 2.
Since |U12 | = 4 but there are no elements of order 4, U12 is
not cyclic.
(3) U14 = {[1], [3], [5], [9], [11], [13]}.
[1] is the identity and has order 1.
[3]2 = [9], [3]3 = [27] = [13], [3]4 = [3][13] = [39] = [11],
[3]5 = [3][11] = [33] = [5], [3]6 = [3][5] = [15] = [1], so
ord([3]) = 6.
[5]2 = [25] = [11], [5]3 = [5][11] = [55] = [13], [5]4 = [5][13] =
[65] = [9], [5]5 = [5][9] = [45] = [3], [5]6 = [5][3] = [15] = [1], so
ord([5]) = 6.
[9]2 = [81] = [11], [9]3 = [9][11] = [99] = [1], so ord([9]) = 3.
[11]2 = [121] = [9], [11]3 = [11][9] = [99] = [1], so ord([11]) =
3.
[13]2 = [−1]2 = [1], so ord([13]) = 2.
Since |U14 | = 6 and there are elements of order 6, U14 is cyclic.
(4) U15 = {[1], [2], [4], [7], [8], [11], [13], [14]}.
[1] is the identity and has order 1.
[2]2 = [4], [2]3 = [8], [2]4 = [16] = [1], so ord([2]) = 4.
[4]2 = [16] = [1], so ord([4]) = 2.
[7]2 = [49] = [4], [7]3 = [7][4] = [28] = [13], [7]4 = [7][13] =
[91] = [1], so ord([7]) = 4.
[8]2 = [64] = [4], [8]3 = [8][4] = [32] = [2], [8]4 = [8][2] =
[16] = [1], so ord([8]) = 4.
[11]2 = [121] = [1], so ord([11]) = 2.
[13]2 = [−2]2 = [4], [13]3 = [−2][4] = [−8] = [7], [13]4 =
[−2][7] = [−14] = [1], so ord([13]) = 4.
[14]2 = [−1]2 = [1], so ord([14]) = 2.
Since |U15 | = 8 but there are no elements of order 8, U15 is
not cyclic.
3. Prove that [323] ∈ Z/1451Z has a multiplicative inverse and find it
using Euclid’s algorithm.
Solution:
Applying Euclid’s Algorithm:
1451
323
159
5
=
=
=
=
(323 × 4) + 159
(159 × 2) + 5
(5 × 31) + 4
(4 × 1) + 1,
so hcf(323, 1451) = 1, and so [323] does have a multiplicative inverse.
Working backwards through the calculation above:
1 =
=
=
=
=
=
=
5−4
5 − (159 − (5 × 31))
−159 + (5 × 32)
−159 + (323 − (159 × 2)) × 32
(323 × 32) − (159 × 65)
(323 × 32) − (1451 − (323 × 4)) × 65
−(1451 × 65) + (323 × 292),
and so [292] is the multiplicative inverse of [323], since
323 × 292 ≡ 1
(mod 1451).
4. Let (S, ∗) be a set with a binary operation satisfying all the group
axioms except for the existence of inverses. I.e, ∗ is associative, and S
has an identity element e such that e ∗ x = x = x ∗ e for all x ∈ S.
(a) Prove that if G is the set of elements x ∈ S that have an inverse
in S (i.e., such that there is x−1 ∈ S such that xx−1 = e = x−1 x), then
(G, ∗) is a group.
(b) Give some examples of groups you have seen that arise naturally
in this way.
Solution:
(a) If x, y are invertible, then (xy)(y −1 x−1 ) = e = (y −1 x−1 )(xy), so
xy is invertible, with inverse y −1 x−1 . So G is closed.
Associativity holds in G, since it does in S.
The identity e is invertible, being equal to its own inverse, so e ∈ G,
and G has an identity.
If x is invertible, then x−1 is invertible with inverse x, so x ∈ G ⇒
x−1 ∈ G, and so every element of G has an inverse in G. So G is a
group.
(b) Some examples:
• (R \ {0}, ×) in (R, ×).
• ({1, −1}, ×) in (Z, ×).
• (Un , ×) in (Z/nZ, ×).
• GL(n, R), the group of invertible n×n matrices, in (Mn (R), ×).
5. Let p > 3 be a prime.
(a) What is the order of U3p ?
(b) If [a] ∈ U3p prove that a2 ≡ 1 (mod 3) and ai ≡ 1 (mod p) for
some i with 0 < i < p. [Hint: Consider possible orders of [a] considered
as an element of U3 and Up .]
(c) Deduce that U3p is not cyclic.
Solution:
(a) Of the numbers 0, 1, . . . , 3p − 1, p are divisible by 3 and another
two (p and 2p) are diisible by p. So 2p − 2 are coprime to 3p, and so
|U3p | = 2p − 2.
(b) If hcf(a, 3p) = 1 then hcf(a, 3) = 1, and so the congruence class
of a (mod 3) is in U3 , every element of which has order 1 or 2, and so
a2 ≡ 1 (mod 3).
Also hcf(a, p) = 1, and since |Up | = p − 1, every element of Up has
order less than p, and so ai ≡ 1 (mod p) for some 0 < i < p.
(c) Taking i as in (b), aj ≡ 1 (mod 3p) if j is even (so that we
are guaranteed aj ≡ 1 (mod 3)) and divisible by i (so that aj ≡ 1
(mod p)). So if i < p − 1 then aj ≡ 1 (mod 3p) for j = 2i < 2p − 2,
and if i = p − 1 then aj ≡ 1 (mod 3p) for j = p − 1 < 2p − 2. In
either case, ord([a]) < 2p − 2 = |U3p |, and so U3p is not cyclic, having
no element whose order is |U3p |.
6. Prove that if p and q are two odd primes with p 6= q, then Upq is
not cyclic (generalizing question 5).
Solution: Follow the same strategy as in question 5.
Of the integers 0, 1, . . . , pq−1, p are divisible by q, q are divisible by p,
and one (zero) is diisible by both, so |Upq | = pq−p−q+1 = (p−1)(q−1).
Let [a] ∈ Upq . The order np of a as an element of Up is at most
p − 1. The order nq of a as an element of Uq is at most q − 1. So
n = lcm(np , nq ) < (p − 1)(q − 1) since either np < p − 1 or nq <
q − 1 or np = p − 1 and nq = q − 1 in which case lcm(np , nq ) divides
(p − 1)(q − 1)/2.
But an ≡ 1 (mod p) and an ≡ 1 (mod q), and so an ≡ 1 (mod pq)
and so ord([a]) ≤ n < |Upq |, and so Upq is not cyclic.
c
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