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Group Homomorphism

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Definitions

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.

Counterexample: Φ: ℤ → ℤ defined by Φ(n) = n + 1 is not an homomorphism because Φ(n + m) = n + m + 1 ≠ Φ(n) + Φ(m) = n + m + 2.

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:

  1. They are both cycle groups. (ℤ/4ℤ, +mod 4) = ⟨1⟩ and {1, -1, i, -1} = ⟨i⟩ = {i0 = 1, i1 = i, i2 = -1, i3 = i}
  2. Φ carries the identity (Φ(0) = 1), and the generator (Φ(1) = i).
  3. They’re identical groups, both Cayley tables are exactly the same, i.e. up to relabeling the group elements.

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}.

 

Examples

Φ is surjective, Φ(ℤ) = img(Φ) = ℤn because ∀ 0 ≤ m ≤ n-1, [m] ∈ ℤn, 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 ∈ ℤ4 | Φ([m]4) = [m]2 = [0]2} = {[0]4, [2]4} = ⟨[2]4⟩.

A ∈ Ker(det) ↭ det(A) = 1 ↭ A ∈ SL2(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)?

  1. x > 0 ⇒ $\sqrt{x} > 0 ⇒ det(\begin{smallmatrix}\sqrt{x} & 0\\ 0 & \sqrt{x}\end{smallmatrix}) = x$.
  2. x < 0 ⇒ $\sqrt{-x} > 0 ⇒ det(\begin{smallmatrix}0 & \sqrt{-x}\\ \sqrt{-x} & 0\end{smallmatrix}) = x$ ⇒ im(Φ) = Φ(GL2(ℝ)) = ℝx.

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} = sl2(ℝ) = {A ∈ gl2(ℝ) | tr(A) = 0}.

💣 However, Φ: GL2(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. gl2(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:

  1. σ1 and σ2 even, Φ(σ1σ2) =[even + even = even] 0 =[Additive notation] Φ(σ1) + Φ(σ2) = 0 + 0.
  2. σ1 and σ2 odd, Φ(σ1σ2) =[odd + odd = even] 0 =2 Φ(σ1) + Φ(σ2) = 1 + 1.
  3. σ1 even and σ2 odd, Φ(σ1σ2) =[even + odd = even] 1 =2 Φ(σ1) + Φ(σ2) = 0 + 1.
  4. σ1 odd and σ2 even, Φ(σ1σ2) =[odd + even = even] 1 =2 Φ(σ1)Φ(σ2) = 1 + 0.

Φ is surjective, e.g., (12) → 1, () → 0. Ker(Φ) = {σ ∈ Sn | σ is even} = An, 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 ∈ S1 ⇒ z = e = ei2π(2πθ) = Φ(θ2π). r ∈ Ker(Φ) ↭ Φ(r) = 1 ↭ e2πri = 1 ↭ cos(2πr) + isin(2πr) = 1 ↭ cos(2πr) = 1 and sin(2πr) = 0 ↭ r ∈ ℤ ⇒ Ker(Φ) = ℤ.

Properties

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: an = e ⇒ Φ(an) = [Φ(an) = (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.

Examples

Recall that to specify a group homomorphism Φ: ℤm → ℤn 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:ℤ6 to ℤ15 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) ∈ ℤ15, the order of Φ(1) has to divide 15, too, so the options are |Φ(1)| = {1, 3}. Besides, we know that |am| = |⟨am⟩| = $\frac{n}{gcd(m, n)}=\frac{15}{gcd(15, m)}$. Therefore, in ℤ15, these are the options:

  1. The only element in ℤ15 of order 1 is 0, |[0]15| = 1, Φ([1]6) = Φ([0·x]6)x = 1 =[0]15 ⇒ Ker(Φ) = ℤ6, im(Φ) = {[0]6}
  2. The only elements in ℤ15 of order 3 are 5 and 10, |[5]15| = |[10]15| = 3. Φ([1]6) = [5]15, Φ([x]6) =[5x]15, e.g., Φ([2]6) = [10]15, Φ([3]6) = [0]15, Φ([4]6) = [20]15 = [5]15. Ker(Φ) = {[0]6, [3]6} = ⟨[3]6⟩. Im(Φ) = {[0]15, [5]15, [10]15} = ⟨[5]15⟩ ≋ ℤ3. Or alternatively Φ([1]6) = [10]15.

Define Φ:ℤ24 to ℤ18 by Φ([x]24) = [mx]18 for some [m]18 ∈ ℤ18.

|Φ(1)| | |1| ⇒ |Φ(1)| | 24 ⇒ |Φ(1)| = {1, 2, 3, 4, 6, 8, 12}. Since Φ(1) ∈ ℤ18, the order of Φ(1) has to divide 18, too, so the options are |Φ(1)| = {1, 2, 3, 6}.

We know that |am| = |⟨am⟩| = $\frac{n}{gcd(m, n)}=\frac{18}{gcd(18, m)}$. Therefore, in ℤ18, these are the options:

  1. |3| = |15| = 6, Φ([x]24) =[3x]18, Φ([x]24) =[15x]18.
  2. |6| = |12| = 3, Φ([x]24) =[6x]18, Φ([x]24) =[12x]18.
  3. |9| = 2, Φ([x]24) =[9x]18.
  4. |0| = 1, Φ([x]24) =[0]18.

Suppose for the sake of contradiction, Φ: ℤ4 × ℤ4 → ℤ8, Φ surjective ⇒ ∃x ∈ ℤ4 × ℤ4: Φ(x) = 1 ∈ ℤ8 ⇒ |Φ(x)| = 8 ⊥ We know that |a| < ∞, then |Φ(a)| divides |a|. Besides, |Φ(x)| = 8 divides |x| ≤ 4 (∀x ∈ ℤ4 × ℤ4, |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(an) = (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))-1Φ(a) = e’ ⇒[Φ homomorphism ⇒ Φ(an) = (Φ(a))n] Φ(b-1)Φ(a) = e’ ⇒ [Φ preserves group operations] Φ(b-1a) = e’ ⇒ b-1a ∈ Ker(Φ) ⇒ [H ≤ G, aH = bH iff a-1b ∈ H] aKer(Φ) = bKer(Φ).

⇐) Conversely, aKer(Φ) = bKer(Φ) ⇒[H ≤ G, aH = bH iff a-1b ∈ H] b-1a ∈ Ker(Φ) ⇒ Φ(b-1a) = e’ ⇒[Repeating the same argument] (Φ(b))-1Φ(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(gn) = 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: ℤ3 → ℤ6 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(a3b2a) = Φ(a)Φ(a)Φ(a)Φ(b)Φ(b)Φ(a).

Can there be a homomorphism from ℤ16 onto ℤ2 × ℤ2?

Nope. Suppose for the sake of contradiction Φ: ℤ16 → ℤ2 × ℤ2, Φ surjective ⇒ [ℤ16 = ⟨1⟩, is cyclic] Φ(ℤ16) =[By assumption, Φ is onto] ℤ2 × ℤ2 is cyclic ⊥

Important Example

What are the group homomorphisms from ℤ12 to ℤ10? These results have been demonstrated or they will be proven in the following articles.

  1. Ker(Φ) ≤ ℤ12 ⇒[By Lagrange’s Theorem] |Ker(Φ)| divides 12 ⇒ |Ker(Φ)| = 1, 2, 3, 4, 6 or 12.
  2. First Isomorphism Theorem: ℤ12/Ker(Φ) ≋ Φ(ℤ12) and Φ(ℤ12) ≤ ℤ10 ⇒[By Lagrange’s Theorem] |G/Ker(Φ)| = |G|/|Ker(Φ)| = 12/|Ker(Φ)| divides 10 ⇒ 12/|Ker(Φ)| = 1, 2, 5 or 10.
  3. 1 & 2 ⇒ |Ker(Φ)| = 6 or 12.
  4. If |Ker(Φ)| = 12 ⇒ G = Ker(Φ) and Φ is the trivial homomorphism.
  5. If |Ker(Φ)| = 6, Ker(Φ) ≤ ℤ12, the only subgroup that fits the bill is Ker(Φ) = ⟨2⟩ = {0, 2, 4, 6, 8, 10}. |G/Ker(Φ)| = 2, so G/Ker(Φ) ≋ ℤ2. Therefore, G is partitioned in two cosets H and 1 + H = {1, 3, 5, 7, 9, 11}. Φ carries the identity, Φ(0) = 0 ⇒[Φ(a) = Φ(b) ↭ aKer(Φ) = bKer(Φ).] We need to map H to zero, and 1 + H to 1 (e.g., 1 + H = 3 + H ↭ Φ(1) = Φ(3)).

What are the group homomorphisms from D4 to ℤ2?

  1. Ker(Φ) ≤ D4 ⇒[By Lagrange’s Theorem] |Ker(Φ)| divides 8 ⇒ |Ker(Φ)| = 1, 2, 4 or 8.
  2. First Isomorphism Theorem: D4/Ker(Φ) ≋ Φ(D4) and Φ(D4) ≤ ℤ2 ⇒[By Lagrange’s Theorem] |G/Ker(Φ)| = |G|/|Ker(Φ)| = 8/|Ker(Φ)| divides 2.
  3. 1 & 2 ⇒ |Ker(Φ)| = 4 or 8.
  4. If |Ker(Φ)| = 8 ⇒ G = Ker(Φ) and Φ is the trivial homomorphism.
  5. If |Ker(Φ)| = 4, |G/Ker(Φ)| = 2, H = Ker(Φ) ≤ D4.

There are three options (three groups of order 4 in D4):
a) H = ⟨r⟩ = {1, r, r2, r3}. sH = {s, sr, sr2, sr3}, D4 = H ∪ sH. Φ: D4 → ℤ2 and is determined by Φ(r) = 0 and Φ(s) = 1
b) H = ⟨s, r2⟩ = {1, s, r2, sr2}, rH = {r, rs, r3, rsr2} =[rks = sr4-k, 1 ≤ k ≤ 3, (rs)r2 = sr3r2 = sr5 = sr] {r, sr3, r3, sr}. Φ: D4 → ℤ2 and is determined by Φ(r) = 1 and Φ(s) = 0
c) H = ⟨sr, r2⟩ = {1, sr, r2, sr3}, rH = {r, rsr, r3, rsr3} = [rks = sr4-k, 1 ≤ k ≤ 3, (rs)r3 = sr3r3 = sr2; rsr = sr3r = s] {r, s, r3, sr2}. Φ: D4 → ℤ2 and is determined by Φ(r) = Φ(s) = 1. Notice that Φ(r2) = 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(an) = (Φ(a))n] Φ(g)Φ(h)Φ(g-1) =[Φ homomorphism] Φ(ghg-1) ∈ Φ(H) because H ◁ G.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn (Abstract Algebra), and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. Andrew Misseldine: College Algebra and Abstract Algebra.
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