# Abstract Algebra/Group Theory/Homomorphism

## Contents

We are finally making our way into the meat of the theory. In this section we will study structure-preserving maps between groups. This study will open new doors and provide us with a multitude of new theorems.

Up until now we have studied groups at the "element level". Since we are now about to take a step back and study groups at the "homomorphism level", readers should expect a sudden increase in abstraction starting from this section. We will try to ease the reader into this increase by keeping one foot at the "element level" throughout this section.

From here on out the notation $e_G$ will denote the identity element in the group $G$ unless otherwise specified.

## Group homomorphisms

Definition 1: Let $(G,*)$ and $(H,\cdot)$ be groups. A homomorphism from $G$ to $H$ is a function $\phi\,:\, G\rightarrow H$ such that for all $g_1,g_2\in G$,

$\phi(g_1*g_2)=\phi(g_1)\cdot\phi(g_2)$.

Thus, a homomorphism preserves the group structure. We have included the multiplication symbols here to make explicit that multiplication on the left hand side occurs in $G$, and multiplication on the right hand side occurs in $H$.

Already we see that this section is different from the previous ones. Up until now we have, excluding subgroups, only dealt with one group at a time. No more! Let us start by deriving some elementary and immediate consequences of the definition.

Theorem 2: Let $G,H$ be groups and $\phi\,:\, G\rightarrow H$ a homomorphism. Then $\phi(e_G)=e_H$. In other words, the identity is mapped to the identity.

Proof: Let $g\in G$. Then, $\phi(g)=\phi(e_Gg)=\phi(e_G)\phi(g)$, implying that $\phi(e_G)$ is the identity in $H$, proving the theorem.

Theorem 3: Let $G,H$ be groups and $\phi\,:\, G\rightarrow H$ a homomorphism. Then for any $g\in G$, $\phi(g^{-1})=\phi(g)^{-1}$. In other words, inverses are mapped to inverses.

Proof: Let $g\in G$. Then $e_H=\phi(e_G)=\phi(gg^{-1})=\phi(g)\phi(g^{-1})$ implying that $\phi(g^{-1})=\phi(g)^{-1}$, as was to be shown.

Theorem 4: Let $G,G^\prime$ be groups, $\phi\,:\,G\rightarrow G^\prime$ a homomorphism and let $H$ be a subgroup of $G$. Then $\phi(H)=\{\phi(h)\mid h\in H\}$ is a subgroup of $G^\prime$.

Proof: Let $h_1,h_2\in H$. Then $\phi(h_1),\phi(h_2)\in \phi(H)$ and $\phi(h_1)\phi(h_2)^{-1}=\phi(h_1)\phi(h_2^{-1})=\phi(h_1h_2^{-1})$. Since $h_1h_2^{-1}\in H$, $\phi(h_1)\phi(h_2)^{-1}=\phi(h_1h_2^{-1})\in \phi(H)$, and so $\phi(H)$ is a subgroup of $G^\prime$.

Theorem 5: Let $G,G^\prime$ be groups, $\phi\,:\,G\rightarrow G^\prime$ a homomorphism and let $H^{\prime}$ be a subgroup of $G^\prime$. Then $\phi^{-1}(H^\prime)=\{g\in G\mid \phi(g)\in H^\prime\}$ is a subgroup of $G$.

Proof: Let $g_1,g_2\in \phi^{-1}(H^\prime)$. Then $\phi(g_1),\phi(g_2)\in H^\prime$, and since $H^\prime$ is a subgroup, $\phi(g_1)\phi(g_2)^{-1}=\phi(g_1)\phi(g_2^{-1})=\phi(g_1g_2^{-1})\in H^\prime$. But then, $g_1g_2^{-1}\in \phi^{-1}(H^\prime)$, and so $\phi^{-1}(H^\prime)$ is a subgroup of $G$.

From Theorem 4 and Theorem 5 we see that homomorphisms preserve subgroups. Thus we can expect to learn a lot about the subgroup structure of a group $G$ by finding suitable homomorphisms into $G$.

In particular, every homomorphism $\phi\,:\, G\rightarrow G^\prime$ has associated with it two important subgroups.

Definition 6: A homomoprhism is called an isomorphism if it is bijective and its inverse is a homomorphism. Two groups are called isomorphic if there exists an isomorphism between them, and we write $G\approx H$ to denote "$G$ is isomorphic to $H$".

Theorem 7: A bijective homomorphism is an isomorphism.

Proof: Let $G,H$ be groups and let $\phi\,:\,G\rightarrow H$ be a bijective homomorphism. We must show that the inverse $\phi^{-1}$ is a homomorphism. Let $h_1,h_2\in H$. then there exist unique $g_1,g_2\in G$ such that $\phi(g_1)=h_1$ and $\phi(g_2)=h_2$. Then we have $\phi(g_1g_2)=h_1h_2$ since $\phi$ is a homomorphism. Now apply $\phi^{-1}$ to all equations. We obtain $\phi^{-1}(h_1)=g_1$, $\phi^{-1}(h_2)=g_2$ and $\phi^{-1}(h_1h_2)=g_1g_2=\phi^{-1}(h_1)\phi^{-1}(h_2)$, so $\phi^{-1}$ is a homomorphism and thus $\phi$ is an isomorphism.

Definition 8: Let $G,G^\prime$ be groups. A homomorphism that maps every element in $G$ to $e^\prime\in G^\prime$ is called a trivial homomorphism (or zero homomorphism), and is denoted by $0_{GG^\prime}\,:\, G\rightarrow G^\prime$

Definition 9: Let $H$ be a subgroup of a group $G$. Then the homomorphism $i\,:\, H\rightarrow G$ given by $i(h)=h$ is called the inclusion of $H$ into $G$. Let $G^\prime$ be a group isomorphic to a subgroup $H$ of a group $G$. Then the isomorphism $\phi\,:\, G^\prime\rightarrow H$ induces an injective homomorphism $i^\prime\,:\, G^\prime\rightarrow G$ given by $i^\prime(g^\prime)=\phi(g^\prime)$, called an imbedding of $G^\prime$ into $G$. Obviously, $i^\prime=i\circ\phi$.

Definition 10: Let $G,G^\prime$ be groups and $\phi\,:\, G\rightarrow G^\prime$ a homomorphism. Then we define the following subgroups:

i) $\ker\,\phi = \{g\in G\mid \phi(g)=e^\prime\}\leq G$, called the kernel of $\phi$, and
ii) $\mathrm{im}\,\phi = \{g^\prime\in G^\prime\mid (\exists g\in G) \phi(g)=g^\prime\}\leq G^\prime$, called the image of $\phi$.

Theorem 11: The composition of homomorphisms is a homomorphism.

Proof: Let $G,H,K$ be groups and $\phi\,:\,G\rightarrow H$ and $\psi\,:\, H\rightarrow K$ homomorphisms. Then $\psi\circ\phi\,:\, G\rightarrow K$ is a function. We must show it is a homomorphism. Let $g,h\in G$. Then $\psi\circ\phi(gh)=\psi(\phi(gh))=\psi(\phi(g)\phi(h))=\psi(\phi(g))\psi(\phi(h))=\psi\circ\phi(g)\psi\circ\phi(h)$, so $\psi\circ\phi$ is indeed a homomorphisms.

Theorem 12: Composition of homomorphisms is associative.

Proof: This is evident since homomorphisms are functions, and composition of functions is associative.

Corollary 13: The composition of isomorphisms is an isomorphism.

Proof: This is evident from Theorem 11 and since the composition of bijections is a bijection.

Theorem 14: Let $G,H$ be groups and $\phi\,:\,G\rightarrow H$ a homomorphism. Then $\phi$ is injective if and only if $\ker\,\phi = \{e\}\subseteq G$.

Proof: Assume $\ker\,\phi=\{e\}$ and $g_1,g_2\in G$. Then $\phi(g_1)=\phi(g_2)\,\Leftrightarrow \phi(g_1)\phi(g_2)^{-1}=\phi(g_1g_2^{-1})=e_H$, implying that $g_1g_2^{-1}\in\ker\,\phi$. But by assumption then $g_1g_2^{-1}=e\,\Leftrightarrow g_1=g_2$, so $\phi$ is injective. Assume now that $\ker\,\phi\neq\{e\}$ and $g\in G$. Then there exists another element $k\in\ker\,\phi$ such that $k\neq e$. But then $\phi(g)=\phi(kg)$. Since both $g$ and $kg\neq g$ map to $\phi(g)=\phi(kg)$, $\phi$ is not injective, proving the theorem.

Corollary 15: Inclusions are injective.

Proof: The result is immediate. Since $i(h)=h$ for all $h\in H\leq G$, we have $i^{-1}(e)=\{e\}$.

The kernel can be seen to satisfy a universal property. The following theorem explains this, but it is unusually abstract for an elementary treatment of groups, and the reader should not worry if he/she cannot understand it immediately.

Commutative diagram showing the universal property of kernels.

Theorem 16: Let $G,G^\prime$ be groups and $\phi\,:\, G\rightarrow G^\prime$ a group homomorphism. Also let $H$ be a group and $\psi\,:\, H\rightarrow G$ a homomorphism such that $\phi\circ\psi=0_{HG^\prime}$. Also let $i\,:\,\ker\,\phi\rightarrow G$ is the inclusion of $\ker\,\phi$ into $G$. Then there exists a unique homomorphism $\bar{\psi}\,:\, H\rightarrow \ker\,\phi$ such that$\psi=i\circ \bar{\psi}$.

Proof: Since $\phi\circ\psi=0_{HG^\prime}$, by definition we must have $\psi(H)\leq\ker\,\phi$, so $\bar{\psi}$ exists. The commutativity $\psi=i\circ\bar{\psi}$ then forces $\bar{\psi}(h)=\psi(h)\in\ker\,\phi$, so $\bar{\psi}$ is unique.

Definition 17: A commutative diagram is a pictorial presentation of a network of functions. Commutativity means that when several routes of function composition from one object lead to the same destination, the two compositions are equal as functions. As an example, the commutative diagram on the right describes the situation in Theorem 16. In the commutative diagrams (or diagrams for short, we will not show diagrams which no not commute) shown in this chapter on groups, all functions are implicitly assumed to be group homomorphisms. Monomorphisms in diagrams are often emphasized by hooked arrows. In addition, epimorphims are often emphasized by double headed arrows. That an inclusion is a monomorphism will be proven shortly.

Remark 18: From the commutative diagram on the right, the kernel can be defined completely without reference to elements. Indeed, Theorem 16 would become the definition, and our Definition 10 i) would become a theorem. We will not entertain this line of thought in this book, but the advanced reader is welcome to work it out for him/herself.

## Automorphism Groups

In this subsection we will take a look at the homomorphisms from a group to itself.

Definition 19: A homomorphism from a group $G$ to itself is called an endomorphism of $G$. An endomorphism which is also an isomorphism is called an automorphism. The set of all endomorphisms of $G$ is denoted $\mathrm{End}(G)$, while the set of all automorphisms of $G$ is denoted $\mathrm{Aut}(G)$.

Theorem 20: $\mathrm{End}(G)$ is a monoid under composition of homomorphisms. Also, $\mathrm{Aut}(G)$ is a submonoid which is also a group.

Proof: We only have to confirm that $\mathrm{End}(G)$ is closed and has an identity, which we know is true. For $\mathrm{Aut}(G)$, the identity homomorphism $\mathrm{id}_G\,:\,G\rightarrow G$ is an isomorphism and the composition of isomorphisms is an isomorphism. Thus $\mathrm{Aut}(G)$ is a submonoid. To show it is a group, note that the inverse of an automorphism is an automorphism, so $\mathrm{Aut}(G)$ is indeed a group.

## Groups with Operators

An endomorphism of a group can be thought of as a unary operator on that group. This motivates the following definition:

Definition 21: Let $G$ be a group and $\Omega\subset\mathrm{End}(G)$. Then the pair $(G,\Omega)$ is called a group with operators. $\Omega$ is called the operator domain and its elements are called the homotheties of $G$. For any $\omega\in\Omega$, we introduce the shorthand $\omega(g)=g^\omega$ for all $g\in G$. Thus the fact that the homotheties of $G$ are endomorphisms can be expressed thus: for all $a,b\in G$ and $\omega\in\Omega$, $(ab)^\omega=a^\omega b^\omega$.

Example 22: For any group $G$, the pair $(G,\emptyset)$ is trivially a group with operators.

Lemma 23: Let $(G,\Omega)$ be a group with operators. Then $\Omega$ can be extended to a submonoid $\Omega^\prime$ of $\mathrm{End}(G)$ such that the structure of $(G,\Omega^\prime)$ is identical to $(G,\Omega)$.

Proof: Let $\Omega^\prime$ include the identity endomorphism and let $\Omega$ be a generating set. Then $\Omega^\prime$ is closed under compositions and is a monoid. Since any element of $\Omega^\prime$ is expressible as a (possibly empty) composition of elements in $\Omega$, the structures are identical.

In the following, we assume that the operator domain is always a monoid. If it is not, we can extend it to one by Lemma 23.

Definition 24: Let $(G,\Omega)$ and $(H,\Omega)$ be groups with operators with the same operator domain. Then a homomorphism $\phi\,:\,(G,\Omega)\rightarrow (H,\Omega)$ is a group homomorphism $\phi\,:\, G\rightarrow H$ such that for all $\omega\in\Omega$ and $g\in G$, we have $\phi(g^\omega)=\phi(g)^\omega$.

Definition 25: Let $(G,\Omega)$ be a group with operators and $H$ a subgroup of $G$. Then $H$ is called a stable subgroup (or a $\Omega$-invariant subgroup) if for all $h\in H$ and $\omega\in\Omega$, $h^\omega\in H$. We say that $H$ respects the homotheties of $G$. In this case $(H,\Omega)$ is a sub-group with operators.

Example 26: Let $V$ be a vector space over the field $F$. If we by $V_+$ denote the underlying abelian group under addition, then $V=(V_+,F)$ is a group with operators, where for any $k\in F$ and $v\in V_+$, we define $v^k=kv$. Then the stable subgroups are precisely the linear subspaces of $V$ (show this).

## Problems

Problem 1: Show that there is no nontrivial homomorphism from $\mathbb{Z}_5$ to $S_3$.