Group Homomorphisms

As in the case of rings, we can have a homomorphism between groups

7.10.1 Definition. Let (G, *) and (G ', ⋅) two groups. A function φ: G ⟶ G' is said to be a homomorphism if

An homomorphism is thus a function between groups that preserves the operation.

By way of analogy with ring theory, we define two subgroups of G and G' the kernel of φ and the image, as follows

It is easy to show that Ker φ is a subgroup of G, whilst Im φ is a subgroup of G'.

7.10.2 Proposition. Let φ be a homomorphism from the group G to the group G'. If e denotes the identity in G, and er denotes the identity in G', then

1. φ(e) = e',

2. φ(g⁻¹) = [φ(g)]⁻¹, for all g in G.

Proof. Parts (a) and (b) are proved just as in the ring case.□

If G = G', the homomorphism φ is an endomorphism. A homomorphism φ is called an epimorphism if φ is onto, and a monomorphism if φ is one-to-one. If φ is both on-to and injective is an isomorphism.

Proposition 7.10.3. Let φ an homomorphism between two groups G and G', then φ is a monomorphism if and only if Ker φ = {e}.

Let φ an homomorphism between two groups G and G', with kernel K = Ker φ. Since K is a subgroup of G we can construct its right cosets modulo K in the following way

Thus two element a,b ∈ G are related through ρ_r modulo Ker φ, iff they possess the same image through φ. The cardinality of Imφ is thus the index of K in G.

Example 7.10.4.

1. Consider the following mapping

   φ: (GL_n(ℝ), ⋅) ⟶ (ℝ \{0, }, ⋅)
     A ⟼ det A

   This is a homomorphism whose kernel is given by SL_n(ℝ) = {A ∈ GL_n(ℝ) | det A = 1}. The image of φ is made by all real numbers different from zero. Two matrices are in the same right coset only if they have the same image through φ i.e. only oif they have the same determinant. This is because A ρ B ⇐⇒ AB⁻¹ ∈ in SL_n(ℝ) and det(AB⁻¹) = det A/det B = 1.

2. The mapping

   φ: (ℝ \{0}, ⋅) ⟶(ℝ \{0}, ⋅)
     φ(x) = 1/|x|

   is a homomorphism, such that Ker φ = {±1}, Imφ = ℝ⁺.

Proposition 7.10.5. Let φ a homomorphism between two groups G and G'. Then:

1. If G is finite, the order of Imφ divides the order of G (and the order of G' as well if G' is also finite).

2. If G is cyclic, then Imφ is also cyclic.

3. If g has finite order, then the order of φ(g) divides the order of g.

Proof.

1. As we showed in proposition 7.10.3, |Imφ| is equal to the index of K in G, which is a divisor of G by Lagrange Theorem.

2. If G = ⟨g⟩ = {g^k| k ∈ ℤ}, then Imφ = {φ(g^k) | k ∈ ℤ} = {φ(g)^k | k ∈ ℤ} = ⟨φ(g)⟩.

3. Let g an element of order n. Then

   φ(g)ⁿ = φ(gⁿ) = φ(e) = e'

   Then, if h is the order of φ(g), from n = hq + r with 0 ≤ r < h we have

4. Let g an element of order n. Then

   e' = φ(g)ⁿ = φ(g^h)^q ⋅ φ(g)^r

   implies that φ(g)^r = e', thus r = 0, hence h divides n   □.

For homomorphisms having as domain a cyclic group G = ⟨g⟩, every homomorphisms is defined by knowing the image of the generator g. We can have two cases: one of a cyclic finite group of order n the other for an infinite cyclic group.

• If G = ⟨g⟩ is infinite, the image of the generator g can be an arbitrary element of G', hence the homomorphisms of an infinite cyclic group in a group G' are as many as the elements of G'.

• If G = ⟨g | gⁿ = 1⟩ is cyclic of order n, the images of g though the homomorphism φ are such that the order of φ(g) is a divisor of n. Conversely given an element of G' whose order is a divisor of n there exists only one homomorphism mapping g in such an element. Thus the homomorphisms of a finite cyclic group of order n are as many as the elements of G' a divisor of n as order.

As a consequence of point a) of Proposition 7.10.5, the only homomorphism from ℤ₈ to ℤ₅ is the null homomorphism. This can also be noticed as follow: if φ is a homomorphism from ℤ₈ to ℤ₅, then |Imφ| must divide both 8 and 5, thus |Imφ| = 1.

Exercises

1. Prove that if φ is an isomorphism between two groups G and G', then for each g ∈ G the order g is equal to the order φ(g).

2. What is the number of isomorphisms between two cyclic groups with the same order G and G'?

3. Find all the group homomorphisms from (ℚ,+) into (ℤ,+).

4. Let φ a homomorphism between two groups G and G'. Prove that if G is abelian than Im φ is abelian too.

5. Let G be a group and consider the map inv: G ⟶ G given by inv(g) = g⁻¹. When is this a group homomorphism?

Solutions

1. Let n the order of g. By Proposition 7.10.4, we know that the order φ(g) divides n. We suppose by contradiction that the order of φ(g) is m, 0 < m < n. Then e = φ(g)^m = φ(g^m): owing to injectivity we must have g^m = e, which is absurd.  ■

2. If ϕ: G → G′ is an isomorphism, then consider where an arbitrary element of G is sent. An arbitrary element of G is of the form g^k, so ϕ(g^k) = ϕ(g)^k. Since ϕ is a bijection, the elements of G′ are precisely ϕ(g), ϕ(g)², … , ϕ(g)ⁿ. Thus a generator of G is mapped to a generator of G'. For any isomorphism f(a * a ... * a) = f(a)*f(a) ...*f(a): Once you map a generator somewhere you fix where every other element in the group goes. Thus the number of isomorphisms is equal to the number of generators. Lemma. Let G be cyclic group of order n. Then G contains exactly φ(n) generators. Proof: Let g be a generator of G, so G = {g,...,gⁿ = 1}. Then g^k generates G if and only if g^km = g for some m, which happens when km ≡ 1 (mod n), that is k must be a unit in ℤ_n, thus there are φ(n) values of k for which g^k is a generator.  ∎

3. Suppose there exists a homomorphism f: (ℚ,+) ⟶ (ℤ,+) such that f(x) = k, with k ∈ ℤ, then there exists some q = x/k ∈ ℚ such that f(x/k) = 1, because when you add x/k to itself k times, you need to get x, hence f(x) = k. But

   2x = x + x = f(q/2) + f(q/2) = f(q/2 + q/2) = f(q) = 1

   which is impossible. Therefore there is no such homomorphism f.  ∎

4. Let a',b' ∈ Im φ. Then a' = φ(a), b' = φ(b), for some a,b ∈ G. a'b' = φ(a)φ(b) = φ(ab) = φ(ba) = φ(a)φ(b) = b'a'.   ∎

5. If G is abelian then the map x → x⁻¹ is a homomorphism:

   (ab)⁻¹ = (ba)⁻¹ = a⁻¹b⁻¹

   If x → x⁻¹ is a homomorphism, G is abelian:

   ab = (b⁻¹a⁻¹)⁻¹ = ((ba)⁻¹)⁻¹ = ba ■