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Let $|G|=p$, with $p$ a prime number, and choose $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$ and is greater than one. It is therefore $p$, and $G=⟨g⟩$. Hence a group of prime cardinality is a cyclic group.

The converse is false: the cyclic group $C_4$ has four elements, and $4$ is not a prime number.

The complete classification up to group isomorphism isHere $C_1$ is the trivial group. The prime number cases follow from i. For four elements, an element of order of a group element four generates $C_4$. If there is none, Lagrange's theorem makes every nonidentity element have order of a group element two. Then iv produces a four-element subgroup, which must be all of $G$, giving the Klein four-group.

For six elements, an element of order of a group element six gives $C_6$. Suppose there is none. If there were no element of order of a group element three either, all nonidentity elements would have order of a group element two. Part iv would then produce a subgroup of cardinality four, contradicting Lagrange's theorem. Thus choose $a$ of order of a group element three. There is also an involution $b$: otherwise the five nonidentity elements could be paired into distinct pairs $\{g,g^{-1}\}$, impossible for an odd number of elements.

The subgroup $H=⟨a⟩$ has index of a subgroup two, so it is a normal subgroup. Indeed, for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, and for $g\in H$ both equal $H$. Consequently $ba{b}^{-1}$ is either $a$ or $a^{-1}$. The first possibility makes $a,b$ commute, and $ab$ then has order of a group element six: $(ab{)}^k=e$ forces $a^k=b^{-k}\in ⟨a⟩\cap ⟨b⟩=\{e\}$, so both $3$ and $2$ divide $k$. This contradicts our assumption. ThereforeEvery element is uniquely $a^i$ or $a^{i}b$, with $0\le i<3$, and the relation $ba=a^{-1}b$ determines all their products. Sending $a$ to $(123)$ and $b$ to $(12)$ gives a group isomorphism with the symmetric group $S_3$. This completes the classification of groups of order at most six.

Of these groups, exactly $C_1,C_2,C_3,C_4,C_2\times C_2,S_3$ embed in $S_4$. Concrete subgroups areThere is no $C_5$ subgroup, since $5$ does not divide $24$. There is no $C_6$ subgroup either: the order of a group element in a symmetric group is the least common multiple of the lengths in its disjoint cycle decomposition, and on four letters the only possibilities are $1,2,3,4$.

Write $C_p=⟨g⟩$. A group homomorphism from this cyclic group is determined by the image of $g$, because $\varphi (g^j)=\varphi (g{)}^j$. Each choice $\varphi (g)=g^a$, with $a=0,\dots ,p-1$, gives a well-defined group homomorphism. It is an group automorphism exactly when $g^a$ generates $C_p$.

For a prime number $p$, every $a=1,\dots ,p-1$ is coprime to $p$, so the powers of $g^a$ run through all of $C_p$. The choice $a=0$ gives the trivial group homomorphism and fails to be bijective. Consequently

By Lagrange's theorem, every nonidentity group element has order $5$, $7$, or $35$. If some element has order $35$, its seventh and fifth powers have orders $5$ and $7$, respectively, and we are done.

Otherwise every nonidentity element has order $5$ or $7$. Distinct subgroups of a given prime number order intersect only in the identity: a shared nonidentity element generates both. If there were no element of order $5$, the $34$ nonidentity elements would therefore partition into sets of six, the nonidentity elements of the cyclic subgroups of order seven. This would force $6\mid 34$, a contradiction. If there were no element of order $7$, the analogous partition into sets of four would force $4\mid 34$, again a contradiction. Thus elements of both orders exist.

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

A smallest example is the Klein four-groupwith coordinatewise addition modulo $2$. Every nonidentity element has order of a group element $2$, so none generates all four elements; hence this group is not a cyclic group.

A group of group order $1$ is a cyclic group. A group of group order $2$ or $3$ is also a cyclic group: choose a nonidentity element; by Lagrange's theorem, its order of a group element is a nontrivial divisor of that prime number, so equals the whole group order. Thus no smaller example exists, and the smallest possible group order is $4$.

The sequence does not converge. For $n=2^k$, the largest prime factor is $p(n)=2$, while the sum of prime factors with multiplicity is $s(n)=2k$. This subsequence therefore has ratio $1/k\to 0$.

There are infinitely many prime numbers: if there were only $q_1,\dots ,q_r$, a prime factor of $q_1\cdots q_r+1$ would be absent from the list. Listing the prime numbers in increasing order gives a subsequence of indices $n=q_j$ for which $p(n)=s(n)=q_j$, so the ratio is constantly one. Every subsequence of a convergent sequence has the same limit of a sequence. The two distinct subsequential limits $\boxed{0\text{ and }1}$ rule out convergence.

Neither $1$ nor $-1$ is divisible by a prime number, so both lie outside $U$. On the other hand, $0\in 2\mathbb{Z}\subseteq U$. If an integer $z$ has $|z|>1$, part (a)(i) supplies a prime number $p$ dividing the positive integer $|z|$; the same $p$ divides $z$, so $z\in p\mathbb{Z}\subseteq U$. These cases exhaust $\mathbb{Z}$, proving

The set $S_n$ is nonempty because $n\in S_n$. By the well-ordering principle for the natural numbers, it has a least element $p$. In particular $p>1$ and $p$ divides $n$.

If $p$ were not a prime number, it would have a positive integer divisor $d$ with $1<d<p$. Since $d$ divides $p$ and $p$ divides $n$, divisibility is transitive and $d$ divides $n$. This would put $d$ in $S_n$, contradicting the minimality of $p$. Hence $\boxed{p\text{ is prime}}$.

Suppose there were only finitely many prime numbers, say $p_1,\dots ,p_k$. Work in the infinite group $(\mathbb{Z},+)$ and set $H_i=p_i\mathbb{Z}$. The left cosets are the $p_i$ residue classes modulo $p_i$, so each $H_i$ is a finite-index subgroup, with $|\mathbb{Z}/H_i|=p_i$.

By part (b)(v), $\mathbb{Z}\setminus {\bigcup }_i{p}_i\mathbb{Z}$ is empty or infinite. But part (a)(ii) identifies it as the nonempty two-element set $\{-1,1\}$. This contradiction proves Euclid's theorem:

The Cauchy-Hadamard theorem states $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. At the prime numbers $n=p$, this root equals $p^{2/p}\to 1$. There are infinitely many prime numbers, so these indices form an unbounded subsequence. For every other positive index the root is zero. The limit superior is therefore one, and $\boxed{R=1}$.

Lagrange's theorem states that the group order of a finite group divides the group order; more precisely $|G|=[G:H]|H|$.

If every element squares to the identity element, every element is its own inverse element, and $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian. For distinct nonidentity $a,b$, the four elements $1,a,b,ab$ are distinct: for example $ab=1$ would imply $a=b$, and $ab=a$ would imply $b=1$. Products remain in this set because $a,b$ commute and square to one. Thus they form a Klein four-group subgroup of order four.

Now let $|G|=2p$ with odd prime number $p$. By Lagrange's theorem, every element order is one of $1,2,p,2p$. If an element has order $2p$, its square has order $p$. If no element had order $p$, all nonidentity elements would therefore have order two. There are at least two distinct such elements, producing a subgroup of order four. This contradicts $4\nmid 2p$. Hence there is an element of order $p$.

It suffices to take four players. Both two-player coalitions $\{1,2\}$ and $\{3,4\}$ win, because two is a prime number, while the empty coalition and the four-player coalition lose.

Suppose real weights and quota $q$ represented this game. The empty coalition loses, so $0<q$. The two winning coalitions requireAdding givesso the four-player coalition must win, a contradiction. Hence prime-cardinality games cannot in general be represented as weighted voting games. The same four-player obstruction appears inside every such game with at least four players.

This proof permits negative individual weights, as the printed definition does. An argument based solely on monotonicity would require a nonnegative-weight assumption and would not suffice for that definition.

Wilson's theorem states that an integer $p>1$ is a prime number if and only ifIn particular this holds for every prime, including $p=2$.

Fermat's little theorem says that for a prime number $p$ and any integer $a$,If $p\nmid a$, cancellation gives the equivalent statement $a^{p-1}\equiv 1(\mathrm{mod}p)$.

In field characteristic $p$, the binomial theorem gives $(X-b{)}^p=X^p-b^p$, since the intermediate binomial coefficients are divisible by the prime number $p$. If $b\in F$ is a root of $X^p-a$, then $b^p=a$, and hence

If the complete list of prime numbers were $p_1,\dots ,p_k$, then $N=p_1\cdots p_k+1>1$ would have a prime divisor by the fundamental theorem of arithmetic. No $p_i$ divides it, since $N\equiv 1(\mathrm{mod}{p}_i)$. This contradicts completeness of the list. Hence there are infinitely many prime numbers.

The fundamental theorem of arithmetic says that every integer $n>1$ is a product of prime numbers, uniquely up to reordering. Equivalently, $n={\prod }_p{p}^{a_p}$ for a unique family of nonnegative integer exponents, only finitely many of which are nonzero.