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9
Homomorphisms
Up to now, we have been discussing the groups and their structure. Now we turn to the
discussion of functions from one group to another, which ‘respect’ the group structure.
These functions are called homomorphisms.
9.1
Homomorphisms: definition and examples
Definition 9.1. A map φ from one group G to another group G is a homomorphism if
φ(ab) = φ(a)φ(b)
(9.1)
for all a, b ∈ G.
Remark.
1. Note that in the l.h.s. of (9.1), the product ab is taken in G, whereas in the
r.h.s. the product φ(a)φ(b) is taken in G .
2. Clearly, an isomorphism between two groups is a special example of a homomorphism.
3. For any groups G and G there is always at least one homomorphism φ : G → G ,
namely the trivial homomorphism defined by φ(g) = e for all g ∈ G, where e is the
identity in G . Equation (9.1) then reduces to the true equation e e = e .
Example 9.2.
1. Consider the groups C∗ and R+ under multiplication and the map
φ : C∗ → R+ given by φ(z) = |z|. Since |z1 z2 | = |z1 | |z2 |, the equation (9.1) is
satisfied and φ is a homomorphism.
2. Consider R under addition and the group U of complex numbers z with |z| = 1 under
multiplication. Let φ : R → U be the map φ(x) = ei2πx . Since
φ(x + y) = ei2π(x+y) = ei2πx ei2πy = φ(x)φ(y),
φ is a homomorphism.
3. Let F (R) be the group of all functions g : R → R under addition. Consider the
mapping φ of F (R) to R under addition, defined by φ(g) = g(0) for any g ∈ F (R).
Recall that, by definition, the sum of two functions f and g is the function f + g
whose value at x is f (x) + g(x). Thus we have
φ(f + g) = (f + g)(0) = f (0) + g(0) = φ(f ) + φ(g),
and (9.1) is satisfied, so φ is a homomorphism.
4. Consider the group GL(n, R). Recall that for matrices A, B ∈ GL(n, R) we have
det(AB) = det A det B.
This means that det is a homomorphism from GL(n, R) to R∗ under multiplication.
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5. Fix r ∈ Z and let φr : Z → Z be defined by φr (n) = rn for all n ∈ Z. For all n, m ∈ Z,
we have
φr (m + n) = r(m + n) = rm + rn = φ(m) + φ(n),
so φr is a homomorphism. Note that φ0 is the trivial homomorphism, φ1 is the identity
map, and φ−1 maps Z onto Z. For all other r ∈ Z, the map φr is not onto Z.
6. Fix n ∈ N. Let γ : Z → Zn be the map given by γ(m) = r, where r is the remainder
when m is divided by n. Let us show that γ is a homomorphism. We need to show
that
γ(s + t) = γ(s)⊕γ(t),
where ⊕ is addition modulo n (i.e., the group operation in Zn ). Dividing s and t by
n, we get
s = q1 n + r 1 ,
t = q2 n + r 2 ,
0 ≤ r1 < n,
0 ≤ r2 < n.
Note that γ(s) = r1 and γ(t) = r2 . Dividing r1 + r2 by n, we get
r 1 + r 2 = q3 n + r 3 ,
0 ≤ r3 < n.
(9.2)
Thus,
s + t = (q1 + q2 + q3 )n + r3 ,
and so γ(s + t) = r3 . On the other hand, equation (9.2) shows that γ(s)⊕γ(t) = r3 .
7. Let G1 and G2 be groups. It is easy to see that the map f : G1 × G2 → G1 ,
f : (g1 , g2 ) → g1 , is a homomorphism, called projection homomorphism.
9.2
Properties of homomorphisms
Theorem 9.3. Let φ be a homomorphism of a group G into a group G .
(i) If e is the identity in G, then φ(e) is the identity e in G .
(ii) If a ∈ G, then φ(a−1 ) = φ(a)−1 .
Recall that we have already proven this fact for isomorphisms (Theorem 4.5). The proof
below repeats word for word the proof of Theorem 4.5.
Proof. (i) For any a ∈ G, one has
φ(a)e = φ(a) = φ(ae) = φ(a)φ(e).
Using the cancellation rule, we get e = φ(e).
(ii) For any a ∈ G, one has
e = φ(e) = φ(aa−1 ) = φ(a)φ(a−1 ),
e = φ(e) = φ(a−1 a) = φ(a−1 )φ(a).
This shows that φ(a−1 ) = φ(a)−1 .
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Next, we recall the definition of an image of a set. Let A and B be any two sets, and
f : A → B be a mapping.
For S ⊂ A, f [S] = {y ∈ B | ∃x ∈ S such that y = f (x)} is the image of S.
Theorem 9.4. Let φ be a homomorphism of a group G into a group G . If H is a subgroup
in G, then φ[H] is a subgroup of G .
Proof. Consider the set φ[H]. We shall check the two conditions of Theorem 3.7. In order to
check that φ[H] is closed under the group operation, let φ(a) and φ(b) be any two elements
of φ[H] and consider their product: φ(a)φ(b) = φ(ab). So we see that their product is in
φ[H], and thus φ[H] is closed under the group operation.
Next, for any φ(a) ∈ φ[H], by Theorem 9.3(ii), one has φ(a)−1 = φ(a−1 ) ∈ φ[H].
Corollary 9.5. Let φ be a homomorphism of a group G into a group G .
(i) The image φ[G] of G is a subgroup of G .
(ii) If |G | is finite, then |φ[G]| is finite and is a divisor of |G |.
Proof. (i) follows from Theorem 9.4(i) with H = G. (ii) follows from (i) by the Theorem
of Lagrange.
9.3
The kernel of a homomorphism
First recall the definition of a pre-image (another term is inverse image) of an element
of a set. Let A and B be any two sets, and f : A → B be a mapping. For y ⊂ B,
f −1 (y) = {x ∈ A | f (x) = y} is the pre-image of y.
Let φ be a homomorphism of a group G into a group G , and let e be the identity in
G . The set φ−1 (e ) plays a very important role for the homomorphism φ and has a special
name.
Definition 9.6. Let φ be a homomorphism of a group G into a group G and e be the
identity in G . The set φ−1 (e ) = {x ∈ G | φ(x) = e } is called a kernel of φ, denoted by
Ker(φ).
The kernel of a homomorphism φ : G → G is always non-empty: by Theorem 9.3(i),
the identity of G is always in Ker φ.
Example 9.7.
1. Consider the groups C∗ and R+ under multiplication and the homomorphism φ : C∗ → R+ given by φ(z) = |z|. The kernel of φ is the subgroup U of all
z such that |z| = 1.
2. Consider the group R under addition and the homomorphism φ : R → U given by
φ(x) = ei2πx . The kernel of φ is the subgroup Z of R.
3. Let F (R) be the group of all functions g : R → R under addition. Let φ : F (R) → R
be the homomorphism defined by φ(g) = g(0) for any g ∈ F (R). The kernel of φ is
the subgroup of all g such that g(0) = 0.
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Theorem 9.8. Let φ be a homomorphism of a group G into a group G , and let e be the
identity in G . Then the kernel H = Ker(φ) is a normal subgroup of G. For any a ∈ G,
one has
φ−1 (φ(a)) = {x ∈ G | φ(x) = φ(a)} = aH.
(9.3)
Note that the first equality in (9.3) is simply the definition of the set φ−1 (φ(a)), whereas
the second one is a statement.
Proof. 1. Let us check that H is a subgroup. We shall check the two conditions of Theorem 3.7.
First, suppose a and b are in H, i.e., φ(a) = φ(b) = e . Then φ(ab) = φ(a)φ(b) = e e = e
and so ab ∈ H.
Next, if a ∈ H, then, by Theorem 9.3(ii), φ(a−1 ) = (φ(a))−1 = e −1 = e , and so e ∈ H.
2. In order to prove (9.3), let us prove the two inclusions
aH ⊂ φ−1 (φ(a)),
φ−1 (φ(a)) ⊂ aH.
(9.4)
For the first one, suppose that g ∈ aH, i.e., g = ah with h ∈ H. Then
φ(g) = φ(ah) = φ(a)φ(h) = φ(a)e = φ(a)
and so g ∈ φ−1 (φ(a)). For the second inclusion, suppose that g ∈ φ−1 (φ(a)), i.e., φ(g) =
φ(a). Then
e = (φ(a))−1 φ(g) = φ(a−1 g)
and so a−1 g = h ∈ H. Thus, g = ah and so g ∈ aH.
3. Similarly, to (9.4), one proves the inclusions
Ha ⊂ φ−1 (φ(a)),
φ−1 (φ(a)) ⊂ Ha.
(9.5)
Thus, aH = φ−1 (φ(a)) = Ha, and so the subgroup H is normal.
Corollary 9.9. A group homomorphism φ : G → G is a one-to-one map if and only if
Ker(φ) = {e}.
Proof. If Ker(φ) = {e}, then the elements mapped into φ(a) are precisely the elements of
the left coset a{e} = {a}, which shows that φ is one-to-one.
Conversely, suppose φ is one-to-one. By Theorem 9.3(i), we know that φ(e) = e . Since
φ is one-to-one, we see that e is the only element mapped into e by φ, so Ker(φ) = {e}.
Example 9.10.
1. Consider the homomorphism det : GL(n, R) → R∗ . The kernel of
det is the set of all matrices A ∈ GL(n, R) such that det A = 1. This set is called
special linear group and denoted by SL(n, R). Theorem 9.8 states that SL(n, R) is a
normal subgroup of GL(n, R).
2. Consider the ‘complex-valued version’ of the previous example. Consider the homomorphism det : GL(n, C) → C∗ . The kernel of det is the subgroup SL(n, C),
consisting of all matrices of GL(n, C) with determinant equal to 1. Theorem 9.8
states that SL(n, C) is a normal subgroup of GL(n, C).
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3. Fix r ∈ Z and let the homomorphism φr : Z → Z be defined by φr (n) = rn for all n ∈
Z. The kernel of φ is {0}; thus, φ is one-to-one, which is easy to see straightforwardly.
4. Fix n ∈ N. Let γn : Z → Zn be the homomorphism given by γn (m) = r, where r
is the remainder when m is divided by n. The kernel of γn consists of all m evenly
divided by n. Thus, Ker(γn ) = nZ.
5. Let G1 and G2 be groups and let e1 be the identity in G1 . Consider the homomorphism
f : G1 × G2 → G1 , f : (g1 , g2 ) → g1 . The kernel of f is H = {(e1 , g2 ) | g2 ∈ G2 }.
Theorem 9.8 states that H is a normal subgroup of G1 × G2 . We are already familiar
with this fact.
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