Group theory/Homomorphism theorems/Examples/Section

Theorem

Let ${}G,Q$ and ${}H$ be groups, and let ${}\varphi \colon G\rightarrow H$ be a group homomorphism and ${}\psi \colon G\rightarrow Q$ a surjective group homomorphism. Suppose that

${}\operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,.$

ist. Then there exists a uniquely determined group homomorphism

${\tilde {\varphi }}\colon Q\longrightarrow H$

such that ${}\varphi ={\tilde {\varphi }}\circ \psi$ holds. Put differently, the diagram

${\begin{matrix}G&{\stackrel {\varphi }{\longrightarrow }}&H&\\\!\!\!\!\!\psi \downarrow &\nearrow {\tilde {\varphi }}\!\!\!\!\!&\\Q&&&&\!\!\!\!\!\!\!\!\\\end{matrix}}$

commutes.

Proof

We show firstly the uniqueness. For every Element ${}u\in Q$, there exists some ${}g\in G$ with ${}\psi (g)=u$. The commutativity of the diagram ensures that

${}{\tilde {\varphi }}(u)=\varphi (g)\,$

holds. This means that there exists at most one ${}{\tilde {\varphi }}$.

We have to show that this condition yields a well-defined mapping. Hence, let ${}g,g'\in G$ be two preimages of ${}u$. Then

${}\psi {\left(g'g^{-1}\right)}=uu^{-1}=e_{Q}\,;$

therefore, we have ${}g'g^{-1}\in \operatorname {kern} \psi \subseteq \operatorname {kern} \varphi$. Hence, ${}\varphi (g)=\varphi (g')$, and the mapping is well-defined. Let ${}u,v\in Q$ be given, and let ${}g,h\in G$ be preimages. Then ${}gh$ is a preimage of ${}uv$. Therefore,

${}{\tilde {\varphi }}(uv)=\varphi (gh)=\varphi (g)\varphi (h)={\tilde {\varphi }}(u){\tilde {\varphi }}(v)\,.$

This means that ${}{\tilde {\varphi }}$ is a group homomorphism.

$\Box$

Example

We consider the surjective group homomorphisms

$\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(4)$

and

$\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12).$

We have

${}\operatorname {kern} \psi =\mathbb {Z} \cdot 12\subseteq \mathbb {Z} \cdot 4=\operatorname {kern} \varphi \,.$

Due to fact, there exists a uniquely determined group homomorphism

${\tilde {\varphi }}\colon \mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(4),$

which is compatible with the remainder mappings. The morphism ${}{\tilde {\varphi }}$ sends the remainder of a number after division by ${}12$ to the remainder after division by ${}4$. In particular, the theorem implies that the second remainder does only depend on the first remainder, not on the number itself.

If, to the contrary, we consider

$\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(5)$

and

$\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12),$

then

${}\operatorname {kern} \psi =\mathbb {Z} \cdot 12\not \subseteq \mathbb {Z} \cdot 5=\operatorname {kern} \varphi \,,$

and there does not exists a natural mapping

$\mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(5).$

For example, the numbers ${}1,13,25,37,49$ have modulo ${}12$ the remainder ${}1$ but modulo ${}5$ their remainders are ${}1,3,0,2,4$.

The mapping constructed in this theorem is called induced mapping or induced homomorphism, and the theorem is called the theorem about the induced homomorphism.

Corollary

Let ${}G$ and ${}H$ be groups, and let

$\varphi \colon G\longrightarrow H$

be a surjective group homomorphism. Then there exists a canonical isomorphism

${\tilde {\varphi }}\colon G/\operatorname {kern} \varphi \longrightarrow H.$

Proof

We apply fact to ${}Q=G/\operatorname {kern} \varphi$ and the canonical projection ${}q\colon G\rightarrow G/\operatorname {kern} \varphi$. This induces a group homomorphism

${\tilde {\varphi }}\colon G/\operatorname {kern} \varphi \longrightarrow H$

fulfilling ${}\varphi ={\tilde {\varphi }}\circ q$, which is surjective. Let ${}[x]\in G/\operatorname {kern} \varphi$ and ${}[x]\in \operatorname {kern} {\tilde {\varphi }}$. Then

${}{\tilde {\varphi }}([x])=\varphi (x)=e_{H}\,.$

Therefore, ${}x\in \operatorname {kern} \varphi$. Hence, ${}[x]=e_{Q}$, that is, the kernel of ${}{\tilde {\varphi }}$ is trivial, and so, due to fact, ${}{\tilde {\varphi }}$ is also injective.

$\Box$

Example

Let ${}G$ be a cyclic group with a generator ${}g$. We consider the corresponding group homomorphism

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

in the sense of fact. Since we have a generator, this mapping is surjective. The kernel of this mapping is determined the order of ${}g$, which we denote by ${}k$ (or it is ${}0$, in case the order is ${}\infty$). Due to fact, there exists a canonical isomorphism

${\tilde {\varphi }}\colon \mathbb {Z} /(k)\longrightarrow G.$

In particular, there exists, up to isomorphism, for every ${}k$, exactly one cyclic group, namely ${}\mathbb {Z} /(k)$.

Example

The group homomorphism

$\mathbb {R} \longrightarrow S^{1},t\longmapsto {\begin{pmatrix}\cos t\\\sin t\end{pmatrix}},$

is surjective. Due to the periodicity of the trigonometric functions, the kernel equals ${}\mathbb {Z} 2\pi$. By the isomorphism theorem, there exists a canonical isomorphism

${}\mathbb {R} /\mathbb {Z} 2\pi \cong S^{1}\,.$

Example

The complex exponential function

$(\mathbb {C} ,0,+)\longrightarrow (\mathbb {C} ^{\times },1,\cdot ),z\longmapsto \exp z,$

is a surjective group homomorphism. The kernel is ${}\mathbb {Z} 2\pi {\mathrm {i} }$. Due to the isomorphism theorem, there exists a canonical isomorphism

${}\mathbb {C} /\mathbb {Z} 2\pi {\mathrm {i} }\cong \mathbb {C} ^{\times }\,.$

Example

The determinant

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

is a surjective group homomorphism; its kernel is by definition the special linear group ${}\operatorname {SL} _{n}\!{\left(K\right)}$. Due to the isomorphism theorem, there exists a canonical isomorphism

${}\operatorname {GL} _{n}\!{\left(K\right)}/\operatorname {SL} _{n}\!{\left(K\right)}\cong K^{\times }\,.$

Theorem

Let ${}G$ and ${}H$ be groups, and let

$\varphi \colon G\longrightarrow H$

be a group homomorphism. Then there exists a canonical factorization

$G{\stackrel {q}{\longrightarrow }}G/\operatorname {kern} \varphi {\stackrel {\theta }{\longrightarrow }}\operatorname {Im} \varphi {\stackrel {\iota }{\hookrightarrow }}H,$

where ${}q$ is the canonical projection, ${}\theta$ is a group isomorphism, and ${}\iota$ is the canonical inclusion of the

image group.

Proof

This follows from fact, applied to the image group ${}U=\operatorname {Im} \varphi \subseteq H$.

$\Box$

This statement is often briefly expressed by saying:

image ${}=$ preimage modulo kernel.