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CHAPTER 6
Isomorphisms
A while back, we saw that hai with |hai| = 42 and Z42 had basically the same
structure, and said the groups were “isomorphic,” in some sense the same.
Definition (Group Isomorphism). An isomorphism from a group G to
a 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 and
write G ⇡ G.
Note. The operations in G and G may be di↵erent. The operation to the
left 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
by Theorem 4.1. Thus
m =) an = am
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 previous
example.
72
6. ISOMORPHISMS
Example. Let G be the positive real numbers with multiplication and G
be 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. Given
hai 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 be
the group of positive numbers under multiplication. Define (x) = 2x 1. is
1–1 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
Given any positive y, (log2 y) = 2log2 y = y, so
is 1–1.
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 the
Klein 4 group. Any other group of order 4 must have an element of order 4, so
is 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).
⇢
p
a 2b
Let G = {a + b 2 a, b 2 Q} and H =
a, b 2 Q Show
b a
that G and H are isomorphic under addition. Prove that G and H are closed
under multiplication. Does your isomorphism preserve multiplication as well as
addition? (G and H are examples of rings — a topic we begin in Chapter 12)
Solution.
p
p
Given a + b 2, c + d 2 2 G.
p
p
p
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 2 G and
p
p
p
(a + b 2)(c + d 2) = (ac + 2bd) + (ad + bc) 2 2 G
since (a + c), (b + d), (ac + 2bd), (ad + bc) 2 Q.


a 2b
c 2d
Also, if
,
2 H,
b a
d c



a 2b
c 2d
a + c 2(b + d)
+
=
and
b a
d c
b+d a+c



a 2b c 2d
ac + 2bd 2(ad + bc)
=
2H
b a d c
ad + bc ac + 2bd
since (a + c), (b + d), (ac + 2bd), ad + bc 2 Q.
Thus G and H are closed under both addition and multiplication. Also, G and
H are groups under addition (you can prove this, if you desire).
6. ISOMORPHISMS
75

p
a 2b
Define : G ! H by (a + b 2) =
.
b a


p
p
a 2b
c 2d
Suppose (a + b 2) = (c + d 2). Then
=
=)
b a
d c
p
p
a = c and b = d =) a + b 2 = c + d 2 =) is 1–1.


p
p
a 2b
a 2b
For
2 H, a + b 2 2 G and (a + b 2) =
, so is onto.
b a
b a
p
p
Now suppose a + b 2, c + d 2 2 G. Then

p
p
p
a + c 2(b + d)
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 =
=
b+d a+c



p
p
a + c 2b + 2d
a 2b
c 2d
=
+
= (a + b 2) + (c + d 2).
b+d a+c=
b a
d c
Thus
is an isomorphism under addition.
Also,

p
p
p
ac + 2bd 2(ad + bc)
(a+b 2)(c+d 2) = (ac+2bd)+(ad+bc) 2 =
=
ad + bc ac + 2bd



p
p
ac + 2bd 2ad + 2bc
a 2b c 2d
=
= (a + b 2) (c + d 2) .
ad + bc ac + 2bd
b a d c
So preserves multiplication also. (We will later see that this makes
homomorphism.)
a ring
⇤
76
6. ISOMORPHISMS
Problem (Page 141 # 54). Consider G = hmi and H = hni where m, n 2
Z. These are groups under addition. Show G ⇡ H. Does this isomorphism
also preserve multiplication?
Solution.
n
Define : G ! H by (x) = x. Suppose (x) = (y). Then
m
n
n
x = y =) x = y =) is 1–1.
m
m
m
m
For x 2 H, x = rn where r 2 Z. Then x = rn = rm 2 G. Then
n
n
⇣m ⌘ n ⇣m ⌘
x =
x = x, so is onto. Now, suppose x, y 2 G. Then
n
m n
n
n
n
(x + y) = (x + y) = x + y = (x) + (y),
m
m
m
so addition is preserved and G ⇡ H. But,
⇣ n ⌘⇣ n ⌘
n
(xy) = (xy) 6=
x
y = (x) (y),
m
m
m
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 ↵,
Then (↵ ) = (↵) ( ) = ab, so
1
(ab) =
1
(↵) ( ) =
1
1
(↵) = a and ( ) = b.
(↵ ) = ↵ =
1
Thus
2G3
(↵)
1
( ) =
is an isomorphism and H ⇡ G.
(3) Suppose G ⇡ H and H ⇡ K with : G ! H and
isomorphisms. Then
: G ! K is 1–1 and onto.
For all a, b 2 G,
⇥
⇤
⇥
⇤
⇥
⇤ ⇥
⇤ ⇥
( )(ab) =
(ab) =
(a) (b) =
(a)
(b) = (
Thus G ⇡ K and ⇡ is an equivalence relation on G.
1
(a)
1
(b).
: H ! K the
⇤⇥
)(a) (
⇤
)(b) .
⇤
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 computation in one group, the isomorphism allows us to perform the analagous
computation in the other group.
78
6. ISOMORPHISMS
Example (Conjugation by a). Let G be a group, a 2 G. Define a function
1
a : G ! G by a (x) = axa .
Suppose a(x) = a(y). Then axa
cancellation. Thus a is 1–1.
1
= aya
1
=) x = y by left and right
Now suppose y 2 G. We need to find x 2 G 3 axa
so if x = a 1ya, for then
a (x)
so
a
=
a (a
1
ya) = a(a 1ya)a
a (xy)
a
= axya
1
= axeya
7 (3)
7 (5)
7 (7)
= (aa 1)y(aa 1) = eye = y,
1
= axa 1aya
1
=
a (x) a (y).
is an isomorphism from G onto G.
Recall U (8) = {1, 3, 5, 7}. Consider
7 (1)
= y. But that will be
is onto.
Finally, for all x, y 2 G,
Thus
1
1
7.
Since 7 is its own inverse,
= 7 · 1 · 7 = 1,
= 7 · 3 · 7 = 5 · 7 = 3,
= 7 · 5 · 7 = 3 · 7 = 5,
= 7 · 7 · 7 = 1 · 7 = 7.
Conjugation by 7 turns out to be the identity isomorphism, which is not very
interesting.
6. ISOMORPHISMS
79
Theorem (6.1 — Cayley’s Theorem). Every group is isomorphic to a
group of permutations.
Proof.
Let G be any group. [We need to find a set A and a permutation on A that
forms a group G that is isomorphic to G.] We take the set G and define, for
each 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,
(Tg Th)(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 Tg
Since composition is associative, G is a group.
Now define
so
= Tg 1 .
: G ! G by (g) = Tg 8 g 2 G.
(g) = (h) =) Tg = Th =) Tg (e) = Th(e) =) ge = he =) g = h,
is 1–1. is clearly onto by construction.
Finally, for g, h 2 G,
so
1
is an isomorphism.
(gh) = Tgh = Tg Th = (g) (h),
⇤
80
6. ISOMORPHISMS
Example. U = {1, 3, 5, 7}.




1 3 5 7
1 3 5 7
1 3 5 7
1 3 5 7
T1 =
, T3 =
, T5 =
, T7 =
.
1 3 5 7
3 1 7 5
5 7 1 3
7 5 3 1
Then U (8) = {T1, T3, T5, T7}.
U (8)
1
3
5
7
1
1
3
5
7
3
3
1
7
5
5
5
7
1
3
7
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.
⇤
⇥
⇤n
(2) For all n 2 Z and for all a 2 G, (an) = (a) .
Proof.
⇥
⇤n
For n 2 Z, n 0, (an) = (a) from the definition of an isomorphism and
induction. For n < 0, n > 0, so
⇥
⇤ n
⇥
⇤n
e = (e) = (ana n) = (an) (a n) = (an) (a)
=) (a) = (an).
⇤
(3) For all a, b 2 G, ab = ba ()
Proof.
ab = ba ()
(a) (b) = (b) (a).
(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,
⇥
⇤k
9 ak 2 G 3 (ak ) = b. Thus b = (a) =) 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
hai = G.
(b) =
(a)k =
(ak ). Since
is 1–1, b = ak . Thus
⇤
(5) |a| = | (a)| 8 a 2 G (Isomorphisms preserve order).
Proof.
⇤n
a = e () (a ) = (e) ()
(a) = e.
Thus a has infinite order () (a) has infinite order, and a has finite order
n () (a) has infinite order n.
⇤
n
n
⇥
(6) For a fixed integer k and a fixed b 2 G, xk = b has the same number
of 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
⇥
⇤k
k
(a ) = (b) =) (a) = (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
⇥
⇤k
(a) = (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 elements
of 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) no
isomorphism 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
Now let x, y 2 G. Then
1
(xy) =
1
(x)
1
is. Let g 2 G.
1
(y) ()
Thus operations are preserved and
1
1
(g) = g =)
(xy) =
1
xy =
1
(x)
(x)
1
(y)
1
is onto.
()
(y) = xy.
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
Proof.
1
⇤
(K) = {g 2 G | (g) 2 K}  G.
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 itself
is called an automorphism.
Problem (Psge 140 # 35). Show that the mapping (a + bi) = a bi
is an automorphism of C under addition. Show that preserves complex
multiplication 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 reflection
across a line through the origin or rotation about the origin is an automorphism
under componentwise addition. For example, consider the reflection about the
line y = 2x.
y=2x
P
(a,b)
Consider (a, b) not on y = 2x. The line through (a, b) ? y = 2x is
⌘
1⇣
y b=
x a . To find P (using substitution):
2
⌘
1⇣
5
1
1
2
2
4
2x b =
x a =) x = a + b =) x = a + b =) y = a + b.
2
2
2
5
5
5
5
Thus the direction vector v = P (a, b) takes (a, b) to P .
⇣ 4
2 2
1 ⌘
v=
a + b, a
b .
5
5 5
5
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
is called the inner automorphism of G induced by a.
a (x)
= axa
1
8x 2 G
Example. The action of the inner automorphism of D4 induced by R90 is
given in the following table.
86
6. ISOMORPHISMS
Problem (Page 145 # 45). Suppose g and h induce the same inner automorphism 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 are
isomorphisms. Thus Aut(G) is closed under composition. We know function
composition is associative and that the identity map is the identity automorphism.
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
g,
Now let
h
2 Inn(G) ✓ Aut(G). For all x 2 G,
g h (x)
so
gh
Also,
=
g
1
g h
=
g 1 g (x)
and so also
=
g (hxh
1
) = ghxh 1g
1
= (gh)x(gh)
1
=
gh (x),
2 Inn(G), and Inn(G) is closed under composition.
g 1
=
87
since
g 1 (gxg
gg 1
=
e.
1
) = g 1gxg 1(g 1)
1
= g 1gxg 1g = exe =
e (x),
⇤
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 second
list may have duplicates. That is the case here. For all n 2 Z12 and for all
x 2 Z12,
n (x)
= n + x + ( n) = n + ( n) + x = 0 + x = 0 + x + 0 =
the identity automorphism. Thus Inn(Z12) = { 0}.
0 (x),
⇤
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) = { ",
repetitions.
(1 2 3) ,
(1 3 2) ,
(2 3) ,
(1 3) ,
(1 2) },
but we need to eliminate
(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
(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
(2 3)
3) = (1 3 2)(1 3)(2 3 1) = (2 3). Thus
(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
(1 3)
is distinct.
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 +{z· · · + 1}) = (↵(1) + ↵(1) + · · · + ↵(1)) = k↵(1).
|
{z
}
k terms
k terms
Now |1| = n =) |↵(1)| = n, so ↵(1) is a generator of Zn since 1 is also a
generator. 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 defined
by ↵(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 +{z· · · + 1})
(1) terms
↵(1) + ↵(1) + · · · + ↵(1) = ↵(1) (1) = T (↵)T ( ),
|
{z
}
(1) terms
so Aut(Zn) ⇡ U (n).
⇤
Example. Aut(Z10) ⇡ U (10). The multiplication tables for the two groups
are given below:
The isomorphism is clear.