Download 3.7 Homomorphism Theorems

3.7 Homomorphism Theorems
111
10. Let ' : G ! G0 be an onto homomorphism. Suppose G is cyclic. Prove that G0 is cyclic.
11. Let ' : G ! G0 be a one-to-one homomorphism from the group G to the group G0 . Suppose
that H is a subgroup of G. We know, by Lemma 3.6.8 that '[H] is a subgroup of G0 . Prove
that if '[H] C G0 , then H C G.
12. Let ' : G ! G0 be an onto homomorphism from the group G to the group G0 . Suppose that
H is a subgroup of G. We know, by Lemma 3.6.8 that '[H] is a subgroup of G0 . Prove that
if H C G, then '[H] C G0 .
Exercise Notes: For Exercise 1, we know by Exercise 10 on page 58 that ' is well-defined. For
Exercise 11, apply the result of Exercise 8 on page 34.
3.7
Homomorphism Theorems
3.7.1
The First Homomorphism Theorem
The following theorem presents a method for producing an isomorphism.
Theorem 3.7.1 (First Homomorphism Theorem). Let ' : G ! G0 be an onto homomorphism
from the group (G, ⇤) to the group (G0 , ⇥). Let K = ker('). Then (G/K, ~) ' (G0 , ⇥).
Proof. Let ' : G ! G0 be an onto homomorphism and let K = ker('). Consider the diagram
G
'
/ G0
✏
G/K
where : G ! G/K is the natural homomorphism defined by (a) = [a] where [a] is the equivalence
class of a using the equivalence relation ⇠K defined by x ⇠K y if and only if xy 1 2 K, for all
x, y 2 G (see Exercise 8 on page 110).
To prove that (G/K, ~) ' (G0 , ⇥), we must find a function : G/K ! G0 and prove that is
a one-to-one and onto homomorphism. That is, we need to define a map to complete the above
diagram and obtain
'
/ G0
G
<
✏
G/K
We try using
(1)
(2)
(3)
(4)
Show
Show
Show
Show
([a]) = '(a). We must do the following:
that
that
that
that
is
is
is
is
well-defined.
a homomorphism.
one-to-one.
onto.
(1) First we show that
is well-defined. Suppose that [a] = [b] for a, b 2 G. We must show
that ([a]) = ([b]); that is, (by the definition of ) we must prove that '(a) = '(b). Since
[a] = [b], it follows from Theorem 0.5.4 that a ⇠K b. Thus, ab 1 2 K. Therefore, '(ab 1 ) = e0
because K = ker('), where e0 is the identity element in G0 . Since ' is a homomorphism, we
get that '(a)'(b) 1 = e0 and thus, '(a) = '(b). Therefore, is well-defined.
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Chapter 3. Groups
(2) Now we prove that : G/K ! G0 is a homomorphism. Let [a], [b] 2 G/N be arbitrary. We
show ([a] ~ [b]) = ([a]) ⇥ ([b]) as follows:
([a] ~ [b]) = ([a ⇤ b])
= '(a ⇤ b)
= '(a) ⇥ '(b)
= ([a]) ⇥ ([b])
by the definition of ~
by the definition of
because ' is a homomorphism
by the definition of
.
(3) We shall prove that : G/K ! G0 is a one-to-one function. Let [x], [y] 2 G/K be arbitrary.
Assume that ([x]) = ([y]). We prove that [x] = [y]. Since ([x]) = ([y]) we see that
'(x) = '(y) by the definition of . Since ' is a homomorphism, it follows that '(xy 1 ) = e0 .
Thus, xy 1 2 K. Hence, x ⇠K y. Therefore, [x] = [y] by Theorem 0.5.4. Thus, we have
shown that is one-to-one.
(4) Finally, we prove that : G/K ! G0 is an onto function. Let b0 2 G0 be arbitrary. Since
' : G ! G0 is onto, there is an element a 2 G such that '(a) = b0 . Note that ([a]) = '(a) =
b0 . Therefore, is onto.
This completes the proof of the Theorem.
In the conclusion of Theorem 3.7.1, rather than write (G/K, ~) ' (G0 , ⇥), we may just write
G/K ' G0 .
Problem 1. Given a, b 2 R with a 6= 0, define the function Ta,b : R ! R by Ta,b (x) = ax + b. Let
G be the set of all such functions, that is, let G = {Ta,b : a, b 2 R and a 6= 0}. For any Ta,b 2 G and
Tc,d 2 G we define the binary operation Ta,b Tc,d to be function composition. We saw in Example 5
on page 68 that (G, ) is a group. Let N = {T1,h 2 G : h 2 R}, and let (R⇤ , ·) be the group of all
non-zero real numbers under multiplication; that is, R⇤ = {x 2 R : x 6= 0} and · is multiplication.
Using Theorem 3.7.1, show that G/N ' R⇤ .
Solution. Let (R⇤ , ·) and (G, ) be the groups defined in the above problem. For any Ta,b , Tc,d 2 G,
one can derive the following “composition rule”:
Ta,b Tc,d = Tac,ad+b .
(3.19)
We also have N ✓ G where N = {T1,h 2 G : h 2 R}. We shall prove that (G/N, }) ' (R⇤ , ·). We
do this by first defining an onto function ' : G ! R⇤ . We shall then prove that ' is a homomorphism and that ker(') = N . The desired result will then follow from the First Homomorphism
Theorem 3.7.1.
Define ' (Ta,b ) = a for all Ta,b 2 G. To see that ' is onto R⇤ , let c 2 R⇤ be arbitrary. Since
c 6= 0, we have that Tc,0 2 G. Furthermore, '(Tc,0 ) = c and thus, ' is onto. We shall now prove
that ' : G ! R⇤ is homomorphism. Let Ta,b , Tc,d 2 G be arbitrary. We show that
' (Ta,b Tc,d ) = ' (Ta,b ) · ' (Tc,d )
as follows:
' (Ta,b Tc,d ) = ' (Tac,ad+b )
by the “composition rule” (3.19)
= ac
by the definition of '
= ' (Ta,b ) · ' (Tc,d )
by the definition of '.
Thus, ' (Ta,b Tc,d ) = ' (Ta,b ) · ' (Tc,d ) and therefore, ' : G ! R⇤ is a homomorphism.
3.7 Homomorphism Theorems
113
By the definition of the kernel, we see that
ker(') = {Ta,b 2 G : ' (Ta,b ) = 1}
where 1 is the identity element for R⇤ . We now prove that ker(') = N . Let Tu,v 2 G be arbitrary.
We see that
Tu,v 2 ker(') i↵ ' (Tu,v ) = 1
by the definition of ker(')
i↵ u = 1
by the definition of '
i↵ Tu,v 2 N
by the definition of N .
Therefore, ker(') = N and thus, Theorem 3.7.1 implies that G/N ' R⇤ .
3.7.2
s
The Correspondence Theorem
Theorem 3.7.2 (Correspondence Theorem). Let ' : G ! G0 be an onto homomorphism from the
group (G, ⇤) to the group (G0 , ⇥). Let K = ker('). Let (H 0 , ⇥) be a subgroup of (G0 , ⇥). Define
the subset H ✓ G by H = ' 1 [H 0 ] = {a 2 G : '(a) 2 H 0 }. Then
(1)
(2)
(3)
(4)
(5)
(H, ⇤) is a subgroup of (G, ⇤).
K ✓ H.
(K, ⇤) is a normal subgroup of (H, ⇤); that is, K C H.
(H/K, ~) ' (H 0 , ⇥).
If (H 0 , ⇥) is a normal subgroup of (G0 , ⇥), then (H, ⇤) is also a normal subgroup of (G, ⇤).
Proof. Let ' : G ! G0 be an onto homomorphism and let K = ker('). Let H 0 be a subgroup of
(G0 , ⇥). Define the subset H ✓ G by
H = {a 2 G : '(a) 2 H 0 }.
(1) We first show that H is a subgroup of (G, ⇤). Since H = '
that H 0 is a subgroup of G0 .
1 [H 0 ],
it follows from Lemma 3.6.10
(2) To see that K ✓ H, let x 2 K be arbitrary. Since x 2 K, it follows that '(x) = e0 because
K = ker('). We know that e0 2 H 0 , because H 0 is a subgroup of G0 , by (1). Therefore,
'(x) 2 H 0 and thus, x 2 H.
(3) We now show that K is a normal subgroup of (H, ⇤). Since K is a subgroup of G (by
Theorem 3.6.15) and since K ✓ H, it follows that K is a subgroup of H. In addition, since
K is a normal subgroup of (G, ⇤) (by Theorem 3.6.16) and since K ✓ H, it also follows that
K is a normal subgroup of H (see Exercises 9 and 10 on page 83).
(4) We shall prove that (H/K, ~) ' (H 0 , ⇥). We do this by first defining an onto function
: H ! H 0 . We shall then show that is a homomorphism and ker( ) = K. The desired
result will then follow from the First Homomorphism Theorem 3.7.1.
Define (h) = '(h) for all h 2 H. By the definition of H we see that : H ! H 0 . Because
' is an onto homomorphism, it follows that is also an onto homomorphism (see Exercise 6
on page 110). In addition, since K = ker(') it follows (because K ✓ H) that K = ker( ).
Thus, Theorem 3.7.1 implies that (H/K, ~) ' (H 0 , ⇥).
(5) Finally, we prove that if H 0 is a normal subgroup of (G0 , ⇥), then H is also a normal subgroup
of (G, ⇤). So assume that H 0 is a normal subgroup of G0 . We prove that H is a subgroup of
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Chapter 3. Groups
G. To do this, let a 2 G and h 2 H be arbitrary. Since h 2 H, it follows that '(h) 2 H 0 . To
show that a 1 ha 2 H, we must prove that '(a 1 ha) 2 H 0 . Observe that
'(a
1
ha) = '(a)
1
'(h)'(a).
Now since '(h) 2 H 0 and because H 0 is a normal subgroup of G0 , we have
'(a)
Hence, '(a
1 ha)
2 H 0 and so, a
1 ha
1
'(h)'(a) 2 H 0 .
2 H. Therefore, H is a normal subgroup of G.
The proof of the theorem is complete.
3.7.3
The Second Homomorphism Theorem
Theorem 3.7.3 (Second Homomorphism Theorem). Let (H, ⇤) be a subgroup of the group (G, ⇤).
In addition, let (N, ⇤) be a normal subgroup of the group (G, ⇤). Define the subset HN ✓ G by
HN = {hn : h 2 H and n 2 N }.
Then
(1) (HN, ⇤) is a subgroup of (G, ⇤).
(2) (N, ⇤) is a normal subgroup of (HN, ⇤) and thus, (HN/N, ~) is a quotient group.
(3) (H \ N, ⇤) is a normal subgroup of (H, ⇤) and thus, (H/(H \ N ), ~) is a quotient group.
(4) (H/(H \ N ), ~) ' (HN/N, ~).
Proof. Let H be a subgroup of the group G and let HN be as defined above.
(1) Since e 2 H and e 2 N , it follows that e = ee 2 HN . Let x, y 2 HN . Thus x = hn and
y = h0 n0 for some h, h0 2 H and n, n0 2 N . Thus, xy = (hn)(h0 n0 ) = h(nh0 )n0 . By Lemma
3.2.26, there is a j 2 N such that nh0 = h0 j. Thus, xy = h(h0 j)n0 = (hh0 )(jn0 ). Therefore,
xy 2 HN because hh0 2 H and jn0 2 N . Let x 2 HN . Thus x = hn for some h 2 H and
n 2 N . Since x 1 = (hn) 1 = n 1 h 1 and n 1 2 N , Lemma 3.2.26, there is a m 2 N such
that n 1 h 1 = h 1 m. Since h 1 2 H and m 2 N , we conclude that x 1 = h 1 m 2 HN .
Therefore, HN is a subgroup of G.
(2) First we show that N ✓ HN . Let n 2 N be arbitrary. Since H is a subgroup, we have
that e 2 H. Therefore, n = en is in HN . Now, because N ✓ HN and N is a subgroup of
G, it follows from (1) that N is a subgroup of HN (see Exercise 9 on page 83). Since N
is normal in G, it now follows that N is also normal in HN (see Exercises 10 on page 83).
Thus, (HN/N, ~) is a group by Theorem 3.5.3 using the equivalence relation x ⇠N y if and
only if xy 1 2 N , for all x, y 2 HN .
(3) Clearly we have that H \ N ✓ H. Because H \ N is a subgroup of G, it follows that H \ N
is a subgroup of H. To prove that H \ N is normal in H, let h 2 H and a 2 H \ N be
arbitrary. We must prove that h 1 ah 2 H \ N . Since a 2 N and N is normal in G, it follows
that h 1 ah 2 N . In addition, because a 2 H and h 2 H it also follows that x = h 1 ah 2 H
because H is a subgroup. Therefore, h 1 ah 2 H \ N . Thus, (H/(H \ N ), ~) is a group by
Theorem 3.5.3 using the equivalence relation x ⇠H\N y if and only if xy 1 2 H \ N , for all
x, y 2 H.
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(4) Define the function ' : H ! HN/N by '(h) = [h] for all h 2 H, where [h] is the equivalence
class [h] = {x 2 HN : h ⇠N x} and ⇠N is the equivalence relation defined on HN in (2)
above. One can now easily prove that ' is a homomorphism. We now show that ' is onto.
Let [hn] 2 HN/N be arbitrary where h 2 H and n 2 N . It follows that [hn] = [h], because
hnh 1 2 N (since N is normal and by Exercise 15 on page 83) and thus, hn ⇠N h. We
conclude that '(h) = [hn]. We now prove that ker(') = H \ N . Let k 2 ker('). Thus,
'(k) = [e]; that is, [k] = [e]. Therefore, k ⇠N e and so, k = ke 1 2 N . Hence, because
ker(') ✓ H, it follows that k 2 H \ N . Now let k 2 H \ N . Thus, k 2 N and so,
k = ke 1 2 N . Therefore, [k] = [e] and so, '(k) = [e]. Consequently, k 2 ker(') and we
conclude that ker(') = H \ N . Theorem 3.7.1 (First Homomorphism Theorem) now implies
that (H/(H \ N ), ~) ' (HN/N, ~).
This completes the proof of the Second Homomorphism Theorem.
3.7.4
The Third Homomorphism Theorem
Theorem 3.7.4 (Third Homomorphism Theorem). Let ' : G ! G0 be an onto homomorphism
from the group (G, ⇤) to the group (G0 , ⇥). Let (N 0 , ⇥) be a normal subgroup of (G0 , ⇥). Define the
subset N ✓ G by N = ' 1 [N 0 ] = {a 2 G : '(a) 2 N 0 }. Then
(1) (N, ⇤) is a normal subgroup of (G, ⇤); that is, N C G.
(2) (G/N, ~) ' (G0 /N 0 , ⌦).
Proof. Let ' : G ! G0 be an onto homomorphism. Let (N 0 , ⇥) be a normal subgroup of (G0 , ⇥).
Define the subset N ✓ G by N = {a 2 G : '(a) 2 N 0 }.
(1) The Correspondence Theorem 3.7.2(5) implies that N is a normal subgroup of G.
(2) We prove that (G/N, ~) ' (G0 /N 0 , ⌦). We first defining an onto function : G ! G0 /N 0 .
We shall then prove that is a homomorphism and that ker( ) = N . The desired result will
then follow from the First Homomorphism Theorem 3.7.1.
Define (a) = ['(a)] for all a 2 G. Here [y] is the equivalence class of y 2 G0 , with respect
to the equivalence relation ⇠N 0 on G0 defined by x ⇠N 0 y if and only if xy 1 2 N 0 , for all
x, y 2 G0 . Then (G0 /N 0 , ⌦) is a group, where the binary operation ⌦ on G0 /N 0 is well-defined
by [a] ⌦ [b] = [a ⇥ b] for all a, b 2 G0 .
Because ' is onto, it follows that is onto. We prove that : G ! G0 /N 0 is a homomorphism.
Let a, b 2 G be arbitrary. We show (a ⇤ b) = (a) ⌦ (b) as follows:
(a ⇤ b) = ['(a ⇤ b)]
= ['(a) ⇥ '(b)]
= ['(a)] ⌦ ['(b)]
= (a) ⌦ (a)
by the definition of
because ' is a homomorphism
by the definition of ⌦
by the definition of
.
Thus, : G ! G0 /N 0 is a homomorphism. Finally, we show that ker( ) = N . By the
definition of the kernel, we see that
ker( ) = {a 2 G : (a) = [e0 ]}
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Chapter 3. Groups
where [e0 ] is the identity element for G0 /N 0 . We see that
x 2 ker( ) i↵
(x) = [e0 ]
by definition of ker( )
0
by definition of
0
by Exercise 6 on page 88
i↵ ['(x)] = [e ]
i↵ ['(x)] = N
i↵ '(x) 2 N
i↵ x 2 N
0
by Exercise 6 on page 88
by the definition of N .
Therefore, ker( ) = N and thus, Theorem 3.7.1 implies that (G/N, ~) ' (G0 /N 0 , ⌦).
This completes the proof of the Third Homomorphism Theorem.
Exercises 3.7
1. Consider the groups (Z6 , ) and (Z3 , ). Define ' : Z6 ! Z3 by '([x]6 ) = [x]3 .
(a)
(b)
(c)
(d)
Show that ' is a homomorphism.
Show that ' is onto.
Evaluate K = ker(').
Using Theorem 3.7.1, show that Z6 /K ' Z3 .
2. Let F be the set of all functions of the form f : R ! R. Let f 2 F and g 2 F be arbitrary.
Define the function (f + g) : R ! R by (f + g)(x) = f (x) + g(x). Then (F, +) is a group. Also,
the structure (R, +) is a group. Let N = {f 2 F : f (1) = 0}. Using Theorem 3.7.1, show that
F/N ' R.
3. Let (R⇤ , ·) be the group of all non-zero real numbers under multiplication and let N ✓ R⇤
be defined by N = {1, 1}. Now let (R+ , ·) be the group of positive real numbers under
multiplication. Using Theorem 3.7.1, show that R⇤ /N ' R+ .
Exercise Notes: For Exercise 1, we know by Exercise 9 on page 58 that ' is well-defined. For
Exercises 2–3, review Problem 1 and its solution on page 112.