Section 13 Homomorphisms

13.1 Definition

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=130&selection=123,0,141,8&color=note|p.125]]

A map φ of a group G into a group G′ is a homomorphism if the homomorphism property $\varphi (ab)=\varphi (a)\varphi (b)$ holds for all $a,b\in G$.

trivial homomorphism: $\varphi (g)=e^{\prime }$ for all $g\in G$

Tip

if G is abelian, then G’ must also be abelian:$a^{\prime }{b}^{\prime }=\varphi (a)\varphi (b)=\varphi (ab)=\varphi (ba)=\varphi (b)\varphi (a)=b^{\prime }{a}^{\prime }$

Recall:

• $det(AB)=det(A)det(B)$ on invertible n*n matrices (Determinant (legacy))
• matrix A is invertible if and only if det(A) is nonzero.

This means that det is a homomorphism mapping $GL(n,\mathbb{R})$ (the multiplicative group of all invertible n*n matrices) into the multiplicative group ${\mathbb{R}}^{\ast }$ of non zero real numbers.

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=132&selection=217,0,224,1&color=note|p.127]]

Homomorphisms of a group G into itself are often useful for studying the structure of G

13.7 Example

One can define multiplication by an integer as a homomorphism of integer addition to itself. ${\varphi }_r:\mathbb{Z}\to \mathbb{Z}$ ${\varphi }_r(m+n)=r(m+n)$

13.9 Example

One can also define integral on a certain interval as a homomorphism of function addition to real number addition. $\sigma :F\to \mathbb{R}$ $\sigma (f)={\int }_0^{1}f(x)dx$

13.10 Example: Reduction Modulo n

$\gamma :\mathbb{Z}\to {\mathbb{Z}}_n$ $\gamma (m)=r,\text{where r is the remainder given by the division when m/n}$ Here, it is a many-to-one map

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=133&selection=286,0,354,18&color=note|p.128]]

Composition of group homomorphisms is again a group homomorphism. That is, if φ : G → G′ and γ : G′ → G′′ are both group homomorphisms then their composition (γ ◦ φ) : G → G′′, where (γ ◦ φ)(g) = γ (φ(g)) for g ∈ G, is also a homomorphism. (See Exercise 49.)

12.12 Properties of Homomorphisms

let $\varphi$ be a homomorphism of a group $G$ into a group $G^{\prime }$

• If $e$ is the identity element in $G$, then $\phi (e)$ is the identity element $e^{\prime }$ in$G^{\prime }$.
• If $a\in G$, then $\phi (a^{-1})=\phi (a{)}^{-1}$.
• If H is a subgroup of $G$, then $\phi [H]$ is a subgroup of $G^{\prime }$.
• If $K^{\prime }$ is a subgroup of $G^{\prime }\cap \phi [G]$, then ${\phi }^{-1}[K^{\prime }]$ is a subgroup of $G$.

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=134&selection=57,0,61,56&color=note|p.129]]

Loosely speaking, φ preserves the identity element, inverses, and subgroups.

13.11 Definition Im

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=133&selection=395,0,511,1&color=note|p.128]]

Let $\varphi$ be a mapping of a set $X$ into a set $Y$ , and let $A\subseteq X$ and $B\subseteq Y$ .

• The image $\varphi [A]$ of $A$ in $Y$ under $\varphi$ is $\{\varphi (a)|a\in A\}$. Similar to Definition of Image of group under a function
• The set $\varphi [X]$ is the range of $\varphi$.
• The inverse image ${\varphi }^{-1}[B]$ of $B$ in $X$ is $\{x\in X|\varphi (x)\in B\}$.

13.13 Definition: Kernel

Let $\varphi :G\to G^{\prime }$ $Ker(\varphi )$ is the kernel of $\varphi$, $Ker(\varphi )=\{x\in G|\varphi (x)=e^{\prime }\}$ In other words: The Elements in G that correspond to the identity element in G’

Example: Linear Transformation Let $\varphi :{\mathbb{R}}^n\to {\mathbb{R}}^m$ and A be an $m\times n$ matrix of real numbers: $\varphi (\mathbf{v})=A\mathbf{v}$ $\varphi$ is a homomorphism, because $\varphi (v+w)=A(v+w)=Av+Aw=\varphi (v)+\varphi (w)$ This is known as a linear transformation. $Ker(\varphi )$ is known as the Null Space (Kernel of a Matrix) of A. It consists all $\mathbf{v}\in {\mathbb{R}}^n$ such that $A\mathbf{v}=0$ (the zero vector)

13.15 Theorem

let $H=Ker(\varphi )$, $a\in G$. Then: ${\varphi }^{-1}[\{\varphi (a)\}]=aH=Ha$In other words.

• A coset of $G$ that contains a collapse into an element $\varphi (a)$ after the homomorphism.
• For any $a\in G$, the full set of things that share the same image $\varphi (a)$ is the coset a$Ker(\varphi )$.

This figure shows how the Cosets collapse:

13.16 Example

Consider the group homomorphism

$$\pi :{\mathbb{C}}^{\ast }\to U=\{z\in \mathbb{C}:|z|=1\},\pi (z)=\frac{z}{|z|}.$$

Geometrically, this map projects each nonzero complex number radially onto the unit circle.

13.17 Example

• Let D be the additive group of all differentiable functions mapping $\mathbb{R}$ into $\mathbb{R}$
• Let F be the additive group of all functions mapping $\mathbb{R}$ into $\mathbb{R}$
• $\varphi :D\to F$
• $\varphi (f)=f^{\prime }$ $Ker(\varphi )$= all constant functions, which form a subgroup C of F.

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=136&selection=335,0,343,1&color=note|p.131]]

Thus Ker(φ) consists of all constant functions, which form a subgroup C of F

A better way is to say C is a subgroup of D, since $Ker(\varphi )$ is normally not defined in $G^{\prime }$

Let’s consider an element of F, $x^2$ and all the elements in D that collapse into $x^2$ we know one element is $\frac{x^3}{3}$ . According to 13.15 Theorem all such functions form the coset $\frac{x^3}{3}+C$ . which is exactly the integral of $x^2$ ${\varphi }^{-1}[\{\varphi (a)\}]=aH$ $\varphi (a)=x^2$ ${\varphi }^{-1}[\{x^2\}]=\frac{x^3}{3}+Ker(\varphi )$

This proves that:

NOTE

If two differentiable functions have the same derivative on a connected domain, they differ by a constant (and nothing else).

13.18 Corollary of 13.15

NOTE

$\varphi$ is a one-to-one map if and only if $Ker(\varphi )=\{e\}$

How to show $\varphi :G\to G^{\prime }$ is an Isomorphism

1. Show $\varphi$ is a homomorphism.
2. Show $Ker(\varphi )=e$
3. Show $\varphi$ maps G onto G’

Check	Criterion
Injective?	$Ker\varphi =e_G$
Surjective?	$Im\varphi =G^{\prime }$
Isomorphism?	Both conditions above hold (homomorphism + bijective)

Warning

A bijective homomorphism is an isomorphism in categories where bijective homomorphisms automatically have homomorphic inverses (e.g., groups, modules, vector spaces), but not in general (e.g., topology)

An Counterexample

IMPORTANT

Every fiber of a homomorphism is exactly one coset of the kernel

13.19 Definition normal

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=137&selection=182,0,210,1&color=note|p.132]]

A subgroup H of a group G is normal if its left and right cosets coincide, that is, if g H = Hg for all g ∈ G.

all abelian groups are normal. Trivial.

$Ker(\varphi )$ is always a normal subgroup of $G$

Ways to Prove a Subgroup $H$ Is Normal in $G$

A subgroup (H) is normal (denoted (H \triangleleft G)) if it satisfies any of the following equivalent conditions:

1. Conjugation Closure Prove directly that$\forall g\in G,\forall h\in H:gh{g}^{-1}\in H.$

• Proof Outline:

1. Take an arbitrary ($g\in G$) and ($h\in H$).

2. Compute $gh{g}^{-1}$ and use properties of $H$ (e.g., how generators behave) to rewrite it as an element of H.

3. Since this holds for all g and h, we conclude $gH{g}^{-1}=H$, so H is closed under conjugation by every element of G.

4. Left and Right Cosets Coincide Equivalently, $\forall g\in G:gH=Hg.$

• Proof Idea:
• To show $gH\subseteq Hg$, take any element $gh\in gH$. If H is conjugation-closed, then $gh=(gh{g}^{-1})g$ with $gh{g}^{-1}\in H$, so $gh\in Hg$.
• The reverse inclusion is similar. Hence $gH=Hg$ for all g, and that’s exactly the condition for normality.

3. Kernel of a Group Homomorphism If there exists a group homomorphism $\phi :G⟶K$such that $H=\ker (\phi ),$ then automatically $H◃G$.

• Reason: Every kernel of a homomorphism is a normal subgroup. See Summary Ways to Prove H is a Normal Subgroup - Summary