MAT301H1S Lec5101 Burbulla

Home , Automorphism, Function composition, Homomorphism, Identity function

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms

MAT301H1S Lec5101 Burbulla

Week 5 Lecture Notes

Winter 2020

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms

Chapter 10: Group Homomorphisms

Chapter 6: Isomorphisms

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms What is a Homomorphism?

Deﬁnition: Let G and H be groups and suppose f : G −→ H is a function such that for all x, y ∈ G,

f (xy) = f (x)f (y).

Then f is called a group homomorphism, or just homomorphism for short. In words, we say “f preserves products.”

Note: to calculate xy we are using the operation in G, but to calculate f (x)f (y) we are using the operation in H.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Examples

∗ ∗ 1. f : R −→ R by f (x) = |x| f (xy) = |xy| = |x||y| = f (x)f (y). ∗ ∗ 2 2. f : R −→ R by f (x) = x f (xy) = (xy)2 = x2y 2 = f (x)f (y).

3. f : Z −→ Zn by f (m) = m mod n, with addition. Use the division algorithm to write

m1 = p1n + q1, m2 = p2n + q2.

Then m1 + m2 = (p1 + p2)n + q1 + q2, so

f (m1+m2) = (q1+q2) mod n ≡ q1 mod n+q2 mod n = f (m1)+f (m2)

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Not All Functions are Homomorphisms

2 Let f : R −→ R by f (x) = x . The operation is addition. Then

f (x + y) = (x + y)2 = x2 + 2xy + y 2

but f (x) + f (y) = x2 + y 2.

So it is not true that f (x + y) = f (x) + f (y) for all x, y ∈ R, so f is not a homomorphism.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Some More Interesting Examples

∗ The determinant function, det : GL(n, R) −→ R , is a homomorphism, since

det(AB) = det(A) det(B).

Deﬁne sgn : Sn −→ {1, −1} by

1 if σ is an even permutaton sgn (σ) = −1 if σ is an odd permutaton

You have to check that for all α, β ∈ Sn,

sgn (αβ) = sgn (α) sgn (β),

which boils down to checking four cases, namely even plus even is even, even plus odd is odd, odd plus even is odd, and odd plus odd is even. Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Properties of Homomorphisms: Th 10.1 First Three Parts

Let f : G −→ H be a homomorphism. Then

1. f (eG ) = eH Proof: x = xeG ⇒ f (x) = f (xeG ) = f (x)f (eG ) ⇒ eH = f (eG ). 2. f (xn) = (f (x))n, for all x ∈ G and all n ∈ Z Proof: f (x2) = f (x x) = f (x) f (x) = (f (x))2, etc for positive n. If n = −1, then −1 −1 −1 xx = eG ⇒ f (xx ) = f (eG ) ⇒ f (x)f (x ) = eH ⇒ f (x−1) = (f (x))−1. 3. If |x| is ﬁnite, then |f (x)| divides |x| Proof: let |x| = n. Then n n n x = eG ⇒ f (x ) = f (eG ) ⇒ (f (x)) = eH ⇒ |f (x)| divides n.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Kernels and Images

If f : G −→ H is a homomorphism, deﬁne

1. ker(f ) = {x ∈ G | f (x) = eH } 2. im (f ) = {f (x) | x ∈ G} = f (G) Then ker(f ) ≤ G and im (f ) ≤ H. Proof: (for kernel)

I f (eG ) = eH ⇒ eG ∈ ker(f ). I x, y ∈ ker(f ) −1 −1 −1 ⇒ f (xy ) = f (x)f (y ) = f (x)(f (y)) = eH eH = eH , so xy −1 ∈ ker(f ).

I Proof that im (f ) ≤ H is left as an exercise.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Examples

∗ 1. For det : GL(n, R) −→ R , ker(det) = {A ∈ GL(n, R) | det(A) = 1} = SL(n, R). 2. For sgn : Sn −→ {1, −1} ker(sgn ) = {σ ∈ Sn | sgn (σ) = 1} = An. 3. For f : Z −→ Zn by f (m) = m mod n, with addition,

ker(f ) = {m | m ≡ 0 mod n} = < n >

and im (f ) = {m mod n | m ∈ Z} = Zn.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Proper Deﬁnition of Determinant

Let A = (aij ) be an n × n matrix. Here is the “proper” deﬁnition of det(A): X det(A) = sgn (σ) a1σ(1)a2σ(2) ····· anσ(n). σ∈Sn

For example, if n = 3, the even permutations are (1), (123), (132) and the odd permutations are (12), (23), (31). Then   a11 a12 a13 det  a21 a22 a23  = a31 a32 a33

a11a22a33+a12a23a31+a13a21a32−a12a21a33−a11a23a32−a13a22a31.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Properties of Subgroups Under Homomorphisms

First three parts of Theorem 10.2: let f : G −→ H be a homomorphism and let K be a subgroup of G. Then 1. f (K) = {f (k) | k ∈ K} is a subgroup of H. 2. If K is cyclic then f (K) is cyclic. 3. If K is Abelian then f (K) is Abelian. Proof: 1. left as an exercise. 2. Suppose K = hki. Then x ∈ K ⇒ x = kn. Then

f (x) = f (kn) = (f (k))n,

which means that f (K) = hf (k)i. 3. Suppose xy = yx for all x, y ∈ K. Then

f (xy) = f (yx) ⇔ f (x)f (y) = f (y)f (x),

which means all elements in f (K) commute.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example cos θ − sin θ Let f : −→ GL(2, ) by f (θ) = . f is a R R sin θ cos θ homomorphism because f (α)f (β) = f (α + β):

cos α − sin α cos β − sin β cos(α + β) − sin(α + β) = , sin α cos α sin β cos β sin(α + β) cos(α + β

using appropriate trig identities. Check that ker(f ) = h2πi and im (f ) is the group of 2 × 2 rotation matrices. In particular, R is Abelian, so im (f ) is Abelian and im (f ) is an Abelian subgroup of GL(2, R). If the positive integer n is ﬁxed and 2π 2πk K = = | k ∈ , n n Z

then f (K) = hf (2π/n)i ≤ Dn, consisting of all rotations of a regular n-gon.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms One-to-one and Onto Homomorphisms

Let f : G −→ H be a group homomorphism. Deﬁnition: f is called one-to-one if

f (x1) = f (x2) ⇒ x1 = x2. Deﬁntion: f is called onto if every element in H is in im (f ). Theorem: let f : G −→ H be a group homomorphism.

I f is one-to-one if and only if ker(f ) = {eG } I f is onto if and only if im (f ) = H Proof: (for one-to-one) suppose f is one-to-one: Let x ∈ ker(f ). Then f (x) = eH and f (eG ) = eH . Therefore x = eG . Now suppose ker(f ) = {eG }, and let f (x1) = f (x2). Then

−1 −1 −1 −1 f (x1x2 ) = f (x1)f (x2 ) = f (x1)(f (x2)) = f (x1)(f (x1)) = eH . −1 −1 Thus x1x2 ∈ ker(f ) ⇒ x1x2 = eG ⇒ x1 = x2. Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms What is an Isomorphism?

Let f : G −→ H be a group homomorphism. Deﬁnition: f is called an isomorphism if f is one-to-one and onto. Thus f is an isomorphism if ker(f ) = {eG } and im (f ) = H. Note: every isomorphism f : G −→ H has an inverse f −1 : H −→ G, deﬁned by

f −1(x) = y ⇔ x = f (y),

which is also an isomorphism. Deﬁnition: if there is an isomorphism f : G −→ H, then we say the groups G and H are isomorphic, and we write

G ≈ H.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 1

1 m Let H = ∈ GL(2, ) | m ∈ and deﬁne f : −→ H 0 1 R Z Z by 1 m f (m) = . 0 1 Then f is an isomorphism: 1. f is a homomorphism since

1 m + n 1 m 1 n f (m+n) = = = f (m)f (n) 0 1 0 1 0 1

1 0 2. f is one-to-one since f (m) = ⇒ m = 0 0 1 3. f is onto since obviously im (f ) = H.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 2

Let G = hai be a cyclic group of order n. Then G ≈ Zn. k Proof: for k ∈ Z, deﬁne f : G −→ Zn by f (a ) = k mod n. j k I f is well-deﬁned: a = a ⇒ n divides j − k ⇒ j ≡ k mod n. j k j+k I f is a homomorphism: f (a a ) = f (a ) = (j + k) mod n; and f (aj ) + f (ak ) = j mod n + k mod n = (j + k) mod n.

I f is one-to-one: k k n q f (a ) = 0 ⇒ k ≡ 0 mod n ⇒ k = qn ⇒ a = (a ) = eG . I f is onto: this is obvious since k ∈ Z, so im (f ) = Zn. m Alternate Proof: use g : Zn −→ G deﬁned by g(m) = a . (g = f −1.) This makes the algebra easier since g is obviously well-deﬁned.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Properties of Isomorphisms

Ref: Theorems 6.2 and 6.3. Let f : G −→ H be an isomorphism. Then 1. f −1 : H −→ G is also an isomorphism

2. f (eG ) = eH 3. for all x ∈ G, f (xn) = (f (x))n 4. x, y commute in G if and only if f (x), f (y) commute in H 5. G = hxi if and only if H = hf (x)i 6. for all x ∈ G, |x| = |f (x)| 7. if G and H are ﬁnite groups, then |G| = |H| 8. G is Abelian if and only if H is Abelian 9. G is cyclic if and only if H is cyclic 10. f (Z(G)) = Z(H)

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Some Proofs:

Properties 2 and 3 are true for all homomorphisms. 4. xy = yx ⇒ f (xy) = f (yx) ⇒ f (x)f (y) = f (y)f (x); this is true for any homomorphism. But if f is an isomorphism, then f (x)f (y) = f (y)f (x) ⇒ f (xy) = f (yx) ⇒ xy = yx, since f is one-to-one 6. let |x| = n, |f (x)| = m. For any homomorphism f we know m m divides n. If f is also an isomorphism: (f (x)) = eH ⇒ m m f (x ) = eH ⇒ x = eG , since f is one-to-one. Thus n divides m. So m = n. 8. follows from 4 9. follows from 5 10. x ∈ Z(G) ⇒ xg = gx, for all g ∈ G ⇒ f (x)f (g) = f (g)f (x) ⇒ f (x) ∈ Z(im (f )). So, for any homomorphism, f (Z(G)) ⊂ Z(im (f )). If f is an isomorphism, then im (f ) = H. For the other inclusion, repeat the argument for f −1 : H −→ G.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 3

Show U(8) ≈ {(1), (12)(34), (13)(24), (23)(14)} Solution: U(8) = {1, 3, 5, 7}. Deﬁne f : U(8) −→ {(1), (12)(34), (13)(24), (23)(14)} by

f (1) = (1), f (3) = (12)(34), f (5) = (13)(24), f (7) = (23)(14).

Now consider the multiplication tables of both groups: · 1 3 5 7 ◦ (1) (12)(34) (13)(24) (23)(14) 1 1 3 5 7 (1) (1) (12)(34) (13)(24) (23)(14) 3 3 1 7 5 (12)(34) (12)(34) (1) (23)(14) (13)(24) 5 5 7 1 3 (13)(24) (13)(24) (23)(14) (1) (12)(34) 7 7 5 3 1 (23)(14) (23)(14) (13)(24) (12)(34) (1) Observe that f maps the multiplication table on the left precisely to the multiplication table on the right: i.e. f (xy) = f (x)f (y). We say, isomorphic groups have the same group structure.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 4

Finite groups that are isomorphic have the same order. But groups can have the same order and not be isomorphic. For example, here are three groups of order 12: Z12, D6, and A4, no two of which are isomorphic. There are many ways to show this:

I the largest order of any element in Z12, D6 or A4 is 12, 6, or 3, respectively.

I compare the number of elements of order 2: Z12 has 1, D6 has 7, and A4 has 3. I Z12 is Abelian and cyclic; the biggest cyclic subgroup D6 has is of order 6; the biggest cyclic subgroup A4 has is of order 3. There are only ﬁve non-isomorphic groups of order 12: the three above, plus one other Abelian group and one other non-Abelian group.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Cayley’s Theorem

Theorem 6.1: Every group is isomorphic to a group of permutations. If |G| = n, then G is isomorphic to a subgroup of Sn. Proof: given a group G we shall construct a group of permutations, H, and then show that G ≈ H. For g ∈ G, deﬁne −1 Tg : G −→ G by Tg (x) = gx. Since Tg (g y) = y, Tg is onto; since Tg (x) = Tg (y) ⇒ gx = gy ⇒ x = y, Tg is one-to-one. Thus Tg is a permutation of all the elements of G. Let

H = {Tg | g ∈ G},

with function composition, and deﬁne φ : G −→ H by

φ(g) = Tg .

We claim φ is an isomorphism:

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms

1. φ is a homomorphism: φ(ab) = φ(a) ◦ φ(b). To show two functions are equal you have to evaluate them at a point. (φ(ab))(x) = Tab(x) = abx and (φ(a) ◦ φ(b))(x) = Ta(Tb(x)) = Ta(bx) = abx.

2. φ is one-to-one: φ(a) = φ(b) ⇒ Ta(x) = Tb(x), for all x ∈ G ⇒ ax = bx ⇒ a = b.

3. φ is onto: by deﬁnition of φ. For any Tg ∈ H, φ(g) = Tg . Finally, if |G| = n, then you can number the elements of G as x1, x2,..., xn and each element Tg ∈ H can be considered as a permutation σ ∈ Sn deﬁned by

Tg (xi ) = xσ(i).

Note: Example 3 above showed that U(8) is isomorphic to a subgroup of S4.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms What Is An Automorphism?

Deﬁntion: if f : G −→ G is an isomorphism, of G to itself, then we call f an automorphism of G.

Every group G has at least one automorphism, the identity map i : G −→ G, deﬁned by i(g) = g.

The map f : G −→ G deﬁned by f (x) = x−1 is an automorphism only if G is Abelian:

f (xy) = f (x)f (y) ⇔ (xy)−1 = x−1y −1 ⇔ y −1x−1 = x−1y −1.

This last statement will be true if G is Abelian.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 5

−1 T Deﬁne f : GL(n, R) −→ GL(n, R) by f (A) = (A ) . Then f is an automorphism of GL(n, R). 1. f is a homomorphism: f (AB) = ((AB)−1)T = (B−1A−1)T = (A−1)T (B−1)T = f (A)f (B).

2. f is one-to-one: f (A) = I ⇒ (A−1)T = I ⇒ A−1 = I T = I ⇒ A = I −1 = I

3. f is onto: observe that −1T f (AT )−1 = (AT )−1 = (AT )T = A

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms What Is an Inner Automorphism?

Theorem: for g ∈ G, φg : G −→ G, deﬁned by

−1 φg (x) = gxg

is an automorphism of G. Proof:

1. φg is a homomorphism:

−1 −1 −1 φg (x)φg (y) = (gxg )(gyg ) = gxyg = φg (xy)

−1 −1 2. φ is one-to-one: φg (x) = e ⇒ gxg = e ⇒ x = g g = e. −1 −1 −1 3. φg is onto: for any x ∈ G, φg (g xg) = g(g xg)g = x

Deﬁnition: φg is called an inner automorphism of G, or to be more speciﬁc, the inner automorphism of G induced by g.

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 6

Note 1: if G is Abelian, then φg is always the identity function, −1 −1 because φg (x) = gxg = gg x = x. Note 2: if g ∈ Z(G), then φg is the identity function, because −1 −1 φg (x) = gxg = xgg = x. Thus the only interesting inner automorphisms of a group are φg for g ∈/ Z(G). 2 1 Example: let A = ; then φ : GL(2, ) −→ GL(2, ) is 7 4 A R R

a b a b 2 1 a b 4 −1 φ = A A−1 = A c d c d 7 4 c d −7 2

8a + 4c − 14b − 7d −2a − c + 4b + 2d = 28a + 16c − 49b − 28d −7a − 4c + 14b + 8d

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Groups of Automorphisms

Theorem 6.4: let Aut (G) be the set of all automorphisms of G; let Inn (G) be the set of all inner automorphisms of G. Then both Aut (G) and Inn (G), with operation composition of functions, are groups, and Inn (G) ≤ Aut (G). Proof: (some) the identity map, i(x) = x, is in Aut (G), so it is non-empty. If f , g ∈ Aut (G), we need to show f ◦ g −1 ∈ Aut (G): 1. f ◦ g −1 is a homomorphism: (f ◦ g −1)(xy) = f ((g −1)(xy)) = f (g −1(x)g −1(y)) = f (g −1(x))f (g −1(y)) = (f ◦ g −1)(x)(f ◦ g −1)(y). 2. f ◦ g −1 is one-to-one: f (g −1(x)) = e ⇒ g −1(x) = e ⇒ x = e 3. f ◦ g −1 is onto: for any x ∈ G, check that (f ◦ g −1)(g(f −1(x)) = x. This shows that Aut (G) contains the identity and is closed under composition and inverses. Finally, composition is associative: it’s always true that f ◦ (g ◦ h) = (f ◦ g) ◦ h. (Similarly for Inn (G).)

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla

Chapter 10: Group Homomorphisms Chapter 6: Isomorphisms Example 7

Aut (Z42) ≈ U(42) : let f ∈ Aut (Z42). Since Z42 = h1i, f is completely determined by what it does to 1. Let f (1) = m. Since f is an isomorphism, |f (1)| = |1| = 42, so there are only twelve possibilities for f :

f (1) = 1, 5, 11, 13, 17, 19, 23, 25, 29, 31, 37 or 41.

That is, f (1) = m ⇔ m ∈ U(42). Let fm be the autormorphism of Z42 such that fm(1) = m. Check that the correspondence φ : U(42) −→ Aut (Z42), given by φ(m) = fm, is an isomorphism.

This is a special case of Theorem 6.5: for every positive integer n,

Aut (Zn) ≈ U(n).

Week 5 Lecture Notes MAT301H1S Lec5101 Burbulla