Group Homomorphisms

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

For now, we investigate the properties of groups. Therefore, a group homomorphism between groups will simply be called a homomorphism.

From §Algebraic Structures, ⁋Definition 6, a (group) isomorphism can also be defined, and from this definition and [Set Theory] §Operations Between Functions, ⁋Proposition 5, it is obvious 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 bijective.

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

Proof

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

Choose arbitrary \(y, y'\in A'\). Then since \(f\) is bijective, there exist unique \(x\), \(x'\) 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 \(x\in A\) satisfying \(f(x)=y\). 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 by the argument immediately following §Semigroups, Monoids, and Groups, ⁋Definition 11, a monoid homomorphism between groups preserves inverses, so

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

and thus \(xy^{-1}\in\Eq(f,g)\). Also, after §Semigroups, Monoids, and Groups, ⁋Definition 11 we observed that a monoid homomorphism between groups preserves the identity element, and since the identity element of \(H\) is unique, \(\Eq(f,g)\) is nonempty. Therefore, by §Semigroups, Monoids, and Groups, ⁋Proposition 15, we obtain the desired result.

For this \(\Eq(f,g)\), the inclusion \(i:\Eq(f,g)\rightarrow G\) 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 \(j': G' \rightarrow \Eq(f,g)\) such that \(i\circ j'=j\).

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

Kernel and Image of a Homomorphism

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 composite \(G\rightarrow\{e\}\rightarrow H\).

On the other hand, the condition that 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

That \(f^{-1}(e')=\{e\}\) must hold if \(f\) is injective is obvious.

Conversely, assume that \(f^{-1}(e')=\{e\}\). Suppose \(x,y\in G\) satisfy \(f(x)=f(y)\). Then

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

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

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

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

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, when any magma homomorphism \(f:A\rightarrow A'\) is given, we have verified that its image \(\im f\) is a submagma of \(A'\). (calculation before §Algebraic Structures, ⁋Definition 8) However, since a submagma of a group need not be a subgroup in general, 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, and 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 from

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

we know 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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