Group homomorphism/Introduction/Section

Definition

Let ${}(G,\circ ,e_{G})$ and ${}(H,\circ ,e_{H})$ denote groups. A mapping

$\psi \colon G\longrightarrow H$

is called group homomorphism, if the equality

${}\psi (g\circ g')=\psi (g)\circ \psi (g')\,$

holds for all

${}g,g'\in G$.

The set of all group homomorphisms from ${}G$ to ${}H$ is denoted by

$\operatorname {Hom} _{}{\left(G,H\right)}.$

Linear mappings between vector spaces are in particular group homomorphisms. The following two lemmas follow directly from the definition.

Lemma

Let ${}G$ and ${}H$ denote groups, and let ${}\varphi \colon G\rightarrow H$ be a group homomorphism. Then ${}\varphi (e_{G})=e_{H}$ and ${}(\varphi (g))^{-1}=\varphi {\left(g^{-1}\right)}$ for every

${}g\in G$.

Proof

To prove the first statement, consider

${}\varphi (e_{G})=\varphi (e_{G}e_{G})=\varphi (e_{G})\varphi (e_{G})\,.$

Multiplication with ${}\varphi (e_{G})^{-1}$ yields ${}e_{H}=\varphi (e_{G})$.
To prove the second claim, we use

${}\varphi {\left(g^{-1}\right)}\varphi (g)=\varphi {\left(g^{-1}g\right)}=\varphi (e_{G})=e_{H}\,.$

This means that ${}\varphi {\left(g^{-1}\right)}$ has the property that characterizes the inverse element of ${}\varphi (g)$. Since the inverse element in a group is, due to fact, uniquely determined, we must have ${}\varphi {\left(g^{-1}\right)}=(\varphi (g))^{-1}$.

$\Box$

Lemma

Let ${}F,G,H$ denote

groups. Then the following properties hold.

1. The identity

   $\operatorname {Id} \colon G\longrightarrow G$

   is a group homomorphism.

2. If ${}\varphi :F\rightarrow G$ and ${}\psi :G\rightarrow H$ are group homomorphisms, then the composition ${}\psi \circ \varphi \colon F\rightarrow H$ is a group homomorphism.
3. For a subgroup ${}F\subseteq G$, the inclusion ${}F\hookrightarrow G$ is a group homomorphism.
4. Let ${}\{e\}$ be the trivial group. Then the mapping ${}\{e\}\rightarrow G$ that sends ${}e$ to ${}e_{G}$ is a group homomorphism. Moreover, the (constant) mapping ${}G\rightarrow \{e\}$ is a group homomorphism.

Proof

This is trivial.

$\Box$

Example

Let ${}d\in \mathbb {N}$ be fixed. The mapping

$\mathbb {Z} \longrightarrow \mathbb {Z} ,n\longmapsto dn,$

is a group homomorphism. This follows immediately from the distributive law. For ${}d\geq 1$, this mapping is injective, and the image is the subgroup ${}\mathbb {Z} d\subseteq \mathbb {Z}$. For ${}d=0$, we have the zero mapping. For ${}d=1$, the mapping is the identity. For ${}d\geq 2$, the mapping is not surjective.

Example

Let ${}d\in \mathbb {N} _{+}$. We consider the set

${}\mathbb {Z} /(d)=\{0,1,\ldots ,d-1\}\,,$

together with the addition described in exercise, which makes it a group. The mapping

$\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(d)$

that sends an integer number ${}n$ to its remainder after division by ${}d$ is a group homomorphism. For, if ${}m=ad+r$ and ${}n=bd+s$ are given with ${}0\leq r,s<d$, then

${}m+n=(a+b)d+r+s\,.$

Here, it may happen that ${}r+s\geq d$. In this case,

${}\varphi (m+n)=r+s-d\,,$

and this coincides with the addition of ${}r$ and ${}s$ in ${}\mathbb {Z} /(d)$. This mapping is surjective, but not injective.

Example

For a field ${}K$ and ${}n\in \mathbb {N} _{+}$, the determinant

$\det \colon \operatorname {GL} _{n}\!{\left(K\right)}\longrightarrow K^{\times },M\longmapsto \det M,$

is a group homomorphism. This follows from the multiplication theorem for the determinant and fact.

Example

The assignment

$S_{n}\longrightarrow \{1,-1\},\pi \longmapsto \operatorname {sgn} (\pi ),$

where ${}S_{n}$ denotes the permutation group for ${}n$ elements, is a group homomorphism, due to fact.

Lemma

Let ${}G$ denote a group. Then there is a correspondence between group elements ${}g\in G$ and group homomorphisms ${}\varphi$ from ${}\mathbb {Z}$ to ${}G$, given by

$g\longmapsto (n\mapsto g^{n}){\text{ and }}\varphi \longmapsto \varphi (1).$

Proof

Let ${}g\in G$ be fixed. That the mapping

$\varphi _{g}\colon \mathbb {Z} \longrightarrow G,n\longmapsto g^{n},$

is a group homomorphism, is just a reformulation of the exponential laws. Because of ${}\varphi _{g}(1)=g^{1}=g$, we obtain from the power mapping ${}\varphi _{g}$ the group element back. Moreover, a group homomorphism ${}\varphi \colon \mathbb {Z} \rightarrow G$ is uniquely determined by ${}\varphi (1)$, as ${}\varphi (n)=(\varphi (1))^{n}$ for ${}n$ positive, and ${}\varphi (n)={\left((\varphi (1))^{-1}\right)}^{-n}$ for ${}n$ negative must hold.

$\Box$

This lemma can be stated quickly by saying ${}G\cong \operatorname {Hom} _{}{\left(\mathbb {Z} ,G\right)}$. It is more difficult to characterize the group homomorphisms from a group ${}G$ to ${}\mathbb {Z}$. The group homomorphisms from ${}\mathbb {Z}$ to ${}\mathbb {Z}$ are just the multiplications with a fixed integer number ${}a$, that is,

$\mathbb {Z} \longrightarrow \mathbb {Z} ,x\longmapsto ax.$