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Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

Write $c=(12\cdots n)$, with $n\ge 2$. Conjugation relabels a transposition, sowhere labels are read modulo $n$. In particular the subgroup contains $(12),(23),\dots ,(n-1n)$. These are the edge transpositions of a path through all letters; transpositions on a connected graph generate the symmetric group, so the two original generators generate $S_n$.

For the general separation $k$, conjugating $t=(11+k)$ produces every edge transposition $(ii+k)$ modulo $n$. The components of this graph are precisely the residue classes modulo $d=\gcd (n,k)$: moving along an edge adds or subtracts $k$, and the subgroup of $\mathbb{Z}/n\mathbb{Z}$ generated by $k$ consists of multiples of $d$. Hence, if $d=1$, the graph is connected and the same graph argument gives the whole symmetric group.

For completeness, the graph argument can be proved without assuming a generating-set theorem. On a simple path $v_0,\dots ,v_{\ell }$, put $t_j=(v_{j-1}{v}_j)$. ThenThis is a conjugate of the last edge transposition; connectivity therefore supplies every transposition, and part (a) supplies every permutation.

If $d>1$, there are $d$ residue blocks of size $n/d>1$, since $1\le k<n$. The cycle $c$ permutes these blocks, and $t$ swaps two letters in the same block. Both preserve this block system, so their generated subgroup does too. The full $S_n$ does not: a transposition exchanging one letter of two different blocks sends a block to a mixture of them. Thus the subgroup is proper. This provesThus a cycle and a transposition generate the symmetric group exactly at coprime separation. The obstruction is preservation of a partition, not failure of transitivity: the $n$-cycle already acts transitively even when $d>1$.

The Sylow theorems state that if a finite group has order $|G|=p^{a}m$ with $p$ prime and $p\nmid m$, it has a subgroup of order $p^a$, called a Sylow subgroup. Every p-group subgroup is contained in a Sylow subgroup, and all Sylow subgroups for the same prime are conjugate subgroups. Their number satisfiesMore precisely $n_p=[G:N_G(P)]$ for a Sylow subgroup $P$, where $N_G(P)$ is its normalizer.

First require $x\ne 0$. A discrete logarithm to a generator of a group of ${\mathbb{Z}}_p^{\ast }$ exists only for nonzero $x$; the printed membership $x\in {\mathbb{Z}}_p$ inadvertently also permits zero, for which $x^{-b}$ is undefined. Zero can be rejected before any quantum circuit.

Put $M=p-1$ and write $x=g^y$. Since $g$ is a generator of a group of order $M$, the map $k\mapsto g^k$ is a bijection from ${\mathbb{Z}}_M$ to the multiplicative group of a finite field. Hence every nonzero $c$ has a unique $k\in {\mathbb{Z}}_M$ with $c=g^k$. Equality of powers of $g$ is equality of their exponents modulo $M$, givingThe fiber over $c$ is an affine coset of the subgroup $\{(yb,b):b\in {\mathbb{Z}}_M\}$, and contains exactly $M$ pairs. This identifies the hidden subgroup problem underlying discrete-logarithm Fourier sampling; knowing $k$ numerically is not needed to construct the coset state.

Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

Multiplicativity of the determinant gives $\delta (gh)=\delta (g)\delta (h)$ and $\delta (1)=1$, so $\delta$ is a linear character. Its finite image is a cyclic group of roots of unity. If it contains $-1$, its order is $2m$ and the character of a representation ${\delta }^m$ maps onto $\{\pm 1\}$. The kernel of a group homomorphism is the preimage of its identity, so $\ker ({\delta }^m)$ is a normal subgroup of index $2$. The subgroup $\ker \delta$ itself need not have index $2$.

By Cauchy theorem for groups, a group $E$ of order $2k$ has an involution $g$. In the regular representation, left multiplication by $g$ is a permutation consisting of $k$ disjoint transpositions, so its determinant is $(-1{)}^k=-1$ when $k$ is odd. The preceding argument gives a normal subgroup of index two.

For a nonabelian simple group of order below $80$, Burnside's theorem requires at least three distinct prime factors. An odd order with three factors is at least $3\cdot 5\cdot 7=105$. An even order with exactly one factor of $2$ has a normal subgroup of index $2$. Thus its order must be divisible by $4$ and by at least two odd primes. Below $80$ the only possibility is $4\cdot 3\cdot 5=60$, since the next is $4\cdot 3\cdot 7=84$. Hence $\boxed{|H|=60}$.

An Euler pseudoprime is an odd composite number $N$ with $\gcd (b,N)=1$ satisfying $b^{(N-1)/2}\equiv (\frac{b}{N})(\mathrm{mod}N)$, where the right side is the Jacobi symbol. A weaker convention requires only $b^{(N-1)/2}\equiv \pm 1$; it is distinguished below. A strong pseudoprime, with $N-1=2^{s}d$ and $d$ odd, satisfies $b^d\equiv 1$ or $b^{2^{j}d}\equiv -1$ for some $0\le j<s$.

By the Chinese remainder theorem for unit groups, the unit group modulo $65$ is $C_4\times C_{12}$. Modulo $5$, $b^{32}=1$. Thus the only possible sign in $b^{32}\equiv \pm 1(\mathrm{mod}65)$ is positive. Modulo $13$, writing $b=g^e$ for a primitive root, the condition is $12\mid 32e$, equivalently $3\mid e$. Consequently $b$ has multiplicative order dividing $4$ in each factor. Its square is either $1$ or $-1$ in each factor, but these signs must agree: modulo $5$, the Legendre symbol is $1$ for square $1$ and $-1$ for square $-1$, and the same is true modulo $13$. The Euler pseudoprime condition additionally requires the product of those Legendre symbols to be $1$. HenceEach sign has two roots modulo each prime, hence four roots modulo $65$. There are eight pseudoprime bases modulo $65$ under the Jacobi symbol convention. Under the weaker sign-only convention there are instead sixteen pseudoprime bases, since the two signs of the squares need not agree; the stated equivalence therefore specifically uses the Jacobi symbol convention.

For the strong pseudoprime condition, $s=6,d=1$. Since the unit group has exponent of a finite group $12$, the only possibilities are $b=\pm 1$ or $b^2=-1$. The six pseudoprime bases are $1,8,18,47,57,64$. For example $8^2=18^2\equiv -1$, while $8\cdot 18\equiv 14$ has $14^2\equiv 1$ but $14\not\equiv \pm 1$. Thus the strong-pseudoprime pseudoprime bases are not a subgroup.

The Nielsen–Schreier theorem says that every subgroup of a free group is free. Realize $F_n$ as the fundamental group of the rose $R_n$, the graph with one vertex and $n$ oriented loops. A subgroup $H\le F_n$ of finite index of a subgroup $d$ corresponds to a connected $d$-sheeted covering graph $X\to R_n$. The graph $X$ has $d$ vertices and $dn$ unoriented edges. Choosing a spanning tree leavesedges outside the tree, and these freely generate ${\pi }_1(X)\cong H$. Thus the Nielsen–Schreier formula is

The hidden subgroup problem is to recover a subgroup $K\le G$, usually by giving generators of a group, from an oracle machine providing $f:G\to Y$ with the promiseThus $f$ is constant on each left coset and distinguishes different left cosets. The input includes an effective representation of elements of the group and access to the group operations; the desired efficiency is measured against the number of bits describing an element, rather than necessarily against $|G|$.

In the vector space ${\mathbb{F}}_2^n$, addition and subtraction agree. Put $V=⟨Y⟩$, the vector subspace spanned by $Y$, andPart b gives $4A\subseteq 2A+V=H$. Conversely $2A\subseteq 4A$, because two copies of any fixed element of $A$ sum to zero. Henceand $0\in H$, so $H$ is a subgroup. For any $a_0\in A$, every $a\in A$ satisfies $a-a_0=a+a_0\in H$, and therefore $A\subseteq a_0+H$.

Because $y_1=0$, the dimension of a vector space $V$ is at most $|Y|-1\le 2K^2-2$. Consequentlywhere the last inequality uses $K\ge 1$. This is the requested bound.

A group $G$ is abelian when $xy=yx$ for every $x,y\in G$. It is cyclic when some generator of a group $g$ satisfiesIf $x=g^r$ and $y=g^s$ belong to a cyclic group, thenso every cyclic group is abelian.

The Klein four-group $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.

The condition on proper subgroups does not force $G$ to be abelian. The quaternion groupis nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm 1\}$, and the three cyclic subgroupseach of order four. They are all cyclic, so

A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

The set $S(\mathbb{N})$ consists of all bijections $\mathbb{N}\to \mathbb{N}$. The composition of two bijections is a bijection, composition is associative, the identity map is an identity, and every bijection has an inverse, so $S(\mathbb{N})$ is a group.

An element of $S_{\mathrm{fin}}(\mathbb{N})$ has finite support. The support of a composition is contained in the union of the two supports, and a permutation and its inverse have the same support. The identity has empty support. Hence $S_{\mathrm{fin}}(\mathbb{N})$ is a subgroup.

Let $\sigma$ be a cycle and choose $n$ that it moves. Because $\sigma$ has finite support, the sequencemust repeat. Since $\sigma$ is invertible, its first repetition returns to $n$; let the least positive return time be $l$. The cycle condition says that every moved point occurs in this orbit, so ${\sigma }^l$ fixes every point. No smaller positive power fixes $n$. Thereforewhich is finite.

For $\tau \in S_{\mathrm{fin}}(\mathbb{N})$, partition its finite support into the orbits of the cyclic group $⟨\tau ⟩$. On each orbit, let ${\sigma }_i$ agree with $\tau$ and fix every point outside that orbit. Then each ${\sigma }_i$ is a permutation cycle, their supports are pairwise disjoint, andDisjoint cycles commute because at every point at most one of them acts nontrivially.

Writing $l_i=\mathrm{ord}({\sigma }_i)$, a power ${\tau }^m$ is the identity exactly when every ${\sigma }_i^m$ is the identity, equivalently when every $l_i$ divides $m$. Thus

False. Let $G=C_4$ and $H=C_2\times C_2$, and in each choose a subgroup of order two. Then the chosen normal subgroups are isomorphic and both quotients are $C_2$, while $G$ and $H$ are not isomorphic.