homomorphism

Group Homomorphism

Let $(G,\circ ),(H,\star )$ be groups. A function $\varphi :G\to H$ is a homomorphism if it preserves the group operation:

$$\varphi (x\circ y)=\varphi (x)\star \varphi (y)\text{for all }x,y\in G$$

When clear from context, we often write both operations as $\circ$ or drop the symbol entirely.

Homomorphisms preserve structure

$$\begin{array}{ccc}\circ & \bar{0}& \bar{1}\\ \bar{0}& \bar{0}& \bar{1}\\ \bar{1}& \bar{1}& \bar{0}\end{array}\varphi :\{\begin{array}{l}\bar{0}\mapsto x\\ \bar{1}\mapsto y\end{array}\begin{array}{ccc}\star & x& y\\ x& x& y\\ y& y& x\end{array}$$

Applying $\varphi$ after $\circ$ gives the same result as applying $\varphi$ to each element and then using $\star$.
For example: $\varphi (\bar{0}\circ \bar{1})=\varphi (\bar{1})=y$ and $\varphi (\bar{0})\star \varphi (\bar{1})=x\star y=y$

Properties of Homomorphisms

For any homomorphism $\varphi :(G,\circ )\to (H,\star )$:

The identity $e_G$ maps to the identity $e_H:$ $\varphi (e_G)=e_H$
Proof: $\varphi (e_G)=\varphi (e_G\circ e_G)=\varphi (e_G)\star \varphi (e_G)⟹\varphi (e_G)\star \cancel{\varphi (e_G)\star \varphi (e_G{)}^{-1}}=e_H$

Inverse of $G$ maps to inverse of $H$: $\forall a\in G:\varphi (a^{-1})=\varphi (a{)}^{-1}$

Kernel and Image

For a homomorphism $\varphi :G\to H$, the kernel and image are defined as:

$$\begin{array}{rl}\ker (\varphi )& :=\{g\in G\mid \varphi (g)=e_H\}={\varphi }^{-1}(\{e_H\})\\ \text{im}(\varphi )& :=\{\varphi (g)\in H\mid g\in G\}=\varphi (G)\end{array}$$

The kernel contains all elements that map to the identity in $H$, while the image contains all elements in $H$ that are “reached” by $\varphi$.
Note: The notation ${\varphi }^{-1}(\{e_H\})$ doesn’t mean inverse, but preimage, i.e. the set of all elements “inputs” that map to $e_H$.

Normal subgroup

For a group $(G,\circ )$ and subgroup $U\le G$, the following are equivalent:

1. $U$ is normal: $U⊴G$
2. $\forall a\in G:a^{-1}\circ U\circ a\subseteq U$ ($U$ is closed under conjugation)
3. $\forall a\in G$: $a\circ U=U\circ a$ (left and right cosets form the same partition, i.e. are equal)

For a commutative group, e.g. abelian groups, this is trivially satisfied for any subgroup → all subgroups are normal.

Proof: We show $(2)⟺(3)$: $x\in U⟹a^{-1}\circ x\circ a=y\in U$ “$\Rightarrow$”: (“von links”) $x\circ a=a\circ y$ $x\circ a\in U\circ a\underset{(U⊴G)}{\underbrace{=a\circ U}}⟹\exists z\in U:x\circ a=a\circ z⟹a\circ z=a\circ y⟹z=y\in U$ “$\Leftarrow$”: (“von rechts”) $\forall a\in G:a^{-1}\circ U\circ a\subseteq U,x\in U⟹a\circ x\circ a^{-1}\in U$ (Voraussetzung) $⟹x\circ a\in a\circ U$ $⟹U\circ a\subseteq a\circ U$ And $a\circ x\circ a^{-1}$ $⟹a\circ x\in U\circ a$ $⟹a\circ U\subseteq U\circ a$ → $U⊴G$

Link to original

$\varphi :G\to H$ homomorphism, $\varphi (G)\le H$ and $\ker (\varphi )⊴G$

If we have a homomorphism from $G$ to $H$, then the image of $G$ under $\varphi$ is a subgroup of $H$ and the kernel of $\varphi$ is a normal subgroup of $G$.

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Proof: First show $\ker (\varphi )⊴G$:

1. $\ker (\varphi )$ is non-empty since $e_G\in \ker (\varphi )$
2. $\forall a,b\in \ker (\varphi )$: $\varphi (a\circ b)=\varphi (a)\star \varphi (b)=e_H\star e_H=e_H$ so $a\circ b\in \ker (\varphi )$
3. $\forall a\in \ker (\varphi )$: $\varphi (a^{-1})=\varphi (a{)}^{-1}=e_H^{-1}=e_H$ so $a^{-1}\in \ker (\varphi )$
4. $\forall x\in \ker (\varphi ),a\in G$: $\varphi (a^{-1}\circ x\circ a)=\varphi (a^{-1})\star \varphi (x)\star \varphi (a)=\varphi (a{)}^{-1}\star e_H\star \varphi (a)=e_H$ so $a^{-1}xa\in \ker (\varphi )$

Properties 1-3 show $\ker (\varphi )$ is a subgroup, and property 4 shows it’s normal.

For $\varphi (G)\le H$, similar subgroup test shows it’s closed under operations and contains inverses.

Visual intuition for the above: Imagine $\varphi$ as a "filter" that maps elements from $G$ to $H$. The kernel is the set of elements that "disappear" when passing through the filter, while the image is the set of elements that "make it through". The filter however is "well-behaved" in the sense that it preserves the group structure.

In this example, we have an injective homomorphism that isn’t surjective, embedding a group into a biger group. It is also an isomorphism to a subgroup of the codomain (target group).
(source)
This is a surjective homomorphism that isn’t injective, collapsing elements into a smaller group, i.e. the image is all of the codomain, but the kernel is non-trivial (contains elements other than the identity identity … that map to the identity).

The big picture is not an isomorphism, beceause the mapping is not bijective (not the same number of elements in the domain and codomain). However, there is an isomorphism between the colors in the domain and codomain, which are the cosets of the domain, as the theorem below shows .

First Isomorphism Theorem

For any homomorphism $\varphi :G\to H$:

$$G/\ker (\varphi )\cong \text{im}(\varphi )$$

The set of all cosets of the kernel in the group $G$ is isomorphic to the image of $\varphi$.

This shows that after “collapsing/factoring out” elements that map to the identity (the kernel), the factor group structure exactly matches the image.

Proof $\psi :G/\ker (\varphi )\to \varphi (G)$ is an isomorphism We need to show that

1. $\psi$ is well-defined: $\psi (a\ker (\varphi ))=\varphi (a)$ (can’t have elements from the same coset $G$ point to different elements in $H$)
2. $\psi$ is surjective
3. $\psi$ is injective
4. $\psi$ is a homomorphism

Ad 1: We need to show that if $a\circ \ker \varphi =b\circ \ker \varphi$, then $\psi (a\circ \ker \varphi )=\varphi (a)$ and $\psi (b\circ \ker \varphi )=\varphi (b)$ are the same.
This would mean that there is an $x\in a\circ \ker \varphi ⟹\exists k_1,k_2\in \ker \varphi :x=a\circ k_1=b\circ k_2$
$\varphi (a)=\varphi (a)\star e_H=\varphi (a)\star \varphi (k_1)=\varphi (a\circ k_1)=\varphi (b\circ k_2)=\varphi (b)\star \varphi (k_2)=\varphi (b)\star e_H=\varphi (b)\boxed{}$
Note: $k_1,k_2$ when mapped become the identity in $H$ as they are in the kernel.

Ad 2: Surjectivity is clear as $\psi$ maps all cosets to elements in the image: $x\in \varphi (G):\exists a\in G:\varphi (a)=x⟹\psi (a\circ \ker \varphi )=\varphi (a)=x$
Note: $\psi (a\circ \ker \varphi )=\varphi (a)$ is the definition of $\psi$: $\psi$ maps cosets to elements in the image!

Ad 3: For injectivity we want to show that: $\psi (a\circ \ker \varphi )=\psi (b\circ \ker \varphi )⟺\varphi (a)=\varphi (b)$
In words: The $\psi$-image of a coset is the same as the $\varphi$-image of the representative of that coset (representative = any element in the coset).
$⟹\varphi (a\circ b^{-1})=\varphi (a)\star \varphi (b{)}^{-1}=e_H⟹a\circ b^{-1}\in \ker \varphi ⟹a\in b\circ \ker \varphi ⟹a\circ \ker \varphi =b\circ \ker \varphi$
Note: $a\circ b^{-1}=x\in \ker \varphi ⟹a=x\circ b\in \ker \varphi \circ b⟹a\in \ker \varphi \circ b$ (and we can wrhite $b\circ$ – from the left – since the kernel is normal)
→ If the images are the same the preimages are the same.

Ad 4: $\psi (\underset{(a\circ b)\circ \ker \varphi }{\underbrace{(a\circ \ker \varphi )\circ (b\circ \ker \varphi )}})=\varphi (a\circ b)=\varphi (a)\star \varphi (b)=\psi (a\circ \ker \varphi )\star \psi (b\circ \ker \varphi )\boxed{}$

EXAMPLE

Same colors map to the same colors (homomorphism – we can map first and then calculate or first map and then calculate), but not all colors are reached (not surjective – this doesn’t work in the other direction).

The different colors represent different cosets of the kernel! And we can calculate within this factor group.

Here, any green element times any blue element is red, which we can express as $(a\circ b)\ker (\varphi )$.
Now, applying the first isomism theorem, we map the red coset to the red element in the domain, the green coset to the green element, and the blue coset to the blue element:

$$\psi (\ker \phi )=n,\psi (a\circ \ker \phi )=u,\psi (b\circ \ker \phi )=v$$

then we have an isomorphism $\psi$ between the factor group and the image:

$$\psi :G/\ker (\phi )\to \text{im}(\phi )$$

So calculating with colors in the codomain is isomorth to calculating with cosets in the domain.

The First Isomorphism Theorem with ${\mathbb{Z}}_4$

Consider $\varphi :(\mathbb{Z},+)\to ({\mathbb{Z}}_4,+)$ mapping integers to their remainder mod 4.
The kernel is $4\mathbb{Z}$ (multiples of 4).
The quotient group $\mathbb{Z}/4\mathbb{Z}$ consists of cosets: $\{\bar{0},\bar{1},\bar{2},\bar{3}\}$
The image is all of ${\mathbb{Z}}_4$ which is also $\{\bar{0},\bar{1},\bar{2},\bar{3}\}$
The theorem tells us $\mathbb{Z}/4\mathbb{Z}\cong {\mathbb{Z}}_4$, i.e. arithmetic modulo 4 is the same as working with remainders

The kernel measures how much the homomorphism "collapses" the structure - a trivial kernel ( $\ker (\varphi )=\{e_G\}$) means no collapsing occurs and distinct elements remain distinct (injection). The image tells us how much of $H$ is "covered" by the homomorphism - if $\text{im}(\varphi )=H$, then every element in $H$ is reached (surjection).

Special Types of Homomorphisms

An injective homomorphism is called a monomorphism
A surjective homomorphism is called an epimorphism
A bijective homomorphism is called an isomorphism, denoted $G\cong H$
A homomorphism from a group to itself is called an endomorphism
A bijective endomorphism is called an automorphism

Group homomorphism between congruence classes and roots of unity

Consider the groups $G=(\mathbb{Z},+)$ and $H=\{{\zeta }_m,{\zeta }_m^2,\dots ,{\zeta }_m^{m-1},{\zeta }_m^m=1\}$ where ${\zeta }_m=\sqrt[m]{1}=e^{2\pi i/m}$ is a primitive $m^{\text{th}}$ root of unity (i.e. the first one that’s not just $1$).

Define the map:

$$\begin{array}{rl}\varphi :\mathbb{Z}& \to H\\ n& \mapsto {\zeta }_m^n=e^{2\pi i(n/m)}\end{array}$$

To verify this is a homomorphism, we check if it preserves the group operation (addition in $\mathbb{Z}$ maps to multiplication in $H$):

$$\varphi (l+n)={\zeta }_m^{l+n}={\zeta }_m^l\cdot {\zeta }_m^n=\varphi (l)\varphi (n)$$

This is a surjective homomorphism since it covers all of $H$.
To find the kernel, we solve for all $x$ where:

$$\begin{array}{rl}\ker (\varphi )& =\{x\in \mathbb{Z}\mid \varphi (x)=1\}\\ & =\{x\in \mathbb{Z}\mid {\zeta }_m^x=1\}\\ & =\{x\in \mathbb{Z}\mid e^{2\pi i(x/m)}=1\}\\ & =m\mathbb{Z}\end{array}$$

The last step follows because $e^{2\pi i(x/m)}=1$ exactly when $x/m$ is an integer.

Applying the First Isomorphism Theorem:

$$\mathbb{Z}/m\mathbb{Z}={\mathbb{Z}}_m\cong H$$

Both ${\mathbb{Z}}_m$ and $H$ are cyclic groups of order $m$: In ${\mathbb{Z}}_m$ we add 1 repeatedly modulo $m$, while in $H$ we multiply by ${\zeta }_m$ repeatedly until returning to 1 after $m$ steps.