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False. Let $G=S_3$ and $H=K=⟨(12)⟩$. For $g=(13)$, direct multiplication gives the double cosetIt has four elements, and $4\nmid 6=|S_3|$. This is a concrete example of double-coset sizes not dividing group order; Lagrange's theorem applies to subgroups, not to arbitrary double cosets.

Every $g\in G$ lies in $HgK$, since both subgroups contain the identity. If two double cosets intersect, chooseThen $g^{\prime }=h_2^{-1}{h}_{1}g{k}_1{k}_2^{-1}\in HgK$, so every element of $H{g}^{\prime }K$ lies in $HgK$. Reversing the argument gives equality. Thus the double cosets partition the group.

Unlike single cosets, double cosets can have different cardinalities. For finite $G$, decompose $HgK$ into left cosets of $K$, or apply the orbit-stabilizer theorem to the action of $H$ on $gK$. Its stabiliser subgroup is $H\cap gK{g}^{-1}$, givingThis formula explains the counterexamples in the following parts.

For a finite group $G$, subgroups $H,K$, and a complex group representation $V$ of $H$, the Mackey restriction formula isChoose one representative $g$ of each double coset $KgH$. The conjugate subgroup representation ${}^{g}V$ is a representation of $gH{g}^{-1}$ on the same vector space, with $(gh{g}^{-1})v=hv$. Restriction of a representation keeps the vector space and restricts the acting subgroup. induced representation is $\mathbb{C}G{\otimes }_{\mathbb{C}H}V$, and analogously for each subgroup of $K$. The displayed isomorphism is an isomorphism of $K$-representations. These definitions specify every summand and its subgroup.

The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two double cosetsAn $H$-bimodule-equivariant set map is determined by $u=\Phi (1)$ and $v=\Phi (t)$. Equivariance requires $u\in Z(H)$ andbecause $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi (h)=hu$ and $\Phi (ht{h}^{\prime })=hv{h}^{\prime }$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.

For $n\ge 5$, the center of $H$ and this centralizer are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\dots ,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\ge 3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(12)\}$. The choiceis well defined and satisfies all the printed equivariance conditions, but sends $(34)$ to $(12)$ rather than to the identity. Uniqueness holds for $n\ge 5$; it is false at $n=4$ as printed. Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.

There is an index error in the displayed identity of the original PDF. With its definition of the signed Young permutation module, the tensor identity for induced representations givesBoth superscripts must be $\alpha$. In particular ${\tilde{M}}^{{\lambda }^{\prime }}\cong \mathrm{sgn}\otimes M^{{\lambda }^{\prime }}$, rather than the printed sign twist of $M^{\lambda }$. For $\lambda =(n)$, $n\ge 2$, the printed left side has dimension $n!$, while the printed right side has dimension one. Thus that identity cannot hold as written.

The requested character inner product between $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ nevertheless has a well-defined answer. Let $R$ be the row Young subgroup of a tableau of shape $\lambda$, and let $C$ be its column subgroup, conjugate to $S_{{\lambda }^{\prime }}$. By Frobenius reciprocity and the Mackey restriction formula, the inner product is a sum over $R\backslash S_n/C$ ofThese double cosets are encoded by nonnegative integer matrices $A=(a_{ij})$ with row sums ${\lambda }_i$ and column sums ${\lambda }_j^{\prime }$, recording intersection sizes of row and column blocks. The intersection subgroup is a product of $S_{a_{ij}}$. Its sign representation is trivial exactly when every $a_{ij}\le 1$. Each zero-one matrix contributes one, and every other matrix contributes zero.

There is exactly one zero-one matrix with these margins. Its first row has length ${\lambda }_1$, equal to the number of nonzero column sums, so that row must consist entirely of ones. Delete it, subtract one from every column margin and discard zero columns. The remaining margins are those of $({\lambda }_2,{\lambda }_3,\dots )$ and its conjugate partition; induction forces the rest. The unique matrix is $a_{ij}=1$ precisely for $j\le {\lambda }_i$. ThereforeThis calculation uses the actual two modules requested in the PDF and does not depend on its erroneous displayed isomorphism.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and $G=G{L}_2(\mathbb{Q}{)}^{+}$. The double-coset Hecke algebra consists of $\Gamma$-bi-invariant complex functions on $G$ supported on finitely many double cosets, with convolutionEach double coset has finitely many orbits under left multiplication by $\Gamma$, by rational conjugation of finite-index modular subgroups. Thus the sum is finite and independent of representatives. The characteristic functions of the double cosets form its basis, and its right action on invariant modular forms is $f\ast h={\sum }_{\Gamma x}h(x)f{|}_{k}x$, using the determinant-normalized slash operator.

For a positive integer $n$, let ${\mathcal{M}}_n$ be all integral two-by-two matrices of positive determinant $n$. It is $\Gamma$-bi-invariant. Define $T(n)$ to be its indicator function, equivalently the sum of the distinct double cosets it contains, each with coefficient one. This is the convention consistent with the requested formula. For composite $n$, it need not be the single double coset of $\mathrm{diag}(1,n)$: for instance $2I\in {\mathcal{M}}_4$ cannot lie in that double coset, since multiplying by unimodular integral matrices preserves the greatest common divisor of the entries.

Prove all the needed subgroup facts directly. For a finite-index subgroup $L\subseteq {\mathbb{Z}}^2$, take $a$ to be the least positive first coordinate appearing in $L$ and $d$ the least positive second coordinate on its intersection with the second axis. Euclidean division then shows that the first-coordinate projection is $a\mathbb{Z}$ and $L\cap (\{0\}\times \mathbb{Z})=\{0\}\times d\mathbb{Z}$. Positivity follows, for example, because the finite quotient group kills a nonzero multiple of each coordinate vector. Choose $(a,b)\in L$ and reduce $b$ modulo $d$ to $0\le b<d$. Every vector of $L$ has first coordinate a multiple of $a$, and subtracting that multiple of $(a,b)$ leaves a multiple of $(0,d)$. Therefore these two vectors form a basis of $L$. The parameters $a,d,b$ are unique. Reducing the first coordinate modulo $a$ and then the second modulo $d$ gives precisely $ad$ quotient representatives, so $[{\mathbb{Z}}^2:L]=ad$.

Apply this row Hermite normal form in rank two to the row lattice of an integral matrix $A$ with $\det A=n$. Its row lattice contains $n{\mathbb{Z}}^2$, since $\mathrm{adj}(A)A=nI$, so it has finite index. Its two rows and the displayed two rows are bases of the same row lattice. The two inverse change-of-basis matrices have integer entries, so their determinants are integers whose product is one. Thus the change-of-basis matrix has determinant $\pm 1$; since both orientations are positive, its determinant is one. Thus each orbit under left multiplication by $\Gamma$ has exactly one of the determinant-n matrix representatives for Hecke operatorsThere are ${\sum }_{d\mid n}d$ such representatives, proving finiteness as well as the formula. The subgroup argument is a proof of the relevant Hermite normal form, not an invocation of an unproved lattice classification.

Consequently the normalized Hecke operator isIt preserves $M_k(\Gamma )$: right multiplication by $\Gamma$ permutes the left-multiplication orbits in ${\mathcal{M}}_n$, giving invariance, and cusp holomorphy under rational slash operators gives the holomorphy of each term at every cusp.

For a triangular representative the determinant-normalized slash operator is $f{|}_{k}A=n^{k/2}{d}^{-k}f((az+b)/d)$. Summing the Fourier expansion of a modular form over $b$ kills every index not divisible by $d$, by finite exponential orthogonality. ThusThe Fourier coefficients of a composite-index Hecke operator are thereforeIn particular $a_1(T_{n}f)=a_n(f)$. If $T_{n}f=\lambda f$, comparison of the $q$ coefficients gives

Finally suppose $a_0(f)\ne 0$ and $f$ is a simultaneous eigenfunction. Constant coefficients force ${\lambda }_n={\sigma }_{k-1}(n)$, so $a_n(f)=a_1(f){\sigma }_{k-1}(n)$ for all $n\ge 1$. Weight two cannot occur, by vanishing of weight-two level-one modular forms. For even $k\ge 4$, use the normalized Eisenstein series with its Fourier expansion of a normalized Eisenstein serieswhere $B_k$ is the Bernoulli number. Then $h=f-a_0(f)E_k$ is a cusp form with coefficients $(a_1(f)-a_0(f)c_k){\sigma }_{k-1}(n)$. The Fourier coefficient bound for a cusp form bounds these by $O(n^{k/2})$, but at arbitrarily large primes $p$ their magnitude is $|a_1-a_0{c}_k|(1+p^{k-1})$. Since $k-1>k/2$, the coefficient factor must vanish. All coefficients of $h$ then vanish, including its constant coefficient, so $h=0$ by its cusp expansion and the identity theorem. This proves the noncuspidal level-one Hecke eigenform characterization:

The decomposition group of the chosen prime is its stabilizer under the Galois group:The transitivity of the Galois action on primes identifies the primes of $M$ above $\mathfrak{p}$ with $G/D$ through $gD\mapsto g\mathfrak{P}$.

Put $H=\mathrm{Gal}(M/L)$. Contraction of prime ideals to ${\mathcal{O}}_L$ is constant on every $H$-orbit, because every element of $H$ fixes $L$ pointwise. It is surjective onto the primes of $L$ above $\mathfrak{p}$, since prime ideals extend in the integral extension ${\mathcal{O}}_M/{\mathcal{O}}_L$. Moreover, two primes of $M$ contracting to the same prime of $L$ are conjugate by $H$: the extension $M/L$ is Galois and the same transitivity theorem applies to it.

Thus the fibers of contraction are exactly the $H$-orbits on $G/D$. These orbits are the double cosets $H\backslash G/D$, proving the primes in an intermediate field as double cosets correspondenceExplicitly, replacing $g$ by $hgd$ does not change the contracted prime. Conversely, if $g\mathfrak{P}$ and $g^{\prime }\mathfrak{P}$ contract to the same prime, some $h\in H$ sends the former to the latter, so $g^{\prime -1}hg\in D$ and $H{g}^{\prime }D=HgD$. This also verifies the stated order of the two sides of the double coset.

The Noncommutative Ruzsa triangle inequality says that nonempty finite subsets $B,C,D$ of a group satisfyThe Ruzsa covering lemma says that if $|CD|\le K|D|$, then some $X\subseteq C$ with $|X|\le K$ satisfies

Now let $A$ be a symmetric subset of a group, so $A^{-1}=A$ and $1\in A$. Apply the triangle inequality with the middle set $A$ to obtain, for $m\ge 4$,The hypothesis therefore gives $|A^m|\le K|A^{m-1}|$. Starting from $|A^3|\le K|A|$ proves

In particular $|A^5|\le K^3|A|$. Apply the covering lemma to $C=A^4$ and $D=A$. There is $X\subseteq A^4$ with $|X|\le K^3$ such thatThe set $A^2$ is symmetric and contains the identity element, so this inclusion is exactly the covering condition showing that

Small doubling alone is insufficient in a noncommutative group. Let $H$ be a finite group, let $G=H\ast ⟨x⟩$ be its free product with an infinite cyclic group, and putThen $A$ is symmetric and $|A^2|\le 5|H|+4=O(|A|)$, while $A^3$ contains the double coset $HxH$. Distinct pairs $(h_1,h_2)\in H^2$ give distinct reduced words $h_{1}x{h}_2$, so $|HxH|=|H{|}^2$. Letting $|H|\to \infty$ proves the small doubling does not control tripling in a noncommutative group phenomenon.

The Mackey's theorem, or Mackey restriction formula, says that if $H,K\le G$, $W$ is a complex $K$-representation, and $D$ is a set of representatives for the double cosets $H\backslash G/K$, thenHere ${}^{x}W$ is the representation of the conjugate subgroup $xK{x}^{-1}$ defined by

For the proof, use the group-algebra modelThe partition of $G$ into double cosets gives the direct-sum decomposition of $(H,K)$-bimodulesPut $L_x=H\cap xK{x}^{-1}$. For each $x$, defineIf $\ell =xk{x}^{-1}\in L_x$, thenso $T_x$ is well defined. It is visibly $H$-equivariant and surjective, and choosing representatives for $H/L_x$ shows that both sides have dimension $[H:L_x]\dim W$; hence it is an isomorphism. Summing the $T_x$ proves the formula.

The Frobenius reciprocity theorem states that for an $H$-representation $U$ and a $G$-representation $V$,For the corresponding character of a representation, this is

Work with the selected transitive orbit $X\cong G/H$. By Maschke's theorem, decompose its permutation character aswhere the ${\chi }_j$ are distinct nontrivial irreducible characters. Character orthogonality givesConsequently the augmentation summand $V$ is irreducible exactly when this inner product equals $2$.

The double cosets $H\backslash G/H$ are the $H$-orbits on $G/H$. Because $G$ acts transitively on $X$, these in turn correspond to the $G$-orbits on $X\times X$: move the first coordinate to the base point $H$, after which its stabilizer is $H$. One orbit is the diagonal. There are exactly two orbits precisely when $H$ is transitive on $X\setminus \{H\}$, equivalently when the action is two-transitive. Part (ii) therefore provesThis is the irreducible augmentation criterion for a transitive group action.

Finally, transitivity implies that the fixed-vector space $\mathbb{C}[X{]}^G$ consists only of constant vectors and has dimension one. If the nonzero representation $V$ were a trivial representation, it would contribute another fixed vector, a contradiction. Thus $V$ is not the trivial representation.

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

Extend the slash operator for modular forms to positive-determinant matrices byThe double cosethas left-coset representativesTherefore the sum of the corresponding slashes, multiplied by $p^{k/2-1}$, is exactly

Right multiplication by an element of $\Gamma (1)$ permutes these left cosets. The cocycle law for the slash operator consequently givesso $T_p(f)$ is a weight-$k$ level-one modular function. Each displayed summand is holomorphic on $\mathfrak{h}$, hence so is their finite sum. This is the Hecke operator on modular forms.