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Chapter 7
Isomorphisms
You may remember when we were studying cyclic groups, we made the remark
that cyclic groups were similar to Zn . Indeed, we proved that every cyclic group
was Abelian using the fact that addition of integers is Abelian. We expand on
this notion. More speci…cally, we will develop a way to determine if two groups
have similar properties. The advantage of this is that if we could tell that two
groups G1 and G2 have similar properties and we already know all the properties
of G1 , then we would immediately know all the properties of G2 . The tool which
will allow us to do this is called an isomorphism, from the Greek words isos
which mean "same" and morphe which means "form". In this chapter, we will
do the following:
1. De…ne what an isomorphism is.
2. Look at examples of isomorphisms.
3. Prove an isomorphism does what we claim it does (preserves properties).
7.1
De…nitions and Elementary Examples
De…nition 272 (Isomorphism) Let G and H be two groups. We will use
multiplication for the notation of their operations, though the operation on G
may not be the same as the one on H.
1. An isomorphism from G to H is a bijection : G ! H with the property
that (ab) = (a) (b) for every a; b in G. This property means that
preserves the group operations.
2. If there exists an isomorphism between G and H, we say that G and H
are isomorphic and we write G = H. Note that some texts also use ,
even .
3. An isomorphism
: G ! G is called an automorphism, that is an
isomorphism of a group to itself.
77
78
CHAPTER 7. ISOMORPHISMS
Let us clarify a few things in this de…nition.
a and b are in G. When we write ab, the operation involved is the operation
on G.
(a) and (b) are elements in H. When we write
involved is the operation on H.
(a) (b), the operation
The table below shows the various form the operation preserving equality
(ab) = (a) (b) would take depending on the operations on G and H.
Operation on G
Operation on H
+
+
+
+
Operation preserving
(a b) = (a) (b)
(a b) = (a) + (b)
(a + b) = (a) (b)
(a + b) = (a) + (b)
To prove that two groups G and H are isomorphic actually requires four
steps, highlighted below:
1. De…ne a function (mapping)
2. Prove
is an injection that is
: G ! H which will be our candidate.
(a) =
(b) =) a = b.
3. Prove is a surjection that is every element h in H is of the form h =
for some g in G.
4. Prove
preserves the group operations that is
(ab) =
(g)
(a) (b).
Remark 273 In an isomorphism, there are two parts. The bijection part, is
there to ensure that there is a one to one correspondence between the elements
of each group. The operation preserving equality ensures that the operation on
each group is preserved.
Remark 274 To prove two groups are not isomorphic, it is not enough to claim
one did not …nd an isomorphism. One has to prove no isomorphism can exist.
We now look at some examples of isomorphisms.
Example 275 Let G = (R; +) and H = (R+ ; :). We already know both G and
H are groups. We claim G = H. The function is de…ned by: : G ! H such
that (x) = 2x . First, we recognize an exponential function, so is a function.
It is also an injection. Suppose (a) = (b) then 2a = 2b . Taking the logarithm
base 2 on each side gives a = b. It is also onto. If y 2 H, one can …nd x 2 G
such that (x) = y. For this to happen, we need to have 2x = y, solving for x
gives x = log2 y. Finally, if a and b are two elements of G, then
(a + b)
=
2a+b
=
=
2a 2 b
(a) (b)
7.1. DEFINITIONS AND ELEMENTARY EXAMPLES
79
Remark 276 Any exponential function would have worked in the previous example. We had to pick one, so we picked 2x but it could have been ax where
a 2 R, a > 0 and a 6= 1.
Example 277 Any in…nite cyclic group is isomorphic to Z. Let G = hai where
jaj = 1. Then, G = ak j k 2 Z . De…ne : G ! Z by
ak = k. Then, is
i
j
a function from G to Z. It is an injection for if a = a then i = j (see theorem
in the previous chapters). It is also onto. Finally
ak aj
ak+j
=
= k+j
ak +
=
aj
Example 278 Any …nite cyclic group hai of order n is isomorphic to Zn with
the mapping : hai ! Zn de…ned by
ak = k mod n. The proof that such a
mapping is an isomorphism is left as an exercise.
Example 279 Let G = SL (2; R) with matrix multiplication. Recall, this is the
group of 2 2 matrices with real entries and determinant equal to 1. Pick a matrix M from G. De…ne : G ! G by (A) = M AM 1 . is an automorphism.
The proof is left as an exercise.
Example 280 Consider the group (C; +) and de…ne : C ! C by (a + bi) =
a bi. is an automorphism. We leave it to the reader to check is a bijection.
We verify that it preserves operation. Let z and w be two elements of C that is
z = a + bi and w = c + di. Then,
(z + w)
=
(a + c + (b + d) i)
= a+c
=
=
(a
(b + d) i
bi) + (c
di)
(z) + (w)
Example 281 Consider the group C with complex multiplication. The mapping as de…ned above from C ! C is an automorphism. See problems.
Next, we look at two groups which cannot be isomorphic to illustrate how
one might prove no isomorphism can exist between two given groups.
Example 282 (Q; +) is not isomorphic to (Q ; ). As noted earlier, it is not
enough to say one did not …nd an isomorphism to conclude one does not exist.
One has to prove an isomorphism cannot exist. If such an isomorphism
existed, it would be onto. Since 1 2 Q , there would exist x 2 Q such that
x x
(x) = 1. Writing x = + and using the properties of , we see that
2
2
x x
(x)
=
1 ()
+
= 1
2
2
x
x
()
= 1
2
2
h x i2
()
= 1
2
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CHAPTER 7. ISOMORPHISMS
which is not possible in the set of rational numbers.
The next example illustrates that an isomorphism is two parts: bijection
and operation preserving. Not every bijection is an isomorphism.
Example 283 This example illustrates the fact that being a bijection is not
enough to be an isomorphism. Consider the group (R; +) and the mapping
: R ! R de…ned by (x) = x3 . We leave it to the reader to verify
is
a bijection. However,
is not an isomorphism. If it were, we would have
3
(x + y) = (x) + (y). But (x + y) = (x + y) and (x) + (y) = x3 + y 3 .
The two are clearly not equal.
Example 284 This example illustrates the power of the operation preserving
equality (ab) = (a) (b) and why we call it operation preserving. This equality says that if we have two isomorphic groups G and H and we know the
operation table of one of them, say G, then we automatically know the operation
table of H. Suppose that G = U (12) = f1; 5; 7; 11g. Its multiplication table is
given by
G
1
5
7 11
1
1
5
7 11
5
5
1 11 7
7
7 11 1
5
11 11 7
5
1
Let say that H = fa; b; c; dg and the isomorphism : G ! H maps the elements
as follows: (1) = a, (5) = b, (7) = c and (11) = d. Suppose we want to
know what bc is. Then, we have
bc =
(5) (7)
=
(5 7) (operation preserving equality)
=
(11) (from the table of U (12) )
=
d
This way, we can build the operation table for H, simply by knowing that of an
isomorphic group. It would be:
H
a
b
c
d
a
a
b
c
d
b
b
a
d
c
c
c
d
a
b
d
d
c
b
a
From this example, you can see that the identity of H is a, it is the image of
1, the identity of G. We will see this is one of the properties of isomorphisms.
They map the identity of one group into the identity of the other.
We now prove that an isomorphism as de…ned above does indeed preserve
group properties.
7.2. PROPERTIES OF ISOMORPHISMS
7.2
81
Properties of Isomorphisms
Instead of giving a di¤erent theorem for each result, we will group the results
in two main theorems. Since an isomorphism maps the elements of a group into
the elements of another group, we will look at the properties of isomorphisms
related to their action on elements. Since an isomorphism also acts on all the
elements of a group, it acts on the group. We will also look at the properties of
isomorphisms related to their action on groups.
Theorem 285 (Isomorphisms Acting on Group Elements) Let G and H
be two groups. Suppose that : G ! H is an isomorphism. Then the following
is true:
1.
2.
maps the identity of G into the identity of H. In other words, if eG is
the identity of G and eH is the identity of H, then (eG ) = eH .
a
1
= [ (a)]
1
3. For every integer n and every a 2 G,
n
(an ) = [ (a)]
4. For any two elements a and b in G, a and b commute if and only if
and (b) commute.
(a)
5. G = hai () H = h (a)i.
6. Isomorphisms preserve order in other words, for every a in G, jaj = j (a)j.
7. For a …xed integer k and a …xed group element b in G, the equation xk = b
has the same number of solutions in G as does the equation xk = (b) in
H.
8. If G is …nite then G and H have exactly the same number of elements of
every order.
Proof. We prove each part separately.
1. Using the notation of the theorem, we have that eG = eG eG hence
(eG ) = (eG eG ) = (eG ) (eG ). But (eG ) is an element of H
hence (eG ) = eH (eG ). Thus, we have eH (eG ) = (eG ) (eG ).
Cancelling (eG ) gives the desired result.
2. Note that eG = aa 1 . Hence, from part1, eH = (eG ) = aa
(a) a 1 . By the uniqueness of inverses, it follows that a
1
[ (a)] .
1
1
=
=
3. We consider 3 cases: n > 0, n = 0, and n < 0. If n > 0, we can
use mathematical induction. The result is clearly true for n = 2.
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CHAPTER 7. ISOMORPHISMS
2
a2 = (aa) = (a) (a) = [ (a)] . Assume the result is true for
k, show true for k + 1.
ak+1
=
ak a
=
ak
=
=
(a)
k
[ (a)]
(a) by induction hypothesis
k+1
[ (a)]
0
Next, if n = 0, then
a0 = (eG ) = eH = [ (a)] . Finally, if
n < 0, then n = m and m > 0. Now,
(an )
=
m
a
=
a
=
a
1 m
=
h
=
[ (a)]
=
m
1
[ (a)]
[ (a)]
1
m
im
by part 2
n
4. Left as an exercise
5. First, assume G = hai and show H = h (a)i. Note that by closure
k
h (a)i H. If y 2 H then y =
ak = [ (a)] hence y 2 h (a)i
thus H h (a)i. Thus, the two are equal. Next, suppose H = h (a)i
and show G = hai. a 2 G and thus by closure hai
G. If b 2 G
k
then (b) = [ (a)] =
ak hence b = ak since is a bijection.
This means that b 2 hai and thus G
hai. It follows that the two
are equal.
6. Left as an exercise
7. Left as an exercise
8. Left as an exercise
Let us make some further remarks.
Remark 286 If the operation of H is addition, then property 3 becomes (an ) =
n (a). If both the operation of G and H are addition, then property 3 becomes
(na) = n (a).
Remark 287 Property 5 says that an isomorphism maps a generator into a
generator.
7.2. PROPERTIES OF ISOMORPHISMS
83
Remark 288 Property 7 is often used to prove that no isomorphism can exist
between two groups. For example, consider the equation x4 = 1 and the groups
R and C with multiplication. In R this equation has 2 solutions while in C
it has 4. Hence, no isomorphism can exist between the two.
Before we state the next theorem, let us de…ne some concepts.
De…nition 289 (Image and Inverse Image of a Set) Let f : A ! B be a
function. Let E A and H B.
1. The image of G under f , denoted f (E) is de…ned to be:
f (E) = ff (x) j x 2 Eg
2. The inverse image of H under f;denoted f
f
1
1
(H) is de…ned to be:
(H) = fx 2 A j f (x) 2 Hg
Note that this de…nition does not require f to have an inverse. Though, in
our case, since we are dealing with isomorphisms, f will have an inverse.
Theorem 290 (Isomorphisms Acting on Groups) Let G and H be two
groups. Suppose that : G ! H is an isomorphism. Then the following is
true:
1.
1
is an isomorphism from H into G.
2. G is Abelian if and only if H is.
3. G is cyclic if and only if H is.
4. If K is a subgroup of G then
5. If J is a subgroup of H then
(K) is a subgroup of H.
1
(J) is a subgroup of G.
Proof. We proof each part separately.
1. left as an exercise
2. left as an exercise
3. left as an exercise
4. (K) 6= ?, it contains the identity of H, eH from the previous theorem. We can use the test for subgroup. First we check closure
under the operation. Let x and y be two elements of (K), show
xy is also in (K). There exist a and b in K such that (a) = x
and (b) = y. Note that since K is itself a group, ab 2 K. Now,
xy = (a) (b) = (ab) 2 (K) since ab 2 K. Finally, we show
closure under inverses. Let x 2 (K), show x 1 2 (K). There
exists a 2 K such that (a) = x. Since K is a group, a 1 2 K. so
1
a 1 = [ (a)] = x 1 2 (K).
5. left as an exercise
We …nish this chapter with a classic theorem.
84
CHAPTER 7. ISOMORPHISMS
7.3
Cayley’s Theorem
The proof of this theorem is not very di¢ cult, and it is a good exercise in group
theory as it uses many concepts previously studied. Before we state and prove
the theorem, let us establish some preliminary results.
Lemma 291 Let G be a group and pick g 2 G. De…ne Tg : G ! G by Tg (x) =
gx. In other words, it is left multiplication by g. Then, Tg is a permutation on
G.
Proof. To prove Tg is a permutation on G, we must prove it is a bijection.
Injection: Let x and y be elements of G and suppose Tg (x) = Tg (y).
We must prove x = y.
Tg (x)
=
()
Tg (y) () gx = gy
x=y
Bijection: Let y 2 G. We must prove there exists x 2 G such that
y = gx. But since we are in a group, y = gx () x = g 1 y.
Lemma 292 Let G be a group and let H = fTg j g 2 Gg where Tg is de…ned as
in lemma 291. Then, H is a group with function composition.
Proof. We have to verify that (H; ) satis…es the four properties of a group.
Closure: Let Ta and Tb be two elements of H. Show Ta Tb belongs to H.
An element of H is of the form Tg with g 2 G, so we must prove Ta Tb = Tg
for some g in G. Since Ta and Tb are two elements of H, it means that
a and b belong to G hence ab 2 G by closure since G is a group. Ta Tb is
a permutation on G, hence a function, so we see how it acts on elements
of G. If x is an arbitrary element of G, then Ta Tb (x) = Ta (bx) = abx =
(ab) x = Tab (x). Since x was arbitrary, we see that Ta Tb = Tab 2 H since
ab 2 G.
Identity: Te where e is the identity of G is the identity of H since Ta Te =
Tae = Ta and Te Ta = Tea = Ta .
Inverses: Let Ta 2 H. Then Ta Ta
1
Ta 1 a = Te . Thus (Ta ) = Ta 1 .
1
= Taa
1
= Te and Ta 1 Ta =
Associativity: Composition of functions is always associative. You should
know this from previous classes, we also proved it in one of the homework
problems from the previous chapter.
We are now ready to state and prove our main theorem.
7.3. CAYLEY’S THEOREM
85
Theorem 293 (Cayley’s Theorem) Every group is isomorphic to a group of
permutations.
Proof. Let G be a group. We need to …nd a group of permutation G is isomorphic to. In particular, we will also need to …nd the isomorphism. Since G is the
only thing we have to start with, we must construct everything else from G. For
each g 2 G, we de…ne Tg as in lemma 291. Then, by the same lemma, Tg is a
permutation on G. We let H be as in lemma 292. by the same lemma, H is a
group of permutations. We claim G = H. We need to produce an isomorphism
between the two groups. It is not hard to …nd. De…ne : G ! H by (a) = Ta .
We verify that is an injection, a bijection and operation preserving.
Injection: Let a; b be elements of G and suppose that
must prove that a = b.
(a)
=
(a) =
(b), we
(b) () Ta = Tb
()
Ta (e) = Tb (e)
()
ae = be
()
a=b
Surjection:
is a surjection by de…nition since the elements of H are
built from the elements of G.
Operation Preserving: Let a; b be elements of G, we must show that
(ab) = (a) (b). From the proof of the lemma, we recall that Ta Tb = Tab
hence
(ab)
= Tab
= Ta Tb
=
(a) (b)
We illustrate this with an example.
Example 294 Consider U (12) = f1; 5; 7; 11g with multiplication mod 12. We
…nd the group of permutations it is isomorphic to. Recall from the proof that
U (12) is isomorphic to H = fT g j g 2 U (12)g = fT1 ; T5 ; T7 ; T11 g. Let us write
the elements of H in array notation.
T1 =
1
1
5
5
7
7
11
11
T5 =
1
5
5
1
7
11
11
7
T7 =
1
7
5
11
7
1
11
5
86
CHAPTER 7. ISOMORPHISMS
T11 =
1
11
5 7
7 5
11
1
Now, we write both multiplication tables:
U (12)
1
5
7
11
1
1
5
7
11
5
5
1
11
7
7
7
11
1
5
11
11
7
5
1
T7
T7
T11
T1
T5
T11
T11
T7
T5
T1
and
H
T1
T5
T7
T11
T1
T1
T5
T7
T11
T5
T5
T1
T11
T7
We can see that the tables are identical, except for the notation of the elements.
In particular, we can verify (ab) = (a) (b). For example, 5 7 = 11. Hence
on one hand (5 7) = (11) = T11 . Also, (5 7) = (5) (7) = T5 T7 . We
can verify that T5 T7 = T11 .
7.4
Problems
Do the following problems.
1. Prove that the mapping de…ned in example 278 is an isomorphism.
2. Prove that the mapping de…ned in example 283 is a bijection.
3. Prove that the mapping de…ned in example 279 is an isomorphism.
4. Finish proving theorem 285.
5. Finish proving theorem 290.
6. Prove that the mapping de…ned in example 280 is a bijection.
7. Prove that the mapping de…ned in example 281 is an automorphism.
8. Let G be the set of all possible groups. De…ne a relation R on G by
ARB () A = B. Prove this is an equivalence relation.
9. Do # 1, 3, 4, 5, 8, 19, 25, 27, 34 at the end of chapter 6.