Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 6 (Cosets, power maps and primes)

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.

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 $H$.

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.

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.

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.

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.

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

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.

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.