CHAPTER 6

Isomorphisms

A while back, we saw that hai with |hai| = 42 and Z42 had basically the samestructure, and said the groups were “isomorphic,” in some sense the same.

Definition (Group Isomorphism). An isomorphism � from a group G toa group G is a 1–1 mapping from G onto G that preserves the group operation.That is,

�(ab) = �(a)�(b) 8 a, b 2 G.

If there is an isomorphism from G onto G, we say G and G arr isomorphic andwrite G ⇡ G.

Note. The operations in G and G may be di↵erent. The operation to theleft of “=” is in G, while that on the right is in G.

70

6. ISOMORPHISMS 71

To establish an isomorphism:

(1) Define � : G ! G.

(2) Show � is 1–1: assuming �(a) = �(b), show a = b.

(3) Show � is onto: 8 g 2 G. show 9 g 2 G 3�� �(g) = g.

(4) Show �(ab) = �(a)�(b) 8 a, b 2 G.

Example. With |a| = 42, show hai ⇡ Z42.

Proof.

Define � : hai ! Z42 by �(an) = n mod 42.

[Show 1–1.] Suppose �(an) = �(am). Then

n = m mod 42 =) 42|n�m =) an = am

by Theorem 4.1. Thus � is 1–1.

[Show onto.] For all n 2 Z42, an 2 hai and �(an) = n, so � is onto.

[Show operation preservation.] For all an, am 2 hai,�(anam) = �(an+m) = n+m mod 42 = n mod 42+m mod 42 = �(an)+�(am).

Thus, by definition, hai ⇡ Z42. ⇤

Example. Any finite cyclic group hai with |hai| = n is isomorphic to Zn

under �(ak) = k mod n. The proof of this is identical to that of the previousexample.

72 6. ISOMORPHISMS

Example. Let G be the positive real numbers with multiplication and Gbe the group of real numbers with addition. Show G ⇡ G.

Proof.

Define � : G ! G by �(x) = ln(x).

[Show 1–1.] Suppose �(x) = �(y). Then

ln(x) = ln(y) =) eln(x) = eln(y) =) x = y,

so � is 1–1.

[Show onto] Now suppose x 2 G. ex > 0 and �(ex) = ln ex = x, so � is onto.

[Show operation preservation.] Finally, for all x, y 2 G,

�(xy) = ln(xy) = ln x + ln y = �(x) + �(y),

so G ⇡ G.

[Question: is � the only isomorphism?] ⇤

Example. Any infinite cyclic group is isomorphic to Z with addition. Givenhai with |a| = 1, define �(ak) = k. The map is clearly onto. If �(an) = �(am),n = m =) an = am, so � is 1–1. Also,

�(anam) = �(an+m) = n + m = �(n) + �(m),

so hai ⇡ Z.

Consider h2i under addition, the cyclic group of even integers. The h2i ⇡ Z =h1i with � : h2i ! Z defined by �(2n) = n.

6. ISOMORPHISMS 73

Example. Let G be the group of real numbers under addition and G bethe group of positive numbers under multiplication. Define �(x) = 2x�1. � is1–1 and onto but

�(1 + 2) = �(3) = 4 6= 2 = 1 · 2 = �(1)�(2),

so � is not an isomorphism.

Can this mapping be adjusted to make it an isomorphism?

Use �(x) = 2x.

If �(x) = �(y), 2x = 2y =) log2 2x = log2 2y =) x = y, so � is 1–1.

Given any positive y, �(log2 y) = 2log2 y = y, so � is onto.

Also, for all x, y 2 R,�(x + y) = 2x+y = 2x2y = �(x)�(y).

Thus � is an isomorphism.

Example. Are U(8) and U(12) isomorphic?

Solution.

U(8) = {1, 3, 5, 7} is noncyclic with |3| = |5| = |7| = 2.

U(12) = {1, 5, 7, 11} is noncyclic with |5| = |7| = |11| = 2.

Define � : U(8) ! U(12) by 1 ! (1), 3 ! 5, 5 ! 7, 7 ! 11.

� is clearly 1–1 and onto. Regarding operation preservation:

�(3 · 5) = �(7) = 11 and �(3)�(5) = 5 · 7 = 11.

�(3 · 7) = �(5) = 7 and �(3)�(7) = 5 · 11 = 7.

�(5 · 7) = �(3) = 5 and �(5)�(7) = 7 · 11 = 5.

�(1 · 3) = �(3) = 5 and �(1)�(3) = 1 · 5 = 5.

etc.

Since both groups are Abelian, we need only check each pair in a single order.Thus, U(8) ⇡ U(12). ⇤

74 6. ISOMORPHISMS

This group of order 4, with all non-identity element of order 2, is called theKlein 4 group. Any other group of order 4 must have an element of order 4, sois cyclic and isomorphic to Z4.

For orders 1, 2, 3, 5, there are only the cyclic groups Z1, Z2, Z3, and Z5.

Problem (Page 140 # 36).

Let G = {a + bp

2�� a, b 2 Q} and H =

⇢a 2bb a

� ���� a, b 2 Q�

Show

that G and H are isomorphic under addition. Prove that G and H are closedunder multiplication. Does your isomorphism preserve multiplication as well asaddition? (G and H are examples of rings — a topic we begin in Chapter 12)

Solution.

Given a + bp

2, c + dp

2 2 G.

(a + bp

2) + (c + dp

2) = (a + c) + (b + d)p

2 2 G and

(a + bp

2)(c + dp

2) = (ac + 2bd) + (ad + bc)p

2 2 G

since (a + c), (b + d), (ac + 2bd), (ad + bc) 2 Q.

Also, if

a 2bb a

�,

c 2dd c

�2 H,

a 2bb a

�+

c 2dd c

�=

a + c 2(b + d)b + d a + c

�and

a 2bb a

� c 2dd c

�=

ac + 2bd 2(ad + bc)ad + bc ac + 2bd

�2 H

since (a + c), (b + d), (ac + 2bd), ad + bc 2 Q.

Thus G and H are closed under both addition and multiplication. Also, G andH are groups under addition (you can prove this, if you desire).

6. ISOMORPHISMS 75

Define � : G ! H by �(a + bp

2) =

a 2bb a

�.

Suppose �(a + bp

2) = �(c + dp

2). Then

a 2bb a

�=

c 2dd c

�=)

a = c and b = d =) a + bp

2 = c + dp

2 =) � is 1–1.

For

a 2bb a

�2 H, a + b

p2 2 G and �(a + b

p2) =

a 2bb a

�, so � is onto.

Now suppose a + bp

2, c + dp

2 2 G. Then

��(a + b

p2) + (c + d

p2)

�= �

�(a + c) + (b + d)

p2�

=

a + c 2(b + d)b + d a + c

�=

a + c 2b + 2db + d a + c =

�=

a 2bb a

�+

c 2dd c

�= �(a + b

p2) + �(c + d

p2).

Thus � is an isomorphism under addition.

Also,

��(a+b

p2)(c+d

p2)

�= �

�(ac+2bd)+(ad+bc)

p2�

=

ac + 2bd 2(ad + bc)ad + bc ac + 2bd

�=

ac + 2bd 2ad + 2bcad + bc ac + 2bd

�=

a 2bb a

� c 2dd c

�=

��(a + b

p2)

���(c + d

p2)

�.

So � preserves multiplication also. (We will later see that this makes � a ringhomomorphism.) ⇤

76 6. ISOMORPHISMS

Problem (Page 141 # 54). Consider G = hmi and H = hni where m,n 2Z. These are groups under addition. Show G ⇡ H. Does this isomorphismalso preserve multiplication?

Solution.

Define � : G ! H by �(x) =n

mx. Suppose �(x) = �(y). Then

n

mx =

n

my =) x = y =) � is 1–1.

For x 2 H, x = rn where r 2 Z. Thenm

nx =

m

nrn = rm 2 G. Then

�⇣m

nx⌘

=n

m

⇣m

nx⌘

= x, so � is onto. Now, suppose x, y 2 G. Then

�(x + y) =n

m(x + y) =

n

mx +

n

my = �(x) + �(y),

so addition is preserved and G ⇡ H. But,

�(xy) =n

m(xy) 6=

⇣ n

mx⌘⇣ n

my⌘

= �(x)�(y),

so multiplication is not preserved by �. ⇤

6. ISOMORPHISMS 77

Theorem (3). ⇡ is an equivalence relation on the set G of all groups.

Proof.

(1) G ⇡ G. Define � : G ! G by �(x) = x. This is clear.

(2) Suppose G ⇡ H. If � : G ! H is the isomrphism and, for g 2 G,�(g) = h, then ��1 : H ! G where ��1(h) = g is 1–1 and onto.

Suppose a, b 2 H. Since � is onto, 9 ↵,� 2 G 3�� �(↵) = a and �(�) = b.Then �(↵�) = �(↵)�(�) = ab, so

��1(ab) = ��1��(↵)�(�)

�= ��1

��(↵�)

�= ↵� =

��1��(↵)

���1

��(�)

�= ��1(a)��1(b).

Thus ��1 is an isomorphism and H ⇡ G.

(3) Suppose G ⇡ H and H ⇡ K with � : G ! H and : H ! K theisomorphisms. Then � : G ! K is 1–1 and onto.

For all a, b 2 G,

( �)(ab) = ⇥�(ab)

⇤=

⇥�(a)�(b)

⇤=

⇥�(a)

⇤

⇥�(b)

⇤=

⇥( �)(a)

⇤⇥( �)(b)

⇤.

Thus G ⇡ K and ⇡ is an equivalence relation on G. ⇤

Note. Thus, when two groups are isomorphic, they are in some sense equal.They di↵er in that their elements are named di↵erently. Knowing of a com-putation in one group, the isomorphism allows us to perform the analagouscomputation in the other group.

78 6. ISOMORPHISMS

Example (Conjugation by a). Let G be a group, a 2 G. Define a function�a : G ! G by �a(x) = axa�1.

Suppose �a(x) = �a(y). Then axa�1 = aya�1 =) x = y by left and rightcancellation. Thus �a is 1–1.

Now suppose y 2 G. We need to find x 2 G 3�� axa�1 = y. But that will beso if x = a�1ya, for then

�a(x) = �a(a�1ya) = a(a�1ya)a�1 = (aa�1)y(aa�1) = eye = y,

so �a is onto.

Finally, for all x, y 2 G,

�a(xy) = axya�1 = axeya�1 = axa�1aya�1 = �a(x)�a(y).

Thus �a is an isomorphism from G onto G.

Recall U(8) = {1, 3, 5, 7}. Consider �7. Since 7 is its own inverse,

�7(1) = 7 · 1 · 7 = 1,

�7(3) = 7 · 3 · 7 = 5 · 7 = 3,

�7(5) = 7 · 5 · 7 = 3 · 7 = 5,

�7(7) = 7 · 7 · 7 = 1 · 7 = 7.

Conjugation by 7 turns out to be the identity isomorphism, which is not veryinteresting.

6. ISOMORPHISMS 79

Theorem (6.1 — Cayley’s Theorem). Every group is isomorphic to agroup of permutations.

Proof.

Let G be any group. [We need to find a set A and a permutation on A thatforms a group G that is isomorphic to G.] We take the set G and define, foreach g 2 G, the function Tg : G ! G by Tg(x) = gx 8x 2 G.

Now Tg(x) = Tg(y) =) gx = gy =) x = y by left cancellation, so Tg is 1–1.

Given y 2 G, by Theorem 2 (solutions of equations) 9 a unique solution

x 2 G 3�� gx = y or Tg(x) = y. Thus Tg is onto and so is a permutation on G.

Now define G by {Tg|g 2 G}. For any g, h 2 G,

(TgTh)(x) = Tg

�Th(x)

�= Tg(hx) = g(hx) = (gh)(x) = Tgh(x) 8 x 2 G,

so G is closed under composition. Then Te is the identity and T�1g = Tg�1.

Since composition is associative, G is a group.

Now define � : G ! G by �(g) = Tg 8 g 2 G.

�(g) = �(h) =) Tg = Th =) Tg(e) = Th(e) =) ge = he =) g = h,

so � is 1–1. � is clearly onto by construction.

Finally, for g, h 2 G,

�(gh) = Tgh = TgTh = �(g)�(h),

so � is an isomorphism. ⇤

80 6. ISOMORPHISMS

Example. U = {1, 3, 5, 7}.

T1 =

1 3 5 71 3 5 7

�, T3 =

1 3 5 73 1 7 5

�, T5 =

1 3 5 75 7 1 3

�, T7 =

1 3 5 77 5 3 1

�.

Then U(8) = {T1, T3, T5, T7}.

U(8) 1 3 5 71 1 3 5 73 3 1 7 55 5 7 1 37 7 5 3 1

U(8) T1 T3 T5 T7

T1 T1 T3 T5 T7

T3 T3 T1 T7 T5

T5 T5 T7 T1 T3

T7 T7 T5 T3 T1

Theorem (6.2 — Properties of Elements Acting on Elements). Suppose �is an isomorphism from a group G onto a group G. Then

(1) �(e) = e.Proof.

�(e) = �(ee) = �(e)�(e). Also, since �(e) 2 G, �(e) = e�(e) =)e�(e) = �(e)�(e) =) e = �(e) by right cancellation. ⇤

(2) For all n 2 Z and for all a 2 G, �(an) =⇥�(a)

⇤n.

Proof.

For n 2 Z, n � 0, �(an) =⇥�(a)

⇤nfrom the definition of an isomorphism and

induction. For n < 0, �n > 0, so

e = �(e) = �(ana�n) = �(an)�(a�n) = �(an)⇥�(a)

⇤�n=)

⇥�(a)

⇤n= �(an).

⇤

(3) For all a, b 2 G, ab = ba () �(a)�(b) = �(b)�(a).Proof.

ab = ba () �(ab) = �(ba) () �(a)�(b) = �(b)�(a).

⇤

6. ISOMORPHISMS 81

(4) G = hai () G = h�(a)i.Proof.

Let G = hai. By closure, h�(a)i ✓ G. Since � is onto, for any element b 2 G,

9 ak 2 G 3�� �(ak) = b. Thus b =⇥�(a)

⇤k=) b 2 h�(a)i. Thus G = h�(a)i.

Now suppose G = h�(a)i. Clearly, hai ✓ G. For all b 2 G, �(b) 2 h�(a)i.Thus 9 k 2 Z 3�� �(b) = �(a)k = �(ak). Since � is 1–1, b = ak. Thushai = G. ⇤

(5) |a| = |�(a)| 8 a 2 G (Isomorphisms preserve order).

Proof.

an = e () �(an) = �(e) ()⇥�(a)

⇤n= e.

Thus a has infinite order () �(a) has infinite order, and a has finite ordern () �(a) has infinite order n. ⇤

(6) For a fixed integer k and a fixed b 2 G, xk = b has the same numberof solutions in G as does xk = �(b) in G.

Proof.

Suppose a is a solution of xk = b in G, i.e., ak = b. Then

�(ak) = �(b) =)⇥�(a)

⇤k= �(b),

so �(a) is a solution of xk = �(b) in G.

Now suppose y is a solution of xk = �(b) in G. Since � is onto, 9 a 2 G 3���(a) = y. Then⇥

�(a)⇤k

= �(b) () �(ak) = �(b) =) ak = b

since � is 1–1. Thus a is a solution of xk = b in G.

Therefore, there exists a 1–1 correspondence between the sets of solutions. ⇤

(7) If G is finite, then G and G have exactly the same number of elementsof every order.

Proof. Follows directly from property (5). ⇤

82 6. ISOMORPHISMS

Example. Is C⇤ ⇡ R⇤ with multiplication the operation of both groups?Solution.

x2 = �1 has 2 solutions in C⇤, but none in R⇤, so by Theorem 6.2(6) noisomorphism can exist. ⇤

Theorem (6.3 — Properties of Isomorphisms Acting on Groups).

Suppose that � is an isomorphism from a group G onto a group G. Then

(1) ��1 is an isomorphism from G onto G.Proof.

The inverse of � is 1–1 since � is. Let g 2 G. ��1��(g)

�= g =) � is onto.

Now let x, y 2 G. Then

��1(xy) = ��1(x)��1(y) () ����1(xy)

�= �

���1(x)��1(y)

�()

xy = ����1(x)

�����1(y)

�= xy.

Thus operations are preserved and ��1 is an isomorphism. ⇤

(2) G is Abelian () G is Abelian.Proof.

Follows directly from Theorem 6.2(3) since isomorphisms preserve commutivity.⇤

(3) G is cyclic () G is cyclic.Proof.

Follows directly from Theorem 6.2(4) since isomorphisms preserve order. ⇤

6. ISOMORPHISMS 83

(4) If K G, then �(K) = {�(k) | k 2 K} G.Proof.

Follows directly from Theorem 6.2(4) since isomorphisms preserve order. ⇤

(5) If K G, then ��1(K) = {g 2 G | �(g) 2 K} G.Proof.

Follows directly from (1) and (4). ⇤

(6) ��Z(G)

�= Z(G).

Proof.

Follows directly from Theorem 6.2(3) since isomorphisms preserve commutivity.⇤

Definition (Automorphism). An isomorphims from a group G onto itselfis called an automorphism.

Problem (Psge 140 # 35). Show that the mapping �(a + bi) = a � biis an automorphism of C under addition. Show that � preserves complexmultiplication as well. (This means � is an automorphims of C⇤ as well.)

Solution.

� is clearly 1–1 and onto. Suppose a + bi, c + di 2 C.

�⇥(a + bi) + (c + di)

⇤= �

⇥(a + c) + (b + d)i

⇤= (a + c)� (b + d)i

(a� bi) + (c� di) = �(a + bi) + �(c + di),

so � is an automorphism of C under addition.

Now suppose a + bi, c + di 2 C⇤.

�⇥(a + bi)(c + di)

⇤= �

⇥(ac� bd) + (ad + bc)i

⇤= (ac� bd)� (ad + bc)i =

(a� bi)(c� di) = �(a + bi)�(c + di),

so � is an automorphism of C⇤ under multiplication. ⇤

84 6. ISOMORPHISMS

Example (related to Example 10 on Page 135). Consider R2. Any reflectionacross a line through the origin or rotation about the origin is an automorphismunder componentwise addition. For example, consider the reflection about theline y = 2x.

y=2x

P

(a,b)

Consider (a, b) not on y = 2x. The line through (a, b) ? y = 2x is

y � b = �1

2

⇣x� a

⌘. To find P (using substitution):

2x� b = �1

2

⇣x� a

⌘=) 5

2x =

1

2a + b =) x =

1

5a +

2

5b =) y =

2

5a +

4

5b.

Thus the direction vector v = P � (a, b) takes (a, b) to P .

v =⇣� 4

5a +

2

5b,

2

5a� 1

5b⌘.

Define�(a, b) = (b, a).

� is clearly 1–1 and onto.

6. ISOMORPHISMS 85

Finally,

�⇥(a, b) + (c, d)

⇤= �(a + c, b + d) =

(b + d, a + c) = (b, a) + (d, c) = �(a, b) + �(c, d),

so � is an isomorphism.Definition (Inner Automorphism Induced by a).

Let G be a group, a 2 G. The function �a defined by �a(x) = axa�1 8 x 2 Gis called the inner automorphism of G induced by a.

Example. The action of the inner automorphism of D4 induced by R90 isgiven in the following table.

86 6. ISOMORPHISMS

Problem (Page 145 # 45). Suppose g and h induce the same inner auto-morphism of a group G. Prove h�1g 2 Z(G).

Proof. Let x 2 G. Then

�g(x) = �h(x) =) gxg�1 = hxh�1 =) h�1gxg�1h = x =)(h�1g)x(h�1g)�1h = x =) (h�1g)x = x(h�1g) =)

h�1g 2 Z(G) since x is arbitrary. ⇤

Definition.

Aut(G) = {� | � is an automorphism of G} and

Inn(G) = {� | � is an inner automorphism of G}.Theorem (6.4 — Aut(G) and Inn(G) are groups).

Aut(G) and Inn(G) are groups under function composition.

Proof.

From the transitive portion of Theorem 3, compositions of isomorphisms areisomorphisms. Thus Aut(G) is closed under composition. We know functioncomposition is associative and that the identity map is the identity automor-phism.

Suppose ↵ 2 Aut(G). ↵�1 is clearly 1–1 and onto. Now

↵�1(xy) = ↵�1(x)↵�1(y) () ((= by 1–1)

↵⇥↵�1(xy)

⇤= ↵

⇥↵�1(x)↵�1(y)

⇤() xy = ↵

⇥↵�1(x)

⇤↵⇥↵�1(y)

⇤()

xy = xy.

Thus ↵�1 is operation preserving, and Aut(G) is a group.

6. ISOMORPHISMS 87

Now let �g,�h 2 Inn(G) ✓ Aut(G). For all x 2 G,

�g�h(x) = �g(hxh�1) = ghxh�1g�1 = (gh)x(gh)�1 = �gh(x),

so �gh = �g�h 2 Inn(G), and Inn(G) is closed under composition.

Also, ��1g = �g�1 since

�g�1�g(x) = �g�1(gxg�1) = g�1gxg�1(g�1)�1 = g�1gxg�1g = exe = �e(x),

and so also �gg�1 = �e. Thus Inn(G) ✓ Aut(G) by the two-step test. ⇤

Example. Find Inn(Z12).

Solution.

Since Z12 = {0, 1, 2, . . . , 11}, Inn(Z12) = {�0,�1,�2, . . . ,�11}, but the secondlist may have duplicates. That is the case here. For all n 2 Z12 and for allx 2 Z12,

�n(x) = n + x + (�n) = n + (�n) + x = 0 + x = 0 + x + 0 = �0(x),

the identity automorphism. Thus Inn(Z12) = {�0}. ⇤

88 6. ISOMORPHISMS

Example. Find Inn(D3).

Solution.

D3 = {", (1 2 3), (1 3 2), (2 3), (1 3), (1 2)} as permutations. Then

Inn(D3) = {�",�(1 2 3),�(1 3 2),�(2 3),�(1 3),�(1 2)}, but we need to eliminaterepetitions.

�(1 2 3)(1 2 3) = (1 2 3)(1 2 3)(3 2 1) = (1 2 3).

�(1 2 3)(1 3 2) = (1 3 2)(1 2 3)(3 2 1) = (1 3 2).

�(1 2 3)(2 3) = (1 3 2)(2 3)(3 2 1) = (1 3).

�(1 2 3)(1 3) = (1 3 2)(1 3)(3 2 1) = (1 2).

�(1 2 3)(1 2) = (1 3 2)(1 2)(3 2 1) = (2 3). Thus �(1 2 3) is distinct from �".

�(1 3 2)(2 3) = (1 3 2)(2 3)(2 3 1) = (1 2). Thus �(1 3 2) is distinct.

�(2 3)(2 3) = (2 3)(2 3)(2 3) = (2 3).

�(2 3)(1 3) = (2 3)(1 3)(2 3) = (1 2). Thus �(2 3) is distinct

�(1 3)(2 3) = (1 3)(2 3)(1 3) = (1 2).

�(1 3)(1 3) = (1 3)(1 3)(1 3) = (1 3).

�(1 3 2)(1 3) = (1 3 2)(1 3)(2 3 1) = (2 3). Thus �(1 3) is distinct.

�(1 2)(2 3) = (1 2)(2 3)(1 2) = (1 3).

�(1 2)(1 3) = (1 2)(1 3)(1 2) = (2 3). Thus �(1 2) is distinct.

Therefore, there are no duplicates and

Inn(D3) = {�",�(1 2 3),�(1 3 2),�(2 3),�(1 3),�(1 2)}. ⇤

6. ISOMORPHISMS 89

Theorem (6.5 – Aut(Zn) = U(n)). For all n 2 N, Aut(Zn) = U(n).

Proof.

Let ↵ 2 Aut(Zn). Then

↵(k) = ↵(1 + 1 + · · · + 1| {z }k terms

) = (↵(1) + ↵(1) + · · · + ↵(1)| {z }k terms

) = k↵(1).

Now |1| = n =) |↵(1)| = n, so ↵(1) is a generator of Zn since 1 is also agenerator. Now consider

T : Aut(Zn) ! U(n) where T (↵) = ↵(1).

Suppose ↵,� 2 Aut(Zn) and T (↵) = T (�). Then ↵(1) = �(1), so 8 k 2 Zn,↵(k) = k↵(1) = k�(1) = �(k). Thus ↵ = � and T is 1–1.

[To show T is onto.] Now suppose r 2 U(n) and consider ↵ : Zn ! Zn definedby ↵(s) = sr mod n 8 s 2 Zn.

[To show ↵ 2 Aut(Zn).] Suppose ↵(x) = ↵(y). Then xr = yr mod n.

But r�1 exists modulo n =) xrr�1 = yrr�1 mod n =) x·1 = y·1 mod n =)x = y mod n, so ↵ is 1–1.

Suppose x 2 Zn. By Theorem2, 9 s 2 Zn 3�� ↵(s) = sr = x, so ↵ is onto.

Now suppose x, y 2 Zn.

↵(x + y) = (x + y)r mod n = (xr + yr) mod n =

xr mod n + yr mod n = ↵(x) + ↵(y),

so ↵ 2 Aut(Zn).

Since T (↵) = ↵(1) = r, T is onto.

90 6. ISOMORPHISMS

Finally, let ↵,� 2 Aut(Zn). Then

T (↵�) = (↵�)(1) = ↵⇥�(1)

⇤= ↵(1 + 1 + · · · + 1| {z }

�(1) terms

)

↵(1) + ↵(1) + · · · + ↵(1)| {z }�(1) terms

= ↵(1)�(1) = T (↵)T (�),

so Aut(Zn) ⇡ U(n). ⇤

Example. Aut(Z10) ⇡ U(10). The multiplication tables for the two groupsare given below:

The isomorphism is clear.