MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products

Week 7 Lecture Notes

Winter 2020

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products What is an External Direct Product of Groups?

Deﬁnition: Let G1, G2,..., Gn be a ﬁnite collection of groups. The external direct product of G1, G2,..., Gn is the set

G1 ⊕ G2 ⊕ · · · ⊕ Gn = {(g1, g2,..., gn) | gi ∈ Gi , 1 ≤ i ≤ n}

with component-wise ‘multiplication’

0 0 0 0 0 0 (g1, g2,..., gn)(g1, g2,..., gn) = (g1g1, g2g2,..., gngn).

Warning: many other books use the notation G1 × G2 × · · · × Gn instead. (In particular, WeBWorK uses this notation!) Theorem: G1 ⊕ G2 ⊕ · · · ⊕ Gn is a group, its identity element is

(eG1 , eG2 ,..., eGn ), and if each Gi is ﬁnite, then

|G1 ⊕ G2 ⊕ · · · ⊕ Gn| = |G1| |G2| · · · |Gn|.

Proof: exercise left for the reader. Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products Example 1

Consider R with addition. In linear algebra, the Euclidean vector spaces

2 3 n R = R ⊕ R, R = R ⊕ R ⊕ R, R = R ⊕ R ⊕ · · · ⊕ R | {z } n times are examples of external direct products you should be familiar with, although in linear algebra we usually write the elements as columns, instead of rows, and call them vectors:       x1 y1 x1 + y1  x2   y2   x2 + y2  ~x + ~y =   +   =    ...   ...   ...  xn yn xn + yn

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 2

Z3 ⊕ Z4 = {(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1)} ∪ {(0, 2), (1, 2), (2, 2), (0, 3), (1, 3), (2, 3)}. |Z3 ⊕ Z4| = 12. The order of (1, 0) is 3, since (1, 0) + (1, 0) + (1, 0) = (3, 0), and 3 ≡ 0 mod 3. The order of (0, 2) is 2, since (0, 2) + (0, 2) = (0, 4), and 4 ≡ 0 mod 4. The order of (1, 3) is 12: observe 2(1, 3) = (2, 6) 6= (0, 0), 3(1, 3) = (3, 9) 6= (0, 0), and 4(1, 3) = (4, 12) 6= (0, 0), and 6(1, 3) = (6, 18) 6= (0, 0). But the order of (1, 3) must divide 12, so the only choice left is 12.

Thus Z3 ⊕ Z4 ≈ Z12.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products Example 3

U(4) ⊕ U(10) = {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3), (3, 7), (3, 9)}. The product (3, 7)(3, 9) = (9, 63) = (1, 3), since the ﬁrst component is multiplied mod 4, and the second component is multiplied mod 10. |U(4) ⊕ U(10)| = 8, but it is not cyclic, even though U(4) and U(10) are both cyclic. Its not cyclic because it has at least two elements of order 2: both (1, 9) and (3, 9) have order 2 in U(4) ⊕ U(10). (As we saw in Chapter 4, the number of elements of order 2 in a cyclic group of even order must be φ(2) = 1.)

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Some Properties of External Direct Products

Let G1, G2,..., Gn, G, H be ﬁnite groups.

1. The order of (g1, g2,..., gn) ∈ G1 ⊕ G2 ⊕ · · · ⊕ Gn is lcm(|g1|, |g2|,..., |gn|). 2. Let G and H be cyclic groups. Then G ⊕ H is cyclic if and only if |G| and |H| are relatively prime.

3. The external direct product G1 ⊕ G2 ⊕ · · · ⊕ Gn of a ﬁnite number of ﬁnite cyclic groups is cyclic if and only if |Gi | and |Gj | are relatively prime, for each pair i 6= j.

4. Let m = n1n2 ··· nk . Then

Zm ≈ Zn1 ⊕ Zn2 ⊕ · · · ⊕ Znk

if and only if ni and nj are relatively prime for each pair i 6= j.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products Example 4

|Z6 ⊕ U(14)| = 6 × 6 = 36. In Z6 ⊕ U(14) the order of (3, 3) is 6, because |3| = 2 in Z6, |3| = 6 in U(14), and lcm(2, 6) = 6.

|S4 ⊕ U(9)| = 24 × 6 = 144. In S4 ⊕ U(9) the order of ((24)(31), 4) is 6, because |(24)(31)| = 2 in S4, |4| = 3 in U(9), and lcm(2, 3) = 6.

cos(2π/3) − sin(2π/3) In SL(2, ) ⊕ the order of , 1 R Z4 sin(2π/3) cos(2π/3)

is 12, because the matrix has order 3 and |1| = 4 in Z4.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 5

Find the number of elements of order 5 in Z25 ⊕ Z5.

Solution: |(a, b)| = lcm(|a|, |b|) = 5 ⇒ (|a|, |b|) = (1, 5), (5, 1) or (5, 5). If |a| = 5, then a = 5, 10, 15 or 20. If |b| = 1, then b = 0. So in this case there will be 4 × 1 = 4 elements (a, b) of order 5. If |b| = 5, and |a| = 5, then b = 1, 2, 3 or 4, and a is as in the ﬁrst case. In this case there are 4 × 4 = 16 elements (a, b) of order 5. If |a| = 1, then a = 0; if |b| = 5, then b = 1, 2, 3 or 4. In this case there are 4 elements (a, b) of order 5.

All together, there are 24 elements of order 5 in Z25 ⊕ Z5.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products Example 6

Z30 ≈ Z2 ⊕ Z15 ≈ Z2 ⊕ Z3 ⊕ Z5. Consequently,

Z2 ⊕ Z30 ≈ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z5.

Similarly, Z30 ≈ Z3 ⊕ Z10 and consequently

Z2 ⊕ Z30 ≈ Z2 ⊕ Z3 ⊕ Z10 ≈ Z6 ⊕ Z10.

This shows that Z2 ⊕ Z30 ≈ Z6 ⊕ Z10. But Z2 ⊕ Z30 is not isomorphic to Z60, one has an element of order 60, but the other one doesn’t.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products U(n) As An External Direct Product

Theorem 8.3: if s and t are relatively prime, then U(st) ≈ U(s) ⊕ U(t). Proof: (outline) deﬁne f : U(st) −→ U(s) ⊕ U(t) by

f (x) = (x mod s, x mod t).

Check f is a homomorphism; exercise left for the reader. Show f is one-to-one:

x ∈ ker(f ) ⇒ x ≡ 1 mod s and x ≡ 1 mod t.

So s|x − 1 and t|x − 1. Since s and t are relatively prime, this implies st|x − 1. Thus x ≡ 1 mod (st). So f is a one-to-one correspondence between two ﬁntite sets of the same order,

|U(st)| = φ(st) = φ(s)φ(t) = |U(s)| |U(t)|,

so f must be onto as well. Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products Deﬁntion: if k is a divisor of n we deﬁne

Uk (n) = {x ∈ U(n) | x ≡ 1 mod k}.

Observation: If s and t are relatively prime and n = st, then

Us (n) ≈ U(t) and Ut (n) ≈ U(s).

For example:

U(21) = {1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20}, |U(21)| = 12,

U(3) = {1, 2}, |U(3)| = 2; and U(7) = {1, 2, 3, 4, 5, 6}, |U(7)| = 6;

U7(21) = {1, 8} and U3(21) = {1, 4, 10, 13, 16, 19}.

The isomorphisms are f : U7(21) −→ U(3) by f (x) = x mod 3; g : U3(21) −→ U(7) by g(y) = y mod 7, as you can check.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 8: External Direct Products Example 7

As a corollary to Theorem 8.3, we have if m = n1n2 ··· nk and each pair ni , nj is relatively prime for i 6= j, then

U(m) ≈ U(n1) ⊕ U(n2) ⊕ · · · ⊕ U(nk ).

For example:

U(105) ≈ U(7) ⊕ U(15) ≈ U(21) ⊕ U(5) ≈ U(3) ⊕ U(5) ⊕ U(7).

Each of these smaller groups in turn are cyclic,

U(3) ≈ Z2, U(5) ≈ Z4, U(7) ≈ Z6.

So U(105) ≈ Z2 ⊕ Z4 ⊕ Z6 and |U(105)| = 2 × 4 × 6 = 48.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 8: External Direct Products Example 8

Recall that Aut (Z105) ≈ U(105). The calculations in the previous slide thus give us a description of Aut (Z105). For example, consider the question: how many automorphisms of Z105 are there with order 12? We can answer this question more easily by asking, how many elements of order 12 are there in Z2 ⊕ Z4 ⊕ Z6? If (a, b, c) ∈ Z2 ⊕ Z4 ⊕ Z6 has order 12, then lcm(|a|, |b|, |c|) = 12 ⇔

(|a|, |b|, |c|) = (1, 4, 3), (2, 4, 3), (1, 4, 6), or (2, 4, 6).

There are four possibilities in each case, so Aut (Z105) has 16 automorphisms of order 12.

Week 7 Lecture Notes MAT301H1S Lec5101 Burbulla