A **group** $G$ is a monoid,such that for every element $x∈G$ there exists an element $y∈G$ such that $xy=yx=e$.Such an element $y$ is called an **inverse** for $x$.Such an inverse is unique,because if $y'$ is also an inverse for $x$,then

We denote this inverse by $x^{-1}$ (or by $-x$ when the law of composition is written additively).

For any positive integer $n$,we let $x^{-n}=(x^{-1})^n$.Then the usual rules for exponentiation hold for all integers,not only for integers $≥ 0$(as we pointed out for monoids in $\S 1$).

In the definitions of unit elements and inverses,we could define left units and left inverses(in the obvious way).One can easily prove that these are also units and inverses respectively under suitable conditions.Namely:

Let $G$ be a set with an associative law of composition,let $e$ be $a$ left unit for that law,and assume that every element has a left inverse.Then $e$ is a unit,and each left inverse is also an inverse.In particular,$G$ is a group.

To prove this,let $a∈G$ and let $b∈G$ be such that $ba=e$.Then

$$bab=eb=b.$$Multiplying on the left by a left inverse for $b$ yields

$$ab=e,$$or in other words,$b$ is also a right inverse for $a$.One sees also that $a$ is a left inverse for $b$.Furthermore,

$$ae=aba=ea=a,$$whence $e$ is a right unit.

**Example.** Let $G$ be a group and $S$ a nonempty set.The set of maps $M(S,G)$ is itself a group; namely for two maps $f,g$ of $S$ into $G$ we define $fg$ to be the map such that

and we define $f^{-1}$ to be the map such that $f^{-1}(x)=f(x)^{-1}$.It is then trivial to verify that $M(S,G)$ is a group.If $G$ is commutative,so is $M(S,G)$,and when the law of composition in $G$ is written additively,so is the law of composition in $M(S,G)$,so that we would write $f+g$ instead of $fg$,and $-f$ instead of $f^{-1}$.

**Example.** Let $S$ be a non-empty set.Let $G$ be the set of bijective mappings of $S$ onto itself.Then $G$ is a group,the law of composition being ordinary composition of mappings.The unit element of $G$ is the identity of $S$,and the other group properties are trivially verified.The elements of $G$ are called **permutations** of $S$.We also denote $G$ by Perm($S$).For more information on Perm($S$) when $S$ is finite,see $\S 5$ below.

**Example.** Let us assume here the basic notions of linear algebra.Let $k$ be a field and $V$ a vector space over $k$.Let $GL(V)$ denote the set of invertible $k-$ linear maps of $V$ onto itself.Then $GL(V)$ is a group under composition of mappings.Similarly,let $k$ be a field and let $GL(n,k)$ be the set of invertible $n×n$ matrices with components in $k$.Then $GL(n,k)$ is a group under the multiplication of matrices.For $n≥ 2$,this group is not commutative.

**Example. The group of automorphisms.** We recommend that the reader now refer immediately to $\S 11$,where the notion of a category is defined,and where several examples are given.For any object $A$ in a category,its automorphisms form a group denoted by $Aut(A)$.Permutations of a set and the linear automorphisms of a vector space are merely examples of this more general structure.

**Example.** The set of rational numbers forms a group under addition.The set of non-zero rational numbers forms a group under multiplication.Similar statements hold for the real and complex numbers.

**Example. Cyclic groups.** The integers $\bf Z$ form an additive group.A group is defined to be **cyclic** if there exists an element $a∈G$ such that every element of $G$(written multiplicatively)is of the form $a^n$ for some integer $n$.If $G$ is written additively,then every element of a cyclic group is of the form $na$.One calls $a$ **a cyclic generator**.Thus $\bf Z$ is an additive cyclic group with generator $1$,and also with generator $-1$.There are no other generators.Given a positive integer $n$,the $n$-th roots of unity in the complex numbers form a cyclic group of order $n$.In terms of the usual notation,$e^{\frac{2π i}{n}}$ is a generator for this group.So is $e^{\frac{2π ir}{n}}$ with $r∈\bf Z$ and $r$ prime to $n$.A generator for this group is called a **primitive** $n$-th root of unity.

**Example. The direct product.** Let $G_1,G_2$ be groups.Let $G_{1}×G_{2}$ be the direct product as sets,so $G_{1}×G_{2}$ is the set of all pairs $(x_{1},x_{2})$ with $x_{i}∈G_{i}$.We define the product componentwise by

Then $G_{1}×G_{2}$ is a group,whose unit element is $(e_{1},e_{2})$(where $e_{i}$ is the unit element of $G_{i}$).Similarly,for $n$ groups we define $G_1×⋯×G_n$ to be the set of $n-$tuples with $x_{i}∈G_{i}$($i=1,⋯,n$),and componentwise multiplication.Even more generally,let $I$ be a set,and for each $i∈I$,let $G_{i}$ be a group.Let $G=\prod\limits G_{i}$ be the set-theoretic product of the sets $G_{i}$.Then $G$ is the set of all families $(x_{i})_{i∈I}$ with $x_{i}∈G_{i}$.We can define a group structure on $G$ by componentwise multiplication,namely,if $(x_{i})_{i∈I}$ and $(y_{i})_{i∈I}$ are two elements of $G$,we define their product to be $(x_{i}y_{i})_{i∈I}$.We define the inverse of $(x_{i})_{i∈I}$ to be $(x_{i}^{-1})_{i∈I}$.It is then obvious that $G$ is a group called the **direct product** of the family.

Let $G$ be a group.A **subgroup** $H$ of $G$ is a subset of $G$ containing the unit element,and such that $H$ is closed under the law of composition and inverse(i.e. it is a submonoid,such that if $x∈H$ then $x^{-1}∈H$).A subgroup is called **trivial** if it consists of the unit element alone.The intersection of an arbitrary non-empty family of subgroups is a subgroup(trivial verification).

Let $G$ be a group and $S$ a subset of $G$.We shall say that $S$ **generates** $G$,or that $S$ is a set of **generators** for $G$,if every element of $G$ can be expressed as a product of element of $S$ or inverses of elements of $S$,i.e. as a product $x_{1}⋯x_{n}$ where each $x_{i}$ or $x_{i}^{-1}$ is in $S$.It is clear that the set of all such products is a subgroup of $G$(the empty product is the unit element),and is the smallest subgroup of $G$ containing $S$.Thus $S$ generates $G$ if and only the smallest subgroup of $G$ containing $S$ is $G$ itself.If $G$ is generated by $S$,then we write $G=⟨S⟩$.By definition,a cyclic group is a group which has one generator.Given elements $x_{1},⋯,x_{n}∈G$,these elements generate a subgroup $⟨x_{1},⋯,x_{n}⟩$,namely the set of all elements of $G$ of the form

A single element $x∈G$ generates a cyclic subgroup.

**Example.** There are two non-abelian groups of order 8. One is the **group of symmetries of the square**, generated by two elements $σ,τ$ such that

The other is the **quaternion group**,generated by two elements,$i,j$ such that if we put $k=ij$ and $m=i^2$,then

After you know enough facts about groups,you can easily do Exercise $35$.

Let $G,G'$ be monoids.A **monoid-homomorphism** (or simply **homomorphism**)of $G$ into $G'$ is a mapping $f:G→G'$ such that $f(xy)=f(x)f(y)$ for all $x,y∈G$,and mapping the element of $G$ into that of $G'$.If $G,G'$ are groups,a **group-homomorphism** of $G$ into $G'$ is simply a monoid-homomorphism.

We sometimes say: “Let $f:G→G'$ be a group-homomorphism” to mean: “Let $G,G'$ be groups,and let $f$ be a homomorphism from $G$ into $G'$.”

Let $f:G→G'$ be a group-homomorphism.Then

$$f(x^{-1})=f(x)^{-1}$$because if $e,e'$ are the unit elements of $G,G'$ respectively,then

$$e'=f(e)=f(xx^{-1})=f(x)f(x^{-1}).$$Furthermore,if $G,G'$ are groups and $f:G→G'$ is a map such that

$$f(xy)=f(x)f(y)$$for all $x,y$ in $G$,then $f(e)=e'$ because $f(ee)f(e)$ and also $=f(e)f(e)$.Multiplying by the inverse of $f(e)$ shows that $f(e)=e'$.

Let $G,G'$ be monoids.A homomorphism $f:G→G'$ is called an **isomorphism** if there exists a homomorphism $g:G'→G$ such that $f∘g$ and $g∘f$ are the identity mappings(in $G'$ and $G$ respectively).It is trivially verified that $f$ is an isomorphism if and only if $f$ is bijective.The existence of an isomorphism between two groups $G$ and $G'$ is sometimes denoted by $G≈G'$.If $G=G'$,we say that isomorphism is an **automorphism**.A homomorphism of $G$ into itself is also called an **endomorphism**.

**Example.** Let $G$ be a monoid and $x$ an element of $G$.Let $N$ denote the(additive)monoid of integers $≥ 0$.Then the map $f:N→G$ such that $f(n)=x^n$ is a homomorphism.If $G$ is a group,we can extend $f$ to a homomorphism of ℤ into $G$($x^n$ is defined for all $n∈ℤ$,as pointed out previously).

Let $n$ be a fixed integer and let $G$ be a commutative group.Then one verifies easily that the map

$$x↦x^n$$from $G$ into itself is a homomorphism.So is the map $x↦x^{-1}$.The map $x↦x^n$ is called the $n$-th **power map**.

**Example.** Let $I=\lbrace i\rbrace $ be an indexing set,and let $\lbrace G_{i}\rbrace $ be a family of groups.Let $G=\prod\limits G_{i}$ be their direct product.Let

be the projection on the $i$-th factor.Then $p_{i}$ is a homomorphism.

Let $G$ be a group,$S$ a set of generators for $G$,and $G'$ another group.Let $f:S→G'$ be a map.If there exists a homomorphism $\overline{f}$ of $G$ into $G'$ whose restriction to $S$ is $f$,then there is only one.

In other words,$f$ has at most one extension to a homomorphism of $G$ into $G'$.This is obvious,but will be used many times in the sequel.

Let $f:G→G'$ and $g:G'→G$ be two group-homomorphism.Then the composite map $g∘f$ is a group-homomorphism.If $f,g$ are isomorphisms then so is $g∘f$.Furthermore $f^{-1}:G'→G$ is also an isomorphism.In particular,the set of all automorphism of $G$ is itself a group,denoted by $Aut(G)$.

Let $f:G→G'$ be a group-homomorphism.Let $e,e'$ be the respective unit elements of $G,G'$.We define the **kernel** of $f$ to be the subset of $G$ consisting of all $x$ such that $f(x)=e'$.From the definitions,it follows at once that the kernel $H$ of $f$ is a subgrop of $G$.(Let us prove for instance that $H$ is closed under the inverse mapping.Let $x∈H$.Then

Since $f(x)=e'$,we have $f(x^{-1})=e'$,whence $x^{-1}∈H$.)

Let $f:G→G'$ be a group-homomorphism again.Let $H'$ be the **image** of $f$.Then $H'$ is a subgroup of $G'$,because it contains $e'$,and if $f(x),f(y)∈H'$,then $f(xy)=f(x)f(y)$ lies also in $H'$.Furthermore,$f(x^{-1})=f(x)^{-1}$ is in $H'$,and hence $H'$ is a subgroup of $G'$.

The kernel and image of $f$ are sometimes denoted by $Ker f$ and $Im f$.

A homomorphism $f:G→G'$ which establishes an isomorphism between $G$ and its image in $G'$ will also called an **embedding**.

A homomorphism whose kernel is trivial is injective.

To prove this, suppose that the kernel of $f$ is trivial,and let $f(x)=f(y)$ for some $x,y∈G$. Multiplying by $f(y^{-1})$ we obtain

$$f(xy^{-1})=f(x)f(y^{-1})=e'.$$Hence $xy^{-1}$ lies in the kernel,hence $xy^{-1}=e$,and $x=y$.If in particular $f$ is also surjective,then $f$ is an isomorphism.Thus a surjective homomorphism whose kernel is tirvial must be an isomorphism.We note that an injective homomorphism is an embedding.

An injective homomorphism is often denoted by a special arrow,such as

$$f:G\hookrightarrow G'.$$There is a useful criterion for a group to be a direct product of subgroups:

**Proposition 2.1** Let $G$ be a group and let $H,K$ be two subgroups such that $H∩K=e$,$HK=G$,and such that $xy=yx$ for all $x∈H$ and $y∈K$.Then the map

such that $(x,y)↦xy$ is an isomorphism.

**Proof.** It is obviously a homomorphism,which is surjective since $HK=G$. If $(x,y)$ is in its kernel,then $x=y^{-1}$, whence $x$ lies in both $H$ and $K$,and $x=e$,so that $y=e$ also,and our map is isomorphism.

We observe that Proposition $2.1$ generalizes by induction to a finite number of subgroups $H_{1},⋯,H_{n}$ whose elements commute with each other,such that

$$H_{1}⋯H_{n}=G,$$and such that

$$H_{i+1}∩(H_{1}⋯H_{i})=e.$$In that case,$G$ is isomorphic to the direct product

$$H_{1}×⋯×H_{n}.$$Let $G$ be a group and $H$ a subgroup.A **left coset** of $H$ in $G$ is a subset of $G$ of type $aH$,for some element $a$ of $G$.An element of $aH$ is called a **coset representative** of $aH$.The map $x↦ax$ induces a bijection of $H$ onto $aH$.Hence any two left cosets have the same cardinality.

Observe that if $a,b$ are elements of $G$ and $aH,bH$ are cosets having one element in common,then they are equal.Indeed,let $ax=by$ with $x,y∈H$.Then $a=byx^{-1}$.But $yx^{-1}∈H$.Hence $aH=b(yx^{-1})H=bH$,because for any $z∈H$ we have $zH=H$.

We conclude that $G$ is the disjoint union of the left cosets of $H$.A similar remark applies to **right cosets** (i.e. subsets of $G$ of type $Ha$).The number of left cosets of $H$ in $G$ is denoted by $(G:H)$,and is called the (left) **index** of $H$ in $G$. The index of the tirvial subgroup is called the **order** of $G$ and is written $(G:1)$. From the above conclusion,we get:

**Proposition 2.2** Let $G$ be a group and $H$ a subgroup.Then

in the sense that if two of these indices are finite, so is the third and equality holds as stated. If $(G:1)$ is finite,the order of $H$ divides the order of $G$.

More generally,let $H,K$ be subgroups of $G$ and let $H⊃K$.Let $\lbrace x_{i}\rbrace $ be a set of(left)coset representatives of $K$ in $H$ and let $\lbrace y_{i}\rbrace $ be a set of coset representatives of $H$ in $G$.Then we contend that $\lbrace y_{i}x_{i}\rbrace $ is a set of coset representatives of $K$ in $G$.

**Proof.** Note that

Hence

$$G=\underset{i,j}{\bigcup}y_{j}x_{i}K.$$We must show that this union is disjoint,i.e. that the $y_{j}x_{i}$ represent distinct cosets.Suppose

$$y_{j}x_{i}K=y_{j'}x_{i'}K$$for a pair of indices $(j,i)$ and $(j',i')$.Multiplying by $H$ on the right,and noting that $x_{i},x_{i'}$ are in $H$,we get

$$y_{j}H=y_{j'}H,$$whence $y_{j}=y_{j'}$.From this it follows that $x_{i}K=x_{i'}K$ and therfore that $x_{i}=x_{i'}$,as was to be shown.

The formula of Proposition $2.2$ may therefore be generalized by writing

$$(G:K)=(G:H)(H:K),$$with the understanding that if two of the three indices appearing in this formula are finite,then so is the third and the formula holds.

The above results are concerned systematically with left cosets.For the right cosets,see Exercise $10$.

**Example.** A group of prime order is cyclic.Indeed,let $G$ have order $p$ and let $a∈G,a≠e$.Let $H$ be the subgroup generated by $a$.Then $\#(H)$ divides $p$ and $≠1$,so #$(H)=p$ and so $H=G$,which is therefore cyclic.

**Example.** Let $J_{n}=\lbrace 1,⋯,n\rbrace $.Let $S_{n}$ the group of permutations of $J_{n}$.We define a **transposition** to be a permutation $τ$ such that there exist two elements $r≠s$ in $J_{n}$ for which $τ(r)=s,τ(s)=r$,and $τ(k)=k$ for all $k≠r,s$.Note that the transposition generate $S_{n}$.Indeed,say $τ$ is a permutation,$σ(n)=k≠n$.Let $τ$ be the transposition interchanging $k,n$.Then $τ σ$leaves $n$ fixed,and by induction,we can write $τ σ$ as a product of transpositions in Perm($J_{n-1})$,thus proving that transpositions generate $S_{n}$.

Next we note that #$(S_{n})=n!$.Indeed,let $H$ be the subgroup of $S_{n}$ consisting of those elements which leave $n$ fixed.Then $H$ may be identified with $S_{n-1}$.If $σ _{i}(i=1,⋯,n)$is an element of $S_{n}$ such that $σ _{i}(n)=i$,then it is immediately verified that $σ _{1},⋯,σ _{n}$ are coset representatives of $H$.Hence by induction

$$(S_{n}:1)=n(H:1)=n!.$$Observe that for $σ _{i}$ we could have taken the transposition $τ _{i}$,which interchanges $i$ and $n$(except for $i=n$,where we could take $σ _{n}$ to be the identify).