“Wow! So you use Internet Explorer in Windows, you must like it nice and slow,” I told one of my friends in the laboratory. “Wow! So you use vim in Linux, you must like it rough and tough,” Sussanne replied with a smirking smile, Apocalypse, Anawim, #justtothepoint.
In algebra, a homomorphism (from ὁμός, “homo” meaning same and μορφή, “morph”, shape or form) is a structure-preserving map between two algebraic structures of the same type, such as two groups, two rings, or two vector spaces.
In this chapter, we will give a more specific definition. A homomorphism Φ from a group G to a group G’ is a function or a mapping from (G, ∘) to (G', *) that preserves the operations of the groups; this means a map, Φ: G → G’ between both groups that satisfies Φ(a∘b) = Φ(a)*Φ(b) for every pair a, b of elements of G, that is, ∀a, b ∈ G (Figure 1.a. & Figure 1.b.).
💡A group homomorphism is a function, satisfying a few “natural” properties, between two groups that identifies similarities between them. Consider the groups ℤ and ℤ/2ℤ. The integer numbers can be partitioned into two subsets: even numbers and odd numbers. Let’s define Φ: (ℤ, +) → (ℤ/2ℤ, +_{mod 2}), Φ(n) = n mod 2, e.g., Φ(1) = Φ(3) = Φ(any arbitrary even number) = 1, Φ(2) = Φ(4) = Φ(any arbitrary odd number) = 0 (Figure 1.b.). Notice that even + even = even (0 + 0 ≡ 0 mod 2), even + odd = odd (0 + 1 ≡ 1 mod 2), odd + even = odd (1 + 0 ≡ 1 mod 2), and odd + odd = even (0 + 0 ≡ 0 mod 2).
We say that this homomorphism maps elements of ℤ to elements of ℤ/2ℤ. The group from which a function originates is the domain (ℤ in our example). The group into which the function maps is the codomain (ℤ/2ℤ). Φ: (G, ∘) → (G’, *) is an homomorphism if and only if Φ(a∘b) = Φ(a)*Φ(b), ∀a, b∈ G. Note that the operation a ∘ b is occurring in the domain while Φ(a)*Φ(b) occurs in the codomain.
Not every function from one group to another is a homomorphism. The condition Φ(a∘b) = Φ(a)*Φ(b), ∀a, b∈ G means that the map Φ preserve the structure of G.
Next, let’s consider the additive group of integers modulo 4 (ℤ/4ℤ, +_{mod 4}) and the multiplicative group ({1, -1, i, -i}, ·) formed by the fourth roots of unity, and their Cayley tables.
(ℤ/4ℤ, +_{mod 4}) | [0]_{4} | [1]_{4} | [2]_{4} | [3]_{4} |
---|---|---|---|---|
[0]_{4} | [0]_{4} | [1]_{4} | [2]_{4} | [3]_{4} |
[1]_{4} | [1]_{4} | [2]_{4} | [3]_{4} | [0]_{4} |
[2]_{4} | [2]_{4} | [3]_{4} | [0]_{4} | [1]_{4} |
[3]_{4} | [3]_{4} | [0]_{4} | [1]_{4} | [2]_{4} |
1 | -1 | i | -i | |
---|---|---|---|---|
1 | 1 | -1 | i | -i |
-1 | -1 | 1 | -i | i |
i | i | -i | -1 | 1 |
-i | -i | i | 1 | -1 |
Consider Φ : (ℤ/4ℤ, +_{mod 4}) → ({1, -1, i, -i}, ·), Φ(0) = 1, Φ(1) = i, Φ(2) = -i, and Φ(3) = -i. Φ is not just an homomorphism, it is indeed an isomorphism, that is, a bijective (both injective and surjective) homomorphism. An automorphism is an isomorphism from a group to itself.
Futhermore, we can observe the following:
The kernel of a homomorphism Φ is the set of elements in G which are mapped to the identity in G'; that is, Ker(Φ) = {a ∈ G | Φ(x) = e’} where e’ is the identity in G’. The elements in the codomain that the function maps to are called the image of the function, denoted Im(Φ), that is, Im(Φ) = Φ(G) = {Φ(g) | g ∈ G}.
Any isomorphism is a homomorphism. An isomorphism is also onto and one-to-one. The kernel of an isomorphism is the trivial subgroup.
Let Φ: G → G’ be defined by Φ(a) = e’ ∀a ∈ G. Φ(ab) = e’ = e’e’ = Φ(a)Φ(b) ∀a, b ∈ G. This is called the trivial homomorphism and Ker(Φ) = G.
Let Φ: ℤ → ℤ be defined by Φ(n) = 2n, Φ is a homomorphism. Φ(n + m) = 2(n + m) = 2n + 2m = Φ(n) + Φ(m). Ker(Φ) = {n ∈ ℤ: Φ(n) = 2n = 0} = {0}.
Let Φ: ℤ → ℤ_{5} = {[0], [1], [2], [3], [4]}, Φ(n) = n mod 5 = [n]_{mod 5} is a homomorphism, Φ(m + m’) = [m + m’]_{mod 5} = [m]_{mod 5} + [m’]_{mod 5} = Φ(m) + Φ(m’). Ker(Φ) = {n ∈ ℤ| Φ(n) = [0]} = {n ∈ ℤ: n ≡ 0 (mod 5)} = 5ℤ (Figure 1.b., n = 5).
Let’s generalize the previous result. Residue modulo n of an integer. Let Φ: (ℤ, +) → (ℤ_{n}, +_{n}), Φ(m) = [m]_{mod n} is a homomorphism: Φ(m + m’) = [m + m’]_{mod n} = [m]_{mod n} + [m’]_{mod n} = Φ(m) + Φ(m’).
Φ is surjective, Φ(ℤ) = img(Φ) = ℤ_{n} because ∀ 0 ≤ m ≤ n-1, [m] ∈ ℤ_{n}, Φ(m) = [m]. m ∈ Ker(Φ) ↭ Φ(m) = [0] ↭ m ≡ 0 (mod n). Therefore, Ker(Φ) = ⟨n⟩ = nℤ.
Besides, it is an homomorphism, Φ([m]_{4} + [m’]_{4}) = Φ([m + m’]_{mod 4}) = [m + m’]_{mod 2} = [m]_{mod 2} + [m’]_{mod 2} = Φ(m) + Φ(m’). Φ is surjective, because Φ([0]_{4}) = [0]_{2} and Φ([1]_{4}) = [1]_{2}. Ker(Φ) = {[m]_{4} ∈ ℤ_{4} | Φ([m]_{4}) = [m]_{2} = [0]_{2}} = {[0]_{4}, [2]_{4}} = ⟨[2]_{4}⟩.
Let G be any arbitrary group, we define Φ: ℤ → G, Φ(n) = g^{n} where g ∈ G is fixed. Φ is a homomorphism, Φ(n + m) = g^{n+n} = g^{n}g^{m} = Φ(n)Φ(m).
Let G be the general linear group, that is, the group of invertible 2 x 2 matrices with entries from the field of real numbers, and with the group operation being matrix multiplication, G = $GL(2, ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ, ad - bc ≠ 0}} \bigr\}$. The mapping Φ: GL(2, ℝ) → ℝ^{*}, Φ(A) = det(A) is an homomorphism, Φ(AB) = det(AB) = det(A)·det(B)) = Φ(A)Φ(B).
A ∈ Ker(det) ↭ det(A) = 1 ↭ A ∈ SL_{2}(ℝ) = {$GL(2, ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ, ad - bc = 1}} \bigr\}$}, that is, the group of 2 x 2 matrices with determinant one.
What is im(Φ)?
Φ is a homomorphism. $Φ((\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}) + (\begin{smallmatrix}a’ & b’\\ c’ & d’\end{smallmatrix})) = a + d + a’ + d’ = Φ((\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})) + Φ((\begin{smallmatrix}a’ & b’\\ c’ & d’\end{smallmatrix}))$. Φ is onto, ∀a ∈ ℝ, $Φ((\begin{smallmatrix}a & b\\ c & 0\end{smallmatrix}))$ = a. Ker(Φ) = {$(\begin{smallmatrix}a & b\\ c & -a\end{smallmatrix})$ | a, b, c ∈ ℝ} = sl_{2}(ℝ) = {A ∈ gl_{2}(ℝ) | tr(A) = 0}.
💣 However, Φ: GL_{2}(ℝ) → ℝ defined by Φ(A) = tr(A) is not an homomorphism since it maps the identity $Φ((\begin{smallmatrix}1 & 0\\ 0 & 1\end{smallmatrix}))$ to 2, which is not the identity in ℝ. The general linear group GL(2, ℝ) is defined as the group of invertible 2 x 2 matrices with entries from the field of real numbers, and with the group operation being matrix multiplication. gl_{2}(ℝ) is the group of 2 x matrices with entries from the field of real numbers, and with the group operation being matrix addition, so the identity is $(\begin{smallmatrix}0 & 0\\ 0 & 0\end{smallmatrix})$, and tr maps the identity to zero, which is the identity in ℝ.
Φ is an homomorphism because every permutation σ is a product of transpositions, and therefore these are these different options:
Φ is surjective, e.g., (12) → 1, () → 0. Ker(Φ) = {σ ∈ S_{n} | σ is even} = A_{n}, the alternating group of even permutations.
sign(xy) = 1 if x and y are both positive or negative, and -1 if one of them is positive and negative, thus sign(xy) = sign(x)·sign(y). Ker(sign) = ℝ^{+}, the set of all positive real numbers.
Φ is onto, since if z ∈ S^{1} ⇒ z = e^{iθ} = e^{i2π(2πθ)} = Φ(θ2π). r ∈ Ker(Φ) ↭ Φ(r) = 1 ↭ e^{2πri} = 1 ↭ cos(2πr) + isin(2πr) = 1 ↭ cos(2πr) = 1 and sin(2πr) = 0 ↭ r ∈ ℤ ⇒ Ker(Φ) = ℤ.
Let Φ be a homomorphism from a group G to another, G’, and let a, b be two arbitrary elements of G. Let H be a subgroup of G, H ≤ G. Then, the following statements hold true:
Proof:
Φ(e) = Φ(e·e) = Φ(e)Φ(e) ⇒ Φ(e) = Φ(e)Φ(e) ⇒ e’ = Φ(e)Φ(e)^{-1} = (Φ(e)Φ(e))Φ(e)^{-1} = Φ(e)
Proof.
If |a| is finite ⇒ ∃n ∈ ℤ: a^{n} = e ⇒ Φ(a^{n}) = [Φ(a^{n}) = (Φ(a))^{n}] (Φ(a))^{n} = Φ(e) =[Φ carries the identity] e’ where e’ is the identity of G’. Therefore, (Φ(a))^{n} = e’ ⇒ |Φ(a)| divides n ( = |a|) ∎
Last step: Suppose m = |Φ(a)| ⇒ ∃q, r: n = mq + r, 0 ≤ r < m ⇒ (Φ(a))^{r} = (Φ(a))^{n}(Φ(a))^{-mq} = (Φ(a))^{n}((Φ(a))^{m})^{-q} = [(Φ(a))^{n} = e’, m = |Φ(a)|] e’·e’ = e’ ⇒ [m = |Φ(a)| and 0 ≤ r < m] r = 0 ⇒ m | n.
Recall that to specify a group homomorphism Φ: ℤ_{m} → ℤ_{n} is enough to say what the value of Φ(1) is. Remember also that for a group homomorphism Φ : G → G’, Φ(e) = e'.
Define Φ:ℤ_{6} to ℤ_{15} by Φ([x]_{6}) = Φ([1+··+_{x}+··+1]_{6}) = [Φ([1]_{6})+··+_{x}+··+Φ([1]_{6})]_{15} = [Φ([1]_{6})x]_{15} = [mx]_{15}.
|Φ(1)| | |1| ⇒ |Φ(1)| | 6 ⇒ |Φ(1)| = {1, 2, 3, 6}. Since Φ(1) ∈ ℤ_{15}, the order of Φ(1) has to divide 15, too, so the options are |Φ(1)| = {1, 3}. Besides, we know that |a^{m}| = |⟨a^{m}⟩| = $\frac{n}{gcd(m, n)}=\frac{15}{gcd(15, m)}$. Therefore, in ℤ_{15}, these are the options:
Define Φ:ℤ_{24} to ℤ_{18} by Φ([x]_{24}) = [mx]_{18} for some [m]_{18} ∈ ℤ_{18}.
|Φ(1)| | |1| ⇒ |Φ(1)| | 24 ⇒ |Φ(1)| = {1, 2, 3, 4, 6, 8, 12}. Since Φ(1) ∈ ℤ_{18}, the order of Φ(1) has to divide 18, too, so the options are |Φ(1)| = {1, 2, 3, 6}.
We know that |a^{m}| = |⟨a^{m}⟩| = $\frac{n}{gcd(m, n)}=\frac{18}{gcd(18, m)}$. Therefore, in ℤ_{18}, these are the options:
Suppose for the sake of contradiction, Φ: ℤ_{4} × ℤ_{4} → ℤ_{8}, Φ surjective ⇒ ∃x ∈ ℤ_{4} × ℤ_{4}: Φ(x) = 1 ∈ ℤ_{8} ⇒ |Φ(x)| = 8 ⊥ We know that |a| < ∞, then |Φ(a)| divides |a|. Besides, |Φ(x)| = 8 divides |x| ≤ 4 (∀x ∈ ℤ_{4} × ℤ_{4}, |x| = 1, 2 or 4).
Proof.
e ∈ Ker(Φ) because Φ carries the identity, Φ(e) = e'.
∀a, b ∈ Ker(Φ), ab^{-1} ∈ G, Φ(ab^{-1}) = [Φ preserves the group operation] Φ(a)Φ(b^{-1}) = [∀n ∈ ℤ, Φ(a^{n}) = (Φ(a))^{n}. ] Φ(a)Φ(b)^{-1} = [a, b ∈ Ker(Φ)] e’e’^{-1} = e’ ⇒ ab^{-1} ∈ Ker(Φ).∎
Proof.
⇒) Suppose Φ(a) = Φ(b) ⇒ [Multiplying both sides by Φ(b)^{-1}] (Φ(b))^{-1}Φ(a) = e’ ⇒[Φ homomorphism ⇒ Φ(a^{n}) = (Φ(a))^{n}] Φ(b^{-1})Φ(a) = e’ ⇒ [Φ preserves group operations] Φ(b^{-1}a) = e’ ⇒ b^{-1}a ∈ Ker(Φ) ⇒ [H ≤ G, aH = bH iff a^{-1}b ∈ H] aKer(Φ) = bKer(Φ).
⇐) Conversely, aKer(Φ) = bKer(Φ) ⇒[H ≤ G, aH = bH iff a^{-1}b ∈ H] b^{-1}a ∈ Ker(Φ) ⇒ Φ(b^{-1}a) = e’ ⇒[Repeating the same argument] (Φ(b))^{-1}Φ(a) = e’ ⇒ Φ(a) = Φ(b)∎
💡The Kernel (preimage of the identity, i.e., Ker(Φ) = Φ^{-1}{e}) is a subgroup, and all other preimages are cosets of the Kernel. Please read an example that illustrates this fact.
Proof.
Φ(a) = a’. First, let’s prove that Φ^{-1}(a’) ⊆ aKer(Φ).
Let x ∈ Φ^{-1}(a’), x ∈ aKer(Φ)?
x ∈ Φ^{-1}(a’) = {x ∈ G | Φ(x) = a’} ⇒ Φ(x) = a’ = Φ(a) ⇒ [Previous property, Φ(a) = Φ(b) if and only if aKer(Φ) = bKer(Φ)] xKer(Φ) = aKer(Φ) ⇒ [aH = bH iff a ∈ bH] x ∈ aKer(Φ).
Conversely, let’s try to prove that aKer(Φ) ⊆ Φ^{-1}(a’). ∀x ∈ Ker(Φ), ax ∈ Φ^{-1}(a’)?
x ∈ Ker(Φ) ⇒ Φ(x) = e’ ⇒ Φ(ax) =[Φ homomorphism] Φ(a)Φ(x) = a’e’ = a’ ⇒ [By definition] ax ∈ Φ^{-1}(a’)∎
∀g’ ∈ Im(G) ⇒[G = ⟨g⟩] ∃n: Φ(g^{n}) = g’ ⇒ (Φ(g))^{n} = g’. Therefore, ∀g’ ∈ Im(G), ∃n: g’ = (Φ(g))^{n} ⇒ Im(G) = ⟨Φ(g)⟩.
Besides, if we know where a homomorphism maps the generators of G, we can determine where it maps all elements of G, Φ: ℤ_{3} → ℤ_{6} is completely determined by Φ(1), because Φ(n) = Φ(1 + 1 + ··_{n}·· + 1) = Φ(1) + Φ(1) + ··_{n}·· + Φ(1), e.g., Φ(1) = 4 ⇒ Φ(2) = Φ(1) + Φ(1) = 4 +_{mod 6} 4 = 2, Φ(0) = Φ(1 + 1 + 1) = 4 +_{mod 6} 4 +_{mod 6} 4 = 0.
G = ⟨a, b⟩, Φ: G → H. Suppose that we know Φ(a) and Φ(b). Using this information we can determine the image of any element in G, e.g., Φ(a^{3}b^{2}a) = Φ(a)Φ(a)Φ(a)Φ(b)Φ(b)Φ(a).
Can there be a homomorphism from ℤ_{16} onto ℤ_{2} × ℤ_{2}?
Nope. Suppose for the sake of contradiction Φ: ℤ_{16} → ℤ_{2} × ℤ_{2}, Φ surjective ⇒ [ℤ_{16} = ⟨1⟩, is cyclic] Φ(ℤ_{16}) =[By assumption, Φ is onto] ℤ_{2} × ℤ_{2} is cyclic ⊥
What are the group homomorphisms from ℤ_{12} to ℤ_{10}? These results have been demonstrated or they will be proven in the following articles.
What are the group homomorphisms from D_{4} to ℤ_{2}?
There are three options (three groups of order 4 in D_{4}):
a) H = ⟨r⟩ = {1, r, r^{2}, r^{3}}. sH = {s, sr, sr^{2}, sr^{3}}, D_{4} = H ∪ sH. Φ: D_{4} → ℤ_{2} and is determined by Φ(r) = 0 and Φ(s) = 1
b) H = ⟨s, r^{2}⟩ = {1, s, r^{2}, sr^{2}}, rH = {r, rs, r^{3}, rsr^{2}} =[r^{k}s = sr^{4-k}, 1 ≤ k ≤ 3, (rs)r^{2} = sr^{3}r^{2} = sr^{5} = sr] {r, sr^{3}, r^{3}, sr}. Φ: D_{4} → ℤ_{2} and is determined by Φ(r) = 1 and Φ(s) = 0
c) H = ⟨sr, r^{2}⟩ = {1, sr, r^{2}, sr^{3}}, rH = {r, rsr, r^{3}, rsr^{3}} = [r^{k}s = sr^{4-k}, 1 ≤ k ≤ 3, (rs)r^{3} = sr^{3}r^{3} = sr^{2}; rsr = sr^{3}r = s] {r, s, r^{3}, sr^{2}}. Φ: D_{4} → ℤ_{2} and is determined by Φ(r) = Φ(s) = 1. Notice that Φ(r^{2}) = Φ(r) + Φ(r) = 1 +_{mod 2} 1 = 0 and Φ(sr) = Φ(s) + Φ(r) = 1 +_{mod 2} 1 = 0.
Proof. They have already been demonstrated in our article about isomorphisms because we have only used that isomorphisms are operation-preserving mappings.
Proof.
Suppose H ◁ G. Claim: Φ(H) ◁ Φ(G)
∀g ∈ G, ∀h ∈ H, Φ(g)Φ(h)(Φ(g))^{-1} =[Φ(a^{n}) = (Φ(a))^{n}] Φ(g)Φ(h)Φ(g^{-1}) =[Φ homomorphism] Φ(ghg^{-1}) ∈ Φ(H) because H ◁ G.