To prove something is a homomorphism, you must demonstrate that the function in question preserves the algebraic structure of the sets involved.
This "structure-preserving" property is defined by a specific set of rules that vary depending on the algebraic objects you are working with, such as groups, rings, or vector spaces. A successful proof involves verifying that the function's definition satisfies these rules for all elements in its domain.
This article will cover the general procedure for proving a homomorphism, focusing on the most common algebraic structures you are likely to encounter in abstract algebra.
The general procedure for proving a homomorphism
No matter the context, the fundamental steps to prove a homomorphism are:
- Identify the algebraic structures. Clearly state the domain and codomain of the function, and identify their respective algebraic structures and operations (e.g., are they groups with addition, or rings with multiplication and addition?).
- State the definition. Write down the specific homomorphism conditions for the type of algebraic structure you are dealing with.
- Choose arbitrary elements. Let the variables represent arbitrary elements from the domain. This is critical for proving that the property holds for all elements, not just a specific few.
- Verify the conditions. Using the function's rule, algebraically manipulate the expression to show that the function's output preserves the operation(s). The left-hand side of your proof will involve applying the function to the result of an operation in the domain, while the right-hand side will apply the operation to the results of the function in the codomain.
Proving a group homomorphism
A function ϕ∶(G1,*)→(G2,⋅)phi colon open paren cap G sub 1 comma * close paren right arrow open paren cap G sub 2 comma center dot close paren
𝜙∶(𝐺1,*)→(𝐺2,⋅)
is a group homomorphism if and only if for all elements a,b∈G1a comma b is an element of cap G sub 1
𝑎,𝑏∈𝐺1
, the following property holds:ϕ(a*b)=ϕ(a)⋅ϕ(b)phi open paren a * b close paren equals phi open paren a close paren center dot phi open paren b close paren
𝜙(𝑎*𝑏)=𝜙(𝑎)⋅𝜙(𝑏)
The operation **
*
on the left occurs in the group G1cap G sub 1
𝐺1
, while the operation ⋅center dot
⋅
on the right occurs in the group G2cap G sub 2
𝐺2
.
Example: The determinant map
Prove that the determinant function is a group homomorphism from the group of invertible n×nn cross n
𝑛×𝑛
matrices with real entries, GLn(R)cap G cap L sub n open paren the real numbers close paren
𝐺𝐿𝑛(ℝ)
, to the multiplicative group of non-zero real numbers, R×the real numbers raised to the cross power
ℝ×
.
-
Identify the structures:
-
Domain: GLn(R)cap G cap L sub n open paren the real numbers close paren
𝐺𝐿𝑛(ℝ)
(invertible n×nn cross n
𝑛×𝑛
matrices with real entries) under matrix multiplication.
-
Codomain: R×the real numbers raised to the cross power
ℝ×
(non-zero real numbers) under real number multiplication.
-
Function: ϕ(A)=det(A)phi open paren cap A close paren equals det of open paren cap A close paren
𝜙(𝐴)=det(𝐴)
for any matrix A∈GLn(R)cap A is an element of cap G cap L sub n open paren the real numbers close paren
𝐴∈𝐺𝐿𝑛(ℝ)
.
-
-
State the condition: We must show that ϕ(AB)=ϕ(A)ϕ(B)phi open paren cap A cap B close paren equals phi open paren cap A close paren phi open paren cap B close paren
𝜙(𝐴𝐵)=𝜙(𝐴)𝜙(𝐵)
for all A,B∈GLn(R)cap A comma cap B is an element of cap G cap L sub n open paren the real numbers close paren
𝐴,𝐵∈𝐺𝐿𝑛(ℝ)
.
-
Choose arbitrary elements: Let A,Bcap A comma cap B
𝐴,𝐵
be any two matrices in GLn(R)cap G cap L sub n open paren the real numbers close paren
𝐺𝐿𝑛(ℝ)
.
-
Verify the condition:
-
Start with the left-hand side of the equation: ϕ(AB)phi open paren cap A cap B close paren
𝜙(𝐴𝐵)
.
-
By definition of the function ϕphi
𝜙
, this is det(AB)det of open paren cap A cap B close paren
det(𝐴𝐵)
.
-
Recall the property of determinants that for any square matrices Acap A
𝐴
and Bcap B
𝐵
, det(AB)=det(A)det(B)det of open paren cap A cap B close paren equals det of open paren cap A close paren det of open paren cap B close paren
det(𝐴𝐵)=det(𝐴)det(𝐵)
.
-
Since det(A)det(B)det of open paren cap A close paren det of open paren cap B close paren
det(𝐴)det(𝐵)
is exactly ϕ(A)ϕ(B)phi open paren cap A close paren phi open paren cap B close paren
𝜙(𝐴)𝜙(𝐵)
, we have shown that ϕ(AB)=ϕ(A)ϕ(B)phi open paren cap A cap B close paren equals phi open paren cap A close paren phi open paren cap B close paren
𝜙(𝐴𝐵)=𝜙(𝐴)𝜙(𝐵)
.
-
Conclusion: Therefore, the determinant function is a group homomorphism.
-
Key properties of group homomorphisms
When proving a group homomorphism, it is often useful to remember these derived properties:
-
Identity: A homomorphism maps the identity element of the domain to the identity element of the codomain. ϕ(eG1)=eG2phi open paren e sub cap G sub 1 close paren equals e sub cap G sub 2
𝜙(𝑒𝐺1)=𝑒𝐺2
.
-
Inverses: A homomorphism maps the inverse of an element to the inverse of its image. ϕ(a-1)=(ϕ(a))-1phi open paren a to the negative 1 power close paren equals open paren phi open paren a close paren close paren to the negative 1 power
𝜙(𝑎−1)=(𝜙(𝑎))−1
.
Proving a ring homomorphism
A function ϕ∶(R1,+,*)→(R2,⊕,⊗)phi colon open paren cap R sub 1 comma positive comma * close paren right arrow open paren cap R sub 2 comma circled plus comma ⊗ close paren
𝜙∶(𝑅1,+,*)→(𝑅2,⊕,⊗)
is a ring homomorphism if it preserves both the addition and multiplication operations. Additionally, if both rings have a multiplicative identity (unity), the homomorphism must also map the unity of the first ring to the unity of the second.
The three conditions are:
-
**Preserves addition:**ϕ(a+b)=ϕ(a)⊕ϕ(b)phi open paren a plus b close paren equals phi open paren a close paren circled plus phi open paren b close paren
𝜙(𝑎+𝑏)=𝜙(𝑎)⊕𝜙(𝑏)
-
**Preserves multiplication:**ϕ(a*b)=ϕ(a)⊗ϕ(b)phi open paren a * b close paren equals phi open paren a close paren ⊗ phi open paren b close paren
𝜙(𝑎*𝑏)=𝜙(𝑎)⊗𝜙(𝑏)
-
**Preserves unity (if applicable):**ϕ(1R1)=1R2phi open paren 1 sub cap R sub 1 close paren equals 1 sub cap R sub 2
𝜙(1𝑅1)=1𝑅2
Example: The modular map
Prove that the function ϕ∶Z→Znphi colon the integers right arrow the integers sub n
𝜙∶ℤ→ℤ𝑛
defined by ϕ(a)=[a]nphi open paren a close paren equals open bracket a close bracket sub n
𝜙(𝑎)=[𝑎]𝑛
(where [a]nopen bracket a close bracket sub n
[𝑎]𝑛
is the residue class of aa
𝑎
modulo nn
𝑛
) is a ring homomorphism.
-
Identify the structures:
-
Domain: The ring of integers Zthe integers
ℤ
under standard addition and multiplication.
-
Codomain: The ring of integers modulo nn
𝑛
, Znthe integers sub n
ℤ𝑛
, under modular addition and multiplication.
-
-
State the conditions: We must show that ϕ(a+b)=ϕ(a)+ϕ(b)phi open paren a plus b close paren equals phi open paren a close paren plus phi open paren b close paren
𝜙(𝑎+𝑏)=𝜙(𝑎)+𝜙(𝑏)
and ϕ(ab)=ϕ(a)ϕ(b)phi open paren a b close paren equals phi open paren a close paren phi open paren b close paren
𝜙(𝑎𝑏)=𝜙(𝑎)𝜙(𝑏)
for all integers a,b∈Za comma b is an element of the integers
𝑎,𝑏∈ℤ
.
-
Choose arbitrary elements: Let a,ba comma b
𝑎,𝑏
be any two integers.
-
Verify the conditions:
-
Condition 1 (Addition):
-
Left-hand side: ϕ(a+b)=[a+b]nphi open paren a plus b close paren equals open bracket a plus b close bracket sub n
𝜙(𝑎+𝑏)=[𝑎+𝑏]𝑛
.
-
Right-hand side: ϕ(a)+ϕ(b)=[a]n+[b]nphi open paren a close paren plus phi open paren b close paren equals open bracket a close bracket sub n plus open bracket b close bracket sub n
𝜙(𝑎)+𝜙(𝑏)=[𝑎]𝑛+[𝑏]𝑛
.
-
By the rules of modular arithmetic, [a+b]n=[a]n+[b]nopen bracket a plus b close bracket sub n equals open bracket a close bracket sub n plus open bracket b close bracket sub n
[𝑎+𝑏]𝑛=[𝑎]𝑛+[𝑏]𝑛
.
-
So, ϕ(a+b)=ϕ(a)+ϕ(b)phi open paren a plus b close paren equals phi open paren a close paren plus phi open paren b close paren
𝜙(𝑎+𝑏)=𝜙(𝑎)+𝜙(𝑏)
.
-
-
Condition 2 (Multiplication):
-
Left-hand side: ϕ(ab)=[ab]nphi open paren a b close paren equals open bracket a b close bracket sub n
𝜙(𝑎𝑏)=[𝑎𝑏]𝑛
.
-
Right-hand side: ϕ(a)ϕ(b)=[a]n[b]nphi open paren a close paren phi open paren b close paren equals open bracket a close bracket sub n open bracket b close bracket sub n
𝜙(𝑎)𝜙(𝑏)=[𝑎]𝑛[𝑏]𝑛
.
-
By the rules of modular arithmetic, [ab]n=[a]n[b]nopen bracket a b close bracket sub n equals open bracket a close bracket sub n open bracket b close bracket sub n
[𝑎𝑏]𝑛=[𝑎]𝑛[𝑏]𝑛
.
-
So, ϕ(ab)=ϕ(a)ϕ(b)phi open paren a b close paren equals phi open paren a close paren phi open paren b close paren
𝜙(𝑎𝑏)=𝜙(𝑎)𝜙(𝑏)
.
-
-
Conclusion: The function ϕphi
𝜙
preserves both addition and multiplication and is therefore a ring homomorphism.
-
Proving a linear map (homomorphism for vector spaces)
A function T∶V→Wcap T colon cap V right arrow cap W
𝑇∶𝑉→𝑊
is a homomorphism between vector spaces (commonly called a linear map or linear transformation) if it preserves vector addition and scalar multiplication.
The two conditions are:
-
**Preserves vector addition:**T(u⃗+v⃗)=T(u⃗)+T(v⃗)cap T open paren modified u with right arrow above plus modified v with right arrow above close paren equals cap T open paren modified u with right arrow above close paren plus cap T open paren modified v with right arrow above close paren
𝑇(𝑢⃗+𝑣⃗)=𝑇(𝑢⃗)+𝑇(𝑣⃗)
-
**Preserves scalar multiplication:**T(cu⃗)=cT(u⃗)cap T open paren c modified u with right arrow above close paren equals c cap T open paren modified u with right arrow above close paren
𝑇(𝑐𝑢⃗)=𝑐𝑇(𝑢⃗)
for any vector u⃗∈Vmodified u with right arrow above is an element of cap V
𝑢⃗∈𝑉
and scalar cc
𝑐
.
Example: The projection map
Prove that the function P∶R3→R2cap P colon R-3 right arrow R-2
𝑃∶ℝ3→ℝ2
defined by P(x,y,z)=(x,y)cap P open paren x comma y comma z close paren equals open paren x comma y close paren
𝑃(𝑥,𝑦,𝑧)=(𝑥,𝑦)
is a linear map.
-
Identify the structures:
-
Domain: The vector space R3R-3
ℝ3
over the field of real numbers.
-
Codomain: The vector space R2R-2
ℝ2
over the field of real numbers.
-
Function: P(x,y,z)=(x,y)cap P open paren x comma y comma z close paren equals open paren x comma y close paren
𝑃(𝑥,𝑦,𝑧)=(𝑥,𝑦)
.
-
-
State the conditions: We must show that P(u⃗+v⃗)=P(u⃗)+P(v⃗)cap P open paren modified u with right arrow above plus modified v with right arrow above close paren equals cap P open paren modified u with right arrow above close paren plus cap P open paren modified v with right arrow above close paren
𝑃(𝑢⃗+𝑣⃗)=𝑃(𝑢⃗)+𝑃(𝑣⃗)
and P(cu⃗)=cP(u⃗)cap P open paren c modified u with right arrow above close paren equals c cap P open paren modified u with right arrow above close paren
𝑃(𝑐𝑢⃗)=𝑐𝑃(𝑢⃗)
.
-
Choose arbitrary elements: Let u⃗=(u1,u2,u3)modified u with right arrow above equals open paren u sub 1 comma u sub 2 comma u sub 3 close paren
𝑢⃗=(𝑢1,𝑢2,𝑢3)
and v⃗=(v1,v2,v3)modified v with right arrow above equals open paren v sub 1 comma v sub 2 comma v sub 3 close paren
𝑣⃗=(𝑣1,𝑣2,𝑣3)
be any two vectors in R3R-3
ℝ3
, and let cc
𝑐
be any scalar.
-
Verify the conditions:
-
Condition 1 (Vector Addition):
-
Left-hand side: P(u⃗+v⃗)=P((u1,u2,u3)+(v1,v2,v3))=P(u1+v1,u2+v2,u3+v3)=(u1+v1,u2+v2)cap P open paren modified u with right arrow above plus modified v with right arrow above close paren equals cap P open paren open paren u sub 1 comma u sub 2 comma u sub 3 close paren plus open paren v sub 1 comma v sub 2 comma v sub 3 close paren close paren equals cap P open paren u sub 1 plus v sub 1 comma u sub 2 plus v sub 2 comma u sub 3 plus v sub 3 close paren equals open paren u sub 1 plus v sub 1 comma u sub 2 plus v sub 2 close paren
𝑃(𝑢⃗+𝑣⃗)=𝑃((𝑢1,𝑢2,𝑢3)+(𝑣1,𝑣2,𝑣3))=𝑃(𝑢1+𝑣1,𝑢2+𝑣2,𝑢3+𝑣3)=(𝑢1+𝑣1,𝑢2+𝑣2)
.
-
Right-hand side: P(u⃗)+P(v⃗)=P(u1,u2,u3)+P(v1,v2,v3)=(u1,u2)+(v1,v2)=(u1+v1,u2+v2)cap P open paren modified u with right arrow above close paren plus cap P open paren modified v with right arrow above close paren equals cap P open paren u sub 1 comma u sub 2 comma u sub 3 close paren plus cap P open paren v sub 1 comma v sub 2 comma v sub 3 close paren equals open paren u sub 1 comma u sub 2 close paren plus open paren v sub 1 comma v sub 2 close paren equals open paren u sub 1 plus v sub 1 comma u sub 2 plus v sub 2 close paren
𝑃(𝑢⃗)+𝑃(𝑣⃗)=𝑃(𝑢1,𝑢2,𝑢3)+𝑃(𝑣1,𝑣2,𝑣3)=(𝑢1,𝑢2)+(𝑣1,𝑣2)=(𝑢1+𝑣1,𝑢2+𝑣2)
.
-
Since the left and right sides are equal, vector addition is preserved.
-
-
Condition 2 (Scalar Multiplication):
-
Left-hand side: P(cu⃗)=P(c(u1,u2,u3))=P(cu1,cu2,cu3)=(cu1,cu2)cap P open paren c modified u with right arrow above close paren equals cap P open paren c open paren u sub 1 comma u sub 2 comma u sub 3 close paren close paren equals cap P open paren c u sub 1 comma c u sub 2 comma c u sub 3 close paren equals open paren c u sub 1 comma c u sub 2 close paren
𝑃(𝑐𝑢⃗)=𝑃(𝑐(𝑢1,𝑢2,𝑢3))=𝑃(𝑐𝑢1,𝑐𝑢2,𝑐𝑢3)=(𝑐𝑢1,𝑐𝑢2)
.
-
Right-hand side: cP(u⃗)=c(u1,u2)=(cu1,cu2)c cap P open paren modified u with right arrow above close paren equals c open paren u sub 1 comma u sub 2 close paren equals open paren c u sub 1 comma c u sub 2 close paren
𝑐𝑃(𝑢⃗)=𝑐(𝑢1,𝑢2)=(𝑐𝑢1,𝑐𝑢2)
.
-
Since the left and right sides are equal, scalar multiplication is preserved.
-
-
Conclusion: The function Pcap P
𝑃
is a linear map and therefore a homomorphism between vector spaces.
-
A note on isomorphisms
An isomorphism is a special type of homomorphism that is both injective (one-to-one) and surjective (onto). Isomorphisms show that two algebraic structures are fundamentally identical. After proving a map is a homomorphism, you can then test for injectivity (by showing the kernel is trivial) and surjectivity to see if it qualifies as an isomorphism.