Suppose the quotient group $G/A$ is a cyclic group, generated by $gA$. For any $x,y\in G$, there are integers $r,s$ and $a,b\in A$ such thatSince $a,b$ lie in the center of a group, they commute with every element, in particular with each other and with the integer powers of a group element $g$. ThereforeAs $x,y$ were arbitrary, $G$ is an abelian group. This is the cyclic quotient by the center.

The specified set $A$ is the center of a group $Z(G)$. It contains $e$. If $a,b\in A$ and $y\in G$, thenso $ab\in A$. From $ay=ya$, multiplying on the left and right by $a^{-1}$ gives $y{a}^{-1}=a^{-1}y$, so $a^{-1}\in A$. Hence $A$ is a subgroup.

For $g\in G$ and $a\in A$, the defining commutativity gives $ga{g}^{-1}=ag{g}^{-1}=a$. Thus the image of each element of $A$ under conjugation is in $A$, and indeed $gA{g}^{-1}=A$. Consequently $A=Z(G)$ is a normal subgroup of $G$.

The orbit of $x$ under conjugation is its conjugacy classIt contains $x$ by using $g=e$, so the conjugacy classes cover $G$. If $z=gx{g}^{-1}$, then $x=g^{-1}zg$, and the group action identity shows that conjugating either element further gives the same orbit; hence $C(z)=C(x)$. If $C(x)$ and $C(y)$ intersect at $z$, this gives $C(x)=C(z)=C(y)$. Thus distinct conjugacy classes are disjoint and the conjugacy classes form a partition of a set $G$.

Write $g\cdot x=gx{g}^{-1}$. If $e$ is the identity element of $G$, then $e\cdot x=ex{e}^{-1}=x$. Also, for $g,h,x\in G$,The middle equality uses associativity and the inverse element rule $(gh{)}^{-1}=h^{-1}{g}^{-1}$. Thus this is a left group action, the conjugation action of $G$ on itself.

Suppose there were only finitely many prime numbers. There is at least one, namely $2$, so this finite set would have a largest element $q$. The integer $2^q-1$ is greater than $1$ and therefore has a prime factor $p$: for example, its least divisor greater than $1$ must be a prime number, since any proper factor of that divisor would contradict minimality.

By ii, $p>q$, contradicting the choice of $q$. Therefore there are infinitely many prime numbers, proving Euclid's theorem by orders of group elements in finite multiplicative groups.

The multiplicative group of a finite field $p$ has group order $p-1$: its elements are exactly the nonzero residue classes. For any prime factor $p$ of $2^q-1$, i and Lagrange's theorem therefore implySince $q$ and $p-1$ are positive integers, $q\le p-1$, and henceThis applies to every prime factor, including repeated prime factors viewed as the same prime number.

If the prime number $p$ divides $2^q-1$, then $p\ne 2$, because $2^q-1$ is odd. Thus the residue of $2$ is an element of the multiplicative group of a finite field ${\mathbb{Z}}_p$, and $2^q\equiv 1(\mathrm{mod}p)$.

Let $d$ be its order of a group element. The integer division writes $q=ad+r$ with $0\le r<d$. Then $1\equiv 2^q\equiv (2^d{)}^a{2}^r\equiv 2^r(\mathrm{mod}p)$, so minimality of $d$ gives $r=0$. Hence $d$ divides $q$. Since $q$ is a prime number, $d=1$ or $d=q$. The former would give $2\equiv 1(\mathrm{mod}p)$, which would mean $p$ divides $1$, an impossibility. Consequently

If $G=\{e\}$, choose ordinary addition on $\mathbb{Z}$, giving the additive group of integers. Every power function is the unique identity function on the singleton $G$, so the group action axioms hold:Thus ordinary addition supplies the required group operation on $\mathbb{Z}$. Together with ii, existence is equivalent to $G$ being the trivial group.

Suppose an arbitrary binary operation $\star$ makes $\mathbb{Z}$ a group and makes $\rho$ a left group action on $G$. For any group action, each acting element induces a bijection: the function induced by its inverse element is its inverse function.

In particular, the acting element whose underlying integer is $0$ must induce a bijection $G\to G$. But its power function isa constant function. This is injective only when all elements of $G$ equal $e$. ThereforeThis proves the power-map group action obstruction for every possible $\star$; it never assumes that $\star$ is ordinary integer multiplication, or that the integer $0$ is the acting identity element.

Fix $x\in G$. The integer powers of a group element satisfyFor nonnegative integers these rules follow by associativity and counting factors; negative integers are handled by $x^{-r}=(x^{-1}{)}^r$, and opposite signs by cancellation of adjacent inverse elements. All factors are powers of the same element, so no abelian group assumption is involved.

It follows that $z\mapsto \rho (z,x)=x^z$ is a group homomorphism from the additive group of integers to $G$, andThe homomorphism uses ordinary addition on $\mathbb{Z}$; composition of the power functions corresponds to ordinary multiplication of exponents.

Let the finite set $G/H$ contain $r$ distinct left cosets. By i, they form a disjoint partition of a set $G$, and by ii each has $|H|$ elements. If $H$ is also finite, counting this finite disjoint set union givesThus $G$ is a finite group. This is the counting argument behind Lagrange's theorem.

For $W=aH$ and $W^{\prime }=bH$, defineIf $w=ah$ then $f(w)=bh\in bH$, so this function is well defined. Its inverse function is $w^{\prime }\mapsto a{b}^{-1}{w}^{\prime }$, since successive left translation in a group by $b{a}^{-1}$ and $a{b}^{-1}$ cancels. ThusIn particular every left coset has the same cardinality as $H$.

Each left coset $gH$ is nonempty, since it contains $g=ge$, and every $g\in G$ lies in its own left coset. Suppose $aH$ and $bH$ intersect, say $a{h}_1=b{h}_2$ with $h_1,h_2\in H$. ThenSince $H$ is a subgroup, $h_2{h}_1^{-1}\in H$, and $aH\subseteq bH$. The reversed relation $b=a{h}_1{h}_2^{-1}$ gives $bH\subseteq aH$. Thus any two left cosets are either equal or disjoint, and together they cover $G$. Therefore $G/H$ is a partition of a set $G$. No finiteness or normal subgroup assumption is needed.

The shift $s:\mathbb{N}\to \mathbb{N}$, $s(n)=n+1$, is injective, so $s\in I(\mathbb{N})$. It is not surjective: the least natural number is missing from its image of a function, whether $\mathbb{N}$ starts at zero or one. For every $t:\mathbb{N}\to \mathbb{N}$, the image of a function of $s\circ t$ is contained in the image of a function of $s$, so $s\circ t\ne {\mathrm{id}}_{\mathbb{N}}$.

The identity function belongs to $I(\mathbb{N})$ and is its identity element under function composition; any group structure using that operation would have this same identity element. Since $s$ has no right inverse element, $I(\mathbb{N})$ is not a group under function composition. It is a monoid, since injectivity is preserved by function composition.

An injective function from a finite set to itself is surjective: its image of a function contains as many distinct elements as the entire set, so it cannot omit one. Thus every element of $I(X)$ is a bijection, and conversely every bijection is injective. HenceIt is a group under function composition by i.

For any $x$, choose $z$ with $x\ast z=e$. By ii, $z\ast x=e$ as well. Associativity and the right identity element property now giveThus $e$ is also a left identity element. The given binary operation is closed on $X$ and associative, $e$ is a two-sided identity element, and each $x$ has the two-sided inverse element $z$. All group axioms hold, so $(X,\ast )$ is a group. This is the one-sided group axioms criterion.

Suppose $x\ast z=e$. Using only associativity and the given right identity element property,Thus $z\ast x$ is an idempotent element of a semigroup. By i it equals $e$, provingNo left identity element or cancellation law is needed for this deduction.

Suppose $y\ast y=y$. Choose $z$ with $y\ast z=e$, using the given right inverse element property. Multiply the idempotent relation on the right by $z$. By associativity and the given right identity element property,Thus the only idempotent element of a semigroup is $e$. This argument uses no cancellation and does not assume a left identity element.

For each $g\in G$, left translation in a group $L_g:x\mapsto gx$ is a bijection, with inverse function $L_{g^{-1}}$. Thus $\theta (g)=L_g$ belongs to $\mathrm{Sym}(G)$. For $g,h,x\in G$, associativity of the group operation givesHence $\theta$ is a group homomorphism. If $\theta (g)=\theta (h)$, evaluation at the identity element $e$ gives $g=ge=he=h$. ThereforeThis is the left regular representation used in Cayley's theorem.

For a singleton $X$, the symmetric group contains only the identity function. For a two-element set, it contains the identity function and the transposition exchanging the two elements; the latter squares to the identity function. In both cases all elements commute, so the group is Abelian.

If $X$ contains three distinct elements $a,b,c$, define the transpositions $\sigma =(ab)$ and $\tau =(bc)$, fixing every other element of $X$. They are bijections, even when $X$ is infinite. With function composition read from right to left,Thus they do not commute, so the symmetric group is not Abelian. Since $X$ is nonempty, this proves