Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford

Source: Mathematics release, bundle Part C 2007 11.pdf, pages 309–312. This combined paper was sat on 8 June 2010, 2.30–5.30pm, for Mathematics, Mathematics and Computer Science, and Mathematics and Philosophy Part C.

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We prove Lie's theorem in the required form by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise solvability gives $[L,L]\ne L$, so choose a codimension-one ideal $K\supseteq [L,L]$ and $x\in L\setminus K$. By induction there are $v\ne 0$ and a linear functional $\lambda :K\to \mathbb{C}$ with $av=\lambda (a)v$ for all $a\in K$.

Let $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Choose the cyclic basis $v,xv,\dots ,x^{r-1}v$ up to its first linear dependence. The commutator identity $a{x}^j=xa{x}^{j-1}+[a,x]x^{j-1}$, with $[a,x]\in K$, proves inductively thatHence $W$ is invariant under $K$ and $x$, and every $a\in K$ has trace $r\lambda (a)$ on $W$. In particular ${\mathrm{tr}}_W([x,a])=0$ implies $r\lambda ([x,a])=0$. Characteristic zero gives $\lambda ([x,a])=0$.

The simultaneous weight space $V_{\lambda }=\{w:aw=\lambda (a)w\forall a\in K\}$ is nonzero and stable under $x$, becauseOver $\mathbb{C}$, the restriction of $x$ to this nonzero finite-dimensional space has an eigenvector. That vector is an eigenvector of both $x$ and every $a\in K$, hence of every element of $L$.

Let $\pi :L\to L/I$ be the quotient map. A full flag of ideals in $L/I$ lifts by inverse images to a flag from $I$ to $L$, with dimensions increasing by one. Prepending $0<I$ supplies the required full flag in $L$. Thus supersolubility lifts across a one-dimensional ideal.

Over $\mathbb{C}$, a nonzero soluble $L$ has a one-dimensional ideal: apply part (a) to its adjoint image acting on $L$. A common eigenvector spans an adjoint-invariant line, which is an ideal. The quotient is soluble and has smaller dimension, so induction and the lifting result give a full flag of ideals. Therefore a soluble complex Lie algebra is supersoluble.

Conversely this implication needs no field restriction. If $L$ has a full flag, its first nonzero ideal $I$ is one-dimensional and abelian, and $L/I$ has a full flag. By induction $L/I$ is soluble, so some term of the derived series of $L$ lies in $I$; its next derived term is zero. Hence

Let $A$ be the two-dimensional affine Lie algebra with basis $u,v$ and bracket $[u,v]=v$, and putBoth factors have zero centre, so $Z(L)=0$. The line $\mathbb{C}v$ in the second factor is an ideal, making $L$ PSS. But its quotient onto the nonabelian simple algebra ${\mathfrak{sl}}_2$ is not soluble. A quotient of a soluble algebra would be soluble, so $L$ is not soluble.

If $L$ is abelian and nonzero, $L^{\prime }=0<L$. Otherwise choose $x\notin Z=Z(L)$ such that $Z+Fx$ is an ideal. For each $a\in L$ writeThe coefficient $\lambda (a)$ is uniquely defined and linear. The Jacobi identity givesso $\lambda ([a,b])=0$. If $\lambda \ne 0$, then $L^{\prime }\subseteq \ker \lambda <L$.

If $\lambda =0$, the same equation says $[[a,b],x]=0$ for all $a,b$. Hence $L^{\prime }\subseteq C_L(x)$. Since $x$ is not central, its centralizer is proper. Thus in both casesNo algebraic closure or characteristic restriction was used.

Choose a basis $(u,v)$ with $U=Fu$. The representation matrices areThe one-dimensional modules $U,V/U$ have characters $\lambda ,\mu$. If they coincide, these modules are isomorphic and the first alternative holds.

Otherwise choose $h$ with $\delta (h)\ne 0$, where $\delta =\lambda -\mu$. Replacing $v$ by $v-a(h)u/\delta (h)$ makes the upper-right entry of $\rho (h)$ zero. The representation identity on commutators now gives, for every $x\in H$,Iteration yields $a((\mathrm{ad}h{)}^{r}x)=\delta (h{)}^{r}a(x)$. Nilpotence of $H$ implies $(\mathrm{ad}h{)}^{r}x=0$ for sufficiently large $r$, by its lower central series. Since $\delta (h)\ne 0$, it follows that $a(x)=0$ for all $x$. The new line $W=Fv$ is therefore a submodule, andThe argument works over the stated arbitrary field, including positive characteristic.

The flag makes every composition factor one-dimensional, described by a character $\lambda :H\to F$. The Jordan–Hölder theorem for modules for finite-length modules says that the multiset of simple composition factors is independent of the chosen composition series.

Here is a construction valid over every $F$. For $h\in H$, every generalized eigenspace of $\rho (h)$ is an $H$-submodule. Indeed $\mathrm{ad}(h)$ is nilpotent on $H$, say of exponent $c$, andIf $v$ is killed by $(\rho (h)-aI{)}^{N_0}$, take $N\ge N_0+c$. Terms with $j\ge c$ vanish by adjoint nilpotence, and all other terms kill $v$. This proves invariance of the generalized eigenspace.

The flag also makes the characteristic polynomial of every $\rho (h)$ split over $F$. Whenever two composition characters differ, choose an $h$ on which they differ and split the module into the distinct generalized eigenspaces of $\rho (h)$. Repeat within each summand until all its composition factors have the same character. Every nontrivial split lowers dimensions, so the process terminates. Finally combine summands carrying the same character. Jordan–Hölder ensures that the resulting distinct blocks have distinct factor types, givingThis recursive method does not assume that a single element separates every character, which could fail over a finite field.

Writing matrices in $2\times 2$ blocks, the defining relation givesThe trace is automatically zero. The relation is linear, and if $as=-s{a}^T$, $bs=-s{b}^T$, thenThus $L$ is closed under the commutator and is a Lie subalgebra of ${\mathfrak{sl}}_4(F)$.

Its diagonal subalgebra isPut ${\epsilon }_1(D(a,b))=a$, ${\epsilon }_2(D(a,b))=b$. Since $[D,E_{ij}]=(D_{ii}-D_{jj})E_{ij}$, the required block decomposition isThe four lines have respective characters ${\epsilon }_1-{\epsilon }_2$, $-{\epsilon }_1+{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $-{\epsilon }_1-{\epsilon }_2$. They are distinct and nonzero because $\mathrm{char}F\ne 2$. The two-dimensional summand $H$ has two trivial composition factors. These five submodules therefore have exactly the factor-grouping properties of part (b). Their dimensions sum to $6$, the dimension obtained from four free entries in $A$ and one each in the skew-symmetric blocks $B,C$.

A reduced crystallographic root system $\Phi$ is a finite spanning subset of a real Euclidean vector space, excluding zero, such that $\Phi \cap \mathbb{R}\alpha =\{\pm \alpha \}$, each reflection $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ preserves $\Phi$, and all Cartan integers $⟨\beta ,\alpha ⟩=2(\beta ,\alpha )/(\alpha ,\alpha )$ are integers. A base is a vector-space basis $\Delta \subset \Phi$ in which each root has integer coordinates either all nonnegative or all nonpositive. The Weyl group is generated by the root reflections.

For nonparallel roots put $p=⟨\beta ,\alpha ⟩$ and $q=⟨\alpha ,\beta ⟩$. Then $pq=4{\cos }^2\theta$ is an integer strictly below $4$ and at least zero. ThereforeIf $(\alpha ,\beta )>0$, the positive integers $p,q$ have product $1,2$ or $3$, so at least one is one. If $p=1$, reflection gives $\beta -\alpha \in \Phi$ and hence its negative. If $q=1$, reflection directly gives $\alpha -\beta \in \Phi$. This proves the positive-inner-product assertion.

To prove that an $\alpha$-string has no gaps, suppose consecutive occurring integer positions $a<b$ in $\beta +t\alpha$ had $b-a\ge 2$. If $(\beta +a\alpha ,\alpha )<0$, apply the previous result to $\beta +a\alpha$ and $-\alpha$ to obtain the missing root at $a+1$. Thus this inner product must be nonnegative. Similarly the absence of a root at $b-1$ forces $(\beta +b\alpha ,\alpha )\le 0$. But the latter inner product exceeds the former by $(b-a)(\alpha ,\alpha )>0$, impossible. Thus every intermediate integer position occurs.

Write the complete finite string as $\beta -r\alpha ,\dots ,\beta +s\alpha$. Reflection maps position $t$ to $-p-t$, so it interchanges the two endpoints. HenceHere $r,s$ are the maximal endpoint distances, not arbitrary shorter distances mentioned in the existence assertion.

Use the Cartan-matrix convention $C_{ij}=⟨{\alpha }_i,{\alpha }_j⟩$. The displayed matrix is realized bywith the standard orthonormal coordinates. Closure under the two simple reflections givesThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2$. Conversely the standard fact that every root is Weyl-conjugate to a simple root, or successive height reduction by simple reflections, shows these two reflection orbits exhaust the roots. This is the rank-two system $B_2$, also called $C_2$ after exchanging the length convention. With a transposed Cartan-matrix convention, interchange which simple-root label is short; the geometric system and Weyl group are unchanged.

The reflection in $e_2$ flips the second coordinate, while the reflection in $e_1-e_2$ swaps the two coordinates. These generate all eight signed permutation matrices in two coordinates. Thuswhere $D_8$ means the symmetry group of a square.

For a Young diagram cell $(i,j)$, its hook length is one plus the number of cells to its right and below. The hook-length formula isApplying it to the two shapes givesThe shapes here require $n\ge 4$.

A hook tabloid remembers its first-row set and the ordered singleton entries below it. The proposed map keeps that first-row set and merges the remaining entries into one unordered row. Thus it is well-defined on tabloids. Permuting entries before merging gives the same tabloid as merging before permuting, so it is a module homomorphism. Every target tabloid is reached by putting its second-row entries, in any order, into the singleton rows. Therefore it is surjective.

For $r=2$, the strictly dominating partitions of $\lambda =(n-2,1,1)$ are exactly $(n),(n-1,1),(n-2,2)$. Indeed dominance forces the first part to be at least $n-2$, and the possibilities for the remaining sum give precisely those shapes or $\lambda$ itself.

In characteristic zero, Maschke's theorem makes the permutation module semisimple, and Young's rule gives one copy of $S^{\lambda }$ and only strictly dominating additional shapes. The trivial multiplicity is one, since a transitive permutation module has a one-dimensional invariant space. For the standard representation $S^{(n-1,1)}$, Frobenius reciprocity identifies its multiplicity with the dimension of fixed vectors under the row stabilizer $S_{n-2}$. In its realization $\{(z_1,\dots ,z_n):\sum z_i=0\}$, those fixed vectors have the form $(c,\dots ,c,a,b)$ with $(n-2)c+a+b=0$, a two-dimensional space. Thus its multiplicity is two.

Finally $\dim M^{\lambda }=n(n-1)$. Subtracting the dimensions of one $S^{\lambda }$, one trivial module and two standard modules leaves $n(n-3)/2$, exactly one $S^{(n-2,2)}$. ThereforeThe additional multiplicities are $1,2,1$ respectively. If the printed weak summation index is read literally alongside its separate $S^{\lambda }$, then $m_{\lambda ,\lambda }=0$; the total multiplicity of $S^{\lambda }$ is one.

A tabloid is the row-equivalence class $\{t\}$ of a tableau filled bijectively with $1,\dots ,n$: order inside each row is forgotten, but the rows remain distinguished. The permutation module $M^{\lambda }$ has these tabloids as an $F$-basis, with ${\Sigma }_n$ permuting their entries. Let $C_t$ be the column stabilizer of $t$ and defineThe Specht module $S^{\lambda }$ is the submodule spanned by all polytabloids $e_t$.

If a module homomorphism $\varphi :M^{\lambda }\to M^{\mu }$ is nonzero on $S^{\lambda }$, it is nonzero on some spanning $e_t$. Equivariance gives $0\ne \varphi (e_t)={\kappa }_t\varphi (\{t\})$, so ${\kappa }_t{M}^{\mu }\ne 0$. The allowed criterion yields $\lambda \ge \mu$.

In characteristic zero, Maschke's theorem gives an equivariant projection $M^{\lambda }\to S^{\lambda }$. If $S^{\lambda }\cong S^{\mu }$, compose that projection with the isomorphism and the inclusion into $M^{\mu }$. The preceding argument gives $\lambda \ge \mu$. Reversing the roles gives $\mu \ge \lambda$, so antisymmetry of dominance givesFor a map defined only on a Specht module in arbitrary characteristic, dominance need not hold. Take $F={\mathbb{F}}_2$, $n=2$, $\lambda =(1,1)$ and $\mu =(2)$. The first Specht module is the one-dimensional span ofThe transposition fixes $e$, so this is the trivial module. The one-row permutation module $M^{(2)}$ is also trivial and one-dimensional. Sending $e$ to its single tabloid is a nonzero equivariant map, yet $(1,1)\ngeqq (2)$. In this characteristic such a map need not extend to the surrounding permutation module; that is exactly where the earlier proof ceases to apply.

Let ${\pi }^{\lambda }$ be the permutation character of $M^{\lambda }$. Young's rule gives ${\pi }^{\lambda }={\chi }^{\lambda }+{\sum }_{\mu >\lambda }{K}_{\mu \lambda }{\chi }^{\mu }$, with integer Kostka multiplicities. This dominance-triangular system has diagonal entries one, so recursive subtraction expresses every irreducible character as an integer linear combination of permutation characters. The latter count fixed tabloids, proving that all ${\chi }^{\lambda }(g)$ are integers.

For a $k$-cycle, the Murnaghan–Nakayama rule reduces towhere height is the number of occupied rows minus one, and $f^{\nu }$ is the degree of the remaining Specht character at the identity. In particular a full $n$-cycle has character zero unless $\lambda =(n-r,1^r)$ is a hook, in which case the value is $(-1{)}^r$.

There is more than one solution: the trivial representation $\lambda =(6)$ has character $1$ everywhere and the sign representation $\lambda =(1^6)$ has character $\pm 1$ everywhere. ThusIn fact they are the only ones. The full-cycle formula in the preceding discussion forces any candidate to be a hook $(6-r,1^r)$. For $r=1$, the standard character is the number of fixed points minus one, so it vanishes on cycle type $(5,1)$; its sign twist $r=4$ does too. For $r=2$, the hook representation is the exterior square of the standard representation and has character $\frac{1}{2}(\chi (g{)}^2-\chi (g^2))$. At cycle type $(4,1,1)$ both standard-character values are one, so this character is zero. Its sign twist $r=3$ also vanishes there. Only $r=0,5$ remain.

Irreducible character orthogonality gives $⟨\chi ,\chi ⟩=1$. Under the proposed vanishing condition, only the identity contributes, soBut $\chi (1)$ is an integer dimension and $720$ is not a square. Thus no such partition exists.

The full-cycle formula says that precisely the nonhook shapes have zero value on a 6-cycle. These are $(4,2),(3,3),(3,2,1),(2,2,1,1),(2,2,2)$.

Use the standard two-row permutation-character rule ${\pi }^{(6-r,r)}={\chi }^{(6)}+{\chi }^{(5,1)}+\cdots +{\chi }^{(6-r,r)}$. A transposition has four fixed points, so ${\chi }^{(5,1)}=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{2}\right)+1=7$ two-subsets, giving ${\chi }^{(4,2)}=7-1-3=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)=8$ three-subsets, giving ${\chi }^{(3,3)}=8-1-3-3=1$.

Conjugating a partition tensors its character with the sign character. Hence the values on a transposition for the conjugate shapes are $-3$ and $-1$. The shape $(3,2,1)$ is self-conjugate, so its character equals its own sign twist and is zero on every odd permutation. The complete candidate table is therefore:

Thus the unique answer is

First prove the unlettered common-eigenvector assertion, in the stronger form that some nonzero vector is killed by every operator in $W$. This is the linear form of Engel's theorem.

Induct on $\dim W$, with the zero algebra immediate. If $a^r=0$, the operator $\mathrm{ad}a$ on $\mathrm{End}V$ is nilpotent: left and right multiplication commute, and every term in $(L_a-R_a{)}^{2r-1}$ contains a power of $a$ at least $r$. This argument works in every characteristic.

For a proper Lie subalgebra $H<W$, its action on $W/H$ by brackets is therefore by nilpotent operators. The image of this action has dimension at most $\dim H<\dim W$, so induction gives a nonzero class $w+H$ killed by $H$. Thus the normalizer $N_W(H)=\{w:[w,H]\subseteq H\}$ strictly contains $H$. Choose $H$ maximal proper. Its normalizer must be $W$, so $H$ is an ideal. Moreover $W/H$ has dimension one, since otherwise any one-dimensional subalgebra of the quotient would lift to an intermediate proper subalgebra.

Induction applied to $H$ acting on $V$ makes $V_0=\{v:Hv=0\}$ nonzero. This space is $W$-invariant: for $h\in H,x\in W,v\in V_0$, $h(xv)=x(hv)+[h,x]v=0$. Choose $x$ complementing $H$ in $W$. Its restriction to the nonzero space $V_0$ is nilpotent and has a nonzero kernel. A vector in this kernel is killed by $x$ and $H$, hence by all of $W$. This proves the preliminary assertion, with every common eigenvalue equal to zero.

Now apply it to the adjoint action of $L$ on itself. If $L\ne 0$, its common zero space is the center $Z(L)$, which is therefore nonzero. Every induced adjoint map on $L/Z(L)$ remains nilpotent. Induction on dimension makes that quotient nilpotent. A central extension of a nilpotent Lie algebra is nilpotent: if its lower central series vanishes after projection, the next term lies in the center and the following term is zero. Hence $L$ is nilpotent. The zero algebra is immediate.

Every subalgebra of a nilpotent Lie algebra is nilpotent, since its lower central series is contained term by term in the ambient lower central series.

Conversely, if $L$ were not nilpotent, part (a) would give $x$ for which $\mathrm{ad}x$ is not nilpotent. Over an algebraically closed field, such an endomorphism has a nonzero eigenvalue $\lambda$ and eigenvector $y$. The vectors $x,y$ are independent because $[x,x]=0$ while $[x,y]=\lambda y\ne 0$. Their span is a two-dimensional subalgebra, and its lower central series is $Ky$ at every stage after the first. It is therefore not nilpotent, a contradiction. Thus

Polarizing $[x,[x,z]]=0$ givesLet $T=[x,[y,z]]$. Antisymmetry and this identity give $[y,[z,x]]=T$ and $[z,[x,y]]=T$. The Jacobi identity therefore gives $3T=0$. Since the characteristic is not three, $T=0$ for every triple. ConsequentlyNo division by two was used, so the proof also applies in characteristic two with the usual alternating definition of a Lie bracket.

The Killing form is ${\kappa }_L(x,y)=\mathrm{Tr}(\mathrm{ad}x\mathrm{ad}y)$. Write $X=\mathrm{ad}x$, $Y=\mathrm{ad}y$, $Z=\mathrm{ad}z$. Since $\mathrm{ad}[x,y]=[X,Y]$, cyclicity of trace givesIf $x\in I^{\perp }$, $a\in L$ and $y\in I$, invariance gives ${\kappa }_L([a,x],y)={\kappa }_L(x,[y,a])=0$, because $[y,a]\in I$. Thus $I^{\perp }$ is an ideal. For $x,y\in I$, the restrictions of their adjoint maps to $I$ are the adjoint maps of $I$, and their induced maps on $L/I$ vanish. Block traces therefore give ${\kappa }_L{|}_{I\times I}={\kappa }_I$.

If $A$ is an Abelian ideal, $\mathrm{ad}a\mathrm{ad}x$ maps $L$ into $A$ and vanishes on $A$ for $a\in A$. In a basis starting with $A$ it has zero diagonal blocks, hence trace zero. Therefore every Abelian ideal lies in $L^{\perp }$.

Cartan criterion for semisimplicity says that a finite-dimensional complex Lie algebra is semisimple exactly when its Killing form is nondegenerate. To see why nondegeneracy implies semisimplicity, any nonzero solvable ideal has a last nonzero derived term which is an Abelian ideal of $L$. The previous paragraph would put it in the radical of the Killing form, a contradiction.

If $L$ is a semisimple ideal of $M$, the restriction ${\kappa }_M{|}_L={\kappa }_L$ is nondegenerate. Put $J=L^{{\perp }_{{\kappa }_M}}$. This is an ideal of $M$ by the argument above, and nondegeneracy on $L$ gives $L\cap J=0$ and $M=L+J$. One way to check the latter is to restrict the pairing map $M\to L^{\ast }$: its restriction to $L$ is an isomorphism, so every vector of $M$ can be adjusted by an element of $L$ to lie in its kernel $J$. Since both are ideals, $[L,J]\subseteq L\cap J=0$. HenceApply this to the supplied extension by the derivation. If $\delta =0$, it is $\mathrm{ad}0$. Otherwise $D$ and $J$ are one-dimensional, and projection $J\to D$ is an isomorphism. Choose its vector with $D$-component $\delta$, writing it as $j=\delta -x$ for $x\in L$. Because $[J,L]=0$, for every $l\in L$ one has $0=[j,l]=\delta (l)-[x,l]$. ThusThis also handles the sign in the extension bracket explicitly.

A base of a root system is a vector-space basis $\Delta$ of the ambient Euclidean space such that every root is an integer combination of its elements with all coefficients nonnegative or all nonpositive. The corresponding nonzero nonnegative combinations are the positive roots.

If distinct $\alpha ,\beta \in \Delta$ had positive inner product, the allowed root-difference fact would make $\alpha -\beta$ a root. Its simple-root coefficients have opposite signs, contradicting the definition. Hence distinct simple roots have nonpositive inner product.

For a positive root $\gamma =\sum n_{\alpha }\alpha$ which is not simple, the identity $(\gamma ,\gamma )=\sum n_{\alpha }(\gamma ,\alpha )>0$ gives a simple root $\alpha$ with $n_{\alpha }>0$ and $(\gamma ,\alpha )>0$. The allowed fact makes $\gamma -\alpha$ a root. Its coefficients remain nonnegative, and it is nonzero because $\gamma$ is not simple. It has smaller height $\sum n_{\alpha }$. Induction expresses it as a sum of simple roots with each partial sum a root; appending $\alpha$ gives the required expression for $\gamma$.

Finally, reflection in a simple root changes only that root's coefficient:For positive $\gamma \ne \alpha$, at least one other simple coefficient is positive: in a reduced root system the only positive root on the line of $\alpha$ is $\alpha$ itself. That other coefficient is unchanged. The reflected vector is a root, so its coefficients cannot all be nonpositive and must all be nonnegative. Thus ${\sigma }_{\alpha }$ preserves all positive roots except $\alpha$, which it sends to $-\alpha$.

Use the column-coroot convention $C_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, consistent with the printed matrix and the stated shorter first root. It givesThe root string through $\beta$ in the $\alpha$ direction contains $\beta ,\beta +\alpha ,\beta +2\alpha ,\beta +3\alpha$. Reflecting $3\alpha +\beta$ in $\beta$ gives $3\alpha +2\beta$. Together with $\alpha$ and their negatives these giveThere are no further roots. Indeed the simple reflections act on coefficient pairs by $(a,b)\mapsto (-a+3b,b)$ and $(a,b)\mapsto (a,a-b)$; the displayed twelve-element set is closed under both. Every positive nonsimple root can be reflected to a positive root of smaller height: choose a simple root with positive inner product as in (a), and its positive integral Cartan pairing decreases height. Repetition reaches a simple root, so every root lies in an orbit already contained in the list. This is the G2 root system.

With $\alpha =e_1$, the inner product and length give the two reflected possibilities

The Young diagram is $[\lambda ]=\{(i,j):1\le i\le k,1\le j\le {\lambda }_i\}$. The hook length of a box iscounting the box itself, the boxes to its right in the same row, and the boxes below it in the same column. The hook-length formula isThe first implication in the equivalence is immediate: if every hook length is odd, no hook can have length two.

Assume no hook has length two. The last row has length one, since otherwise its penultimate box has hook length two. If two adjacent row lengths are equal, take the last two rows in the maximal block of that length; the last box in the upper of those two rows has hook length two. Thus all row lengths are distinct. If a gap ${\lambda }_i-{\lambda }_{i+1}$ were at least two, the penultimate box in row $i$ would again have hook length two, since there would be no boxes below it. Therefore every successive gap equals one and the final part equals one, proving (2) implies (3).

The final condition makes $\lambda =(k,k-1,\dots ,1)$. Its column lengths are the same staircase, sowhich is odd. This proves (3) implies (1) and completes the equivalence. Such a partition exists exactly at the triangular numbersThe hook formula gives $v_2(f^{\lambda })=v_2(n!)=a$, since its denominator is odd. Thus $2^a$ divides $f^{\lambda }$. Conversely $2^a\mid f^{\gamma }$ implies $v_2(\prod h_{\gamma })=a-v_2(f^{\gamma })\le 0$. The product is an integer, so its valuation is zero and all its hooks are odd. By the equivalence $\gamma$ is a staircase; its size uniquely determines its length, so $\gamma =\lambda$.

For the character assertion, every removable rim hook has length equal to some hook length of the diagram. Here is a direct justification. If a border strip $\lambda /\mu$ occupies rows $i$ through $j$, connectedness and absence of a two-by-two block make adjacent strips of boxes overlap in exactly one column. Hence ${\mu }_l={\lambda }_{l+1}-1$ for $i\le l<j$, so its size is ${\lambda }_i-{\mu }_j+j-i$. The column $c={\mu }_j+1$ has height $j$, giving exactly $h_{\lambda }(i,c)={\lambda }_i-{\mu }_j+j-i$. Therefore every removable rim hook of the staircase has odd length.

A fixed-point-free permutation of power-of-two order has only even cycles. Apply the Murnaghan–Nakayama rule starting with any cycle. It would require removal of a rim hook of that even length, but none exists. The sum is empty, soFinally suppose $\gamma$ has distinct parts and is not $\lambda$. It cannot be another staircase of the same size. Thus some gap between successive rows is at least two, or the last row has length at least two; in either case a horizontal rim domino is removable. No vertical rim domino is removable, since that would require two equal adjacent row lengths. Murnaghan-Nakayama on a transposition consequently gives a nonempty sumall signs being positive and all tableau counts positive. Thus the character is nonzero at every transposition, proving the requested existence.

For a tableau $t$, let $C_t$ be its column stabilizer and $e_t={\sum }_{\sigma \in C_t}\mathrm{sgn}(\sigma )\{\sigma t\}$ its polytabloid. The Specht module is the span of all such polytabloids inside the tabloid permutation module.

A tabloid with second row of size $k$ is determined by that subset, so $\dim M^{(n-k,k)}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The two-row hook formula givesAt $k=3$ this provesBy Maschke's theorem, characteristic-zero permutation modules are completely reducible, so their Specht submodules are summands. The nonzero sum of all three-subset tabloids is fixed by every permutation; its span is a copy of the trivial module $S^{(n)}$.

A permutation sends $v_a$ to $v_{ga}$, $w_{ab}$ to $w_{ga,gb}$ and $x_{abc}$ to $x_{ga,gb,gc}$. Relabeling the summation indices givesLinearity therefore proves both maps are module homomorphisms. The map denoted $\varphi$ in the second displayed source formula is ${\varphi }_2$.