Group Homomorphisms

For the time being, we explore properties of groups. Thus, we will simply refer to group homomorphisms between groups as homomorphisms.

From §Algebraic Structures, ⁋Definition 6, we can also define (group) isomorphisms, and from this definition and [Set Theory] §Operations on Functions, ⁋Proposition 4, it is clear that any isomorphism must be a bijection. In many cases, the converse also holds.

Proposition 1 A magma homomorphism \(f:A\rightarrow A'\) is an isomorphism if and only if \(f\) is a bijection.

If \(A\) has an identity element \(e\) and \(f:A\rightarrow A'\) is a bijection, then \(f(e)\) is the identity element of \(A'\), and consequently \(f^{-1}\) is a magma homomorphism that sends the identity element of \(A'\) to the identity element of \(A\).

Proof

It suffices to show the reverse direction. Since \(f\) is a bijection, there exists an inverse function \(f^{-1}:G'\rightarrow G\) as functions. If \(f^{-1}\) is a homomorphism, then by definition \(f\) is an isomorphism.

Choose arbitrary \(y, y'\in A'\). Since \(f\) is a bijection, there exist unique \(x\) and \(x'\) such that \(f(x)=y\) and \(f(x')=y'\). Now

\[f^{-1}(yy')=f^{-1}(f(x)f(x'))=f^{-1}(f(xx'))=xx'=f^{-1}(y)f^{-1}(y')\]

so \(f^{-1}\) is a homomorphism, and therefore \(f\) is an isomorphism.

On the other hand, if \(f:A\rightarrow A'\) is a bijection, then for any \(y\in A'\), there exists a unique \(x\in A\) such that \(f(x)=y\). Now

\[y=f(x)=f(xe)=f(x)f(e),\qquad y=f(x)=f(ex)=f(e)f(x)\]

so \(f(e)\) is the identity element of \(A'\).

Equalizer of Homomorphisms

The following holds.

Proposition 2 Let group homomorphisms \(f,g:G \rightarrow H\) be given. Then

\[\Eq(f,g)=\{x\in G\mid f(x)=g(x)\}\]

is a subgroup of \(G\).

Proof

If \(x,y\in \Eq(f,g)\), then

\[f(xy^{-1})=f(x)f(y)^{-1}=g(x)g(y)^{-1}=g(xy^{-1})\]

so \(xy^{-1}\in\Eq(f,g)\). Therefore, by §Semigroups, Monoids, Groups, ⁋Proposition 15, we obtain the desired result.

The \(i:\Eq(f,g)\rightarrow G\) defined in this way has the following property.

If a group homomorphism \(j:G' \rightarrow G\) satisfies \(f\circ j=g\circ j\), then there exists a unique homomorphism \(j': G' \rightarrow G\) such that \(i\circ j'=j\).

This is because the image of \(j\) is contained in \(\Eq(f,g)\) by definition. Thus every morphism in \(\Grp\) has an equalizer. ([Category Theory] §Limits, ⁋Example 7) In fact, every morphism in \(\Grp\) also has a coequalizer, but to define this, we first need to define normal subgroups and quotient groups.

Kernel and Image of Homomorphisms

The group \(\{e\}\) is a zero object in the category \(\Grp\). Therefore, for any groups \(G,H\), the zero map \(e:G \rightarrow H\) is defined as the composition \(G\rightarrow\{e\}\rightarrow H\).

Meanwhile, whether a group homomorphism \(f\) is injective can be expressed as follows.

Proposition 3 A homomorphism \(f:G\rightarrow G'\) is injective if and only if \(f^{-1}(e')=\{e\}\).

Proof

If \(f\) is injective, then \(f^{-1}(e')=\{e\}\) is obvious.

Conversely, suppose \(f^{-1}(e')=\{e\}\). Let \(x,y\in G\) satisfying \(f(x)=f(y)\) be given. Then

\[e'=f(x)f(y)^{-1}=f(xy^{-1})\]

and by assumption, \(xy^{-1}=e\). From this, we conclude \(x=y\).

For any homomorphism \(f:G\rightarrow G'\), the set \(f^{-1}(e')\) above shows how far \(f\) is from being injective. This set is named as follows.

Definition 4 The kernel of a homomorphism \(f:G\rightarrow G'\) is defined as the set \(f^{-1}(e')\) and denoted by \(\ker f\).

Then \(f^{-1}(e')\) is not merely a set but becomes a subgroup of \(G\).

Proposition 5 For any homomorphism \(f:G\rightarrow G'\), \(\ker f\) is a subgroup of \(G\).

Proof

By definition, \(\ker f=\Eq(f,e)\).

On the other hand, we verified that for any magma homomorphism \(f:A\rightarrow A'\), its image \(\im f\) is a submagma of \(A'\). (See the calculation before §Algebraic Structures, ⁋Definition 8) However, a submagma of a group is not necessarily a subgroup in general, so the following proposition must be proved separately.

Proposition 6 For any homomorphism \(f:G\rightarrow G'\), \(\im f\) is a subgroup of \(G'\).

Proof

We already know that \(\im f\) is a submagma of \(G'\), so by §Semigroups, Monoids, Groups, ⁋Proposition 15, it suffices to show that \(\im f\) is closed under taking inverses. Let \(y\in\im f\) and suppose \(x\in G\) satisfies \(f(x)=y\). Then

\[f(x^{-1})=f(x)^{-1}=y^{-1}\]

shows that \(y^{-1}\in\im f\).

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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