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.

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 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.

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

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 $-\alpha$.

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.

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

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.

For the characteristic-three Specht module for the two-by-two partition, reduce the matrices modulo three. The vector $e_1+e_2$ is fixed by every displayed transposition: $(12)$ and $(34)$ send it to $e_1-2e_2=e_1+e_2$, and $(23)$ interchanges its two terms. Thus its span is a proper nonzero trivial submodule.

On the quotient impose $e_2=-e_1$. Each generating transposition then acts as minus one, so the quotient is the sign representation. Consequentlyand the two composition factors are trivial and sign, each once. In the $D$ notation they are $D^{(4)}$ and $D^{(2,2)}$ respectively. Indeed the restricted tabloid form has Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$, whose characteristic-three radical is precisely the trivial line, so the simple quotient is $D^{(2,2)}$.

The extension does not split: the minus-one eigenspace of $\rho (12)$ is the line through $e_2$, but $\rho (23)$ sends that line to the line through $e_1$. Thus there is no sign submodule complementing the trivial line, consistent with the failure of semisimplicity.

Choose the two standard tableauxand use their polytabloids $e_1,e_2$, with column-vector coordinates. Writing $[ab\mid cd]$ for the tabloid with those two row sets givesThe coefficients at $[12\mid 34]$ and $[13\mid 24]$ prove independence over every field. Applying the generators directly givesand $(34)$ acts in the same way as $(12)$. Their span is invariant under the generators, contains $e_1$, and contains its whole orbit, hence is the Specht module. Thus it is a basis without needing a characteristic-zero argument. The matrices areThese integral matrices may be reduced in the chosen characteristic.

Give $M^{\lambda }$ the invariant symmetric bilinear form for which its tabloid basis is orthonormal. DefineIn characteristic $p>0$, this quotient is nonzero and simple when $\lambda$ is $p$-regular, meaning that no part occurs $p$ or more times. For other partitions it is zero, so the label “simple module” includes the nonzero/$p$-regular qualification.

The James submodule theorem states that for every $F{\Sigma }_n$-submodule $U\subseteq M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Applied to submodules of $S^{\lambda }$, it says that every proper submodule lies in the radical of the restricted form, giving the simplicity of the displayed nonzero quotient.

The stabilizer of an ordered pair is the subgroup ${\Sigma }_{n-2}$ fixing both entries, so the permutation module is ${\mathrm{Ind}}_{{\Sigma }_{n-2}}^{{\Sigma }_n}1$. Induce in two stages through ${\Sigma }_{n-1}$. The first stage is $S^{(n-1)}\oplus S^{(n-2,1)}$. The characteristic-zero induction branching rule for a symmetric group adds one box in every possible partition position, with multiplicity one. Thus the two second-stage decompositions areBy transitivity of induction this yieldsThe dimensions are $1$, $2(n-1)$, $n(n-3)/2$ and $(n-1)(n-2)/2$, whose sum is $n(n-1)$, checking the decomposition also at $n=4$.

Write $v_{ij}$ for the tabloid with second-row entry $i$ and third-row entry $j$. Equivariance gives $\varphi (v_{ij})=w_i+w_j$ for every distinct pair. For a tableau $t$, the only nontrivial column antisymmetrizer is on the three entries in its first column. In its alternating sum, pair each column permutation with the permutation obtained by interchanging the two singleton-row positions. Their images under $\varphi$ are the same unordered sum of two $w$'s, while their signs are opposite. Hence $\varphi (e_t)=0$ for every polytabloid, proving $\boxed{S^{(n-2,1,1)}\subseteq \ker \varphi }$.

For distinct $i,j,k$,Each summand is an image and two is invertible in characteristic zero. Thus every basis vector of $M^{(n-1,1)}$ is in the image, proving surjectivity. rank-nullity theorem then gives

The permutation module $M^{\lambda }$ has a basis consisting of row tabloids of shape $\lambda$: tableaux are identified when each row has the same set of entries. A permutation acts by applying it to every entry, and the action is extended linearly over $F$. Its dimension is $n!/{\prod }_i{\lambda }_i!$.

Here the two singleton rows form an ordered pair of distinct entries, soA standard tableau of the same shape must put one in its top-left corner. Choose two entries from $2,\dots ,n$ for the two remaining boxes of the first column, in increasing order; all other entries then fill the top row in increasing order. Thus the standard-polytabloid basis gives

The two partition diagrams are transposes, so their characteristic-zero Specht modules differ by tensoring with the sign representation. A transposition has sign minus one. ConsequentlyThe middle character is $n-3$, since the standard representation is the point-permutation representation minus its trivial line, and a transposition fixes $n-2$ points.

Write the hook cycle partition as $\mu =(k,1^{n-k})$, $1\le k\le n$. At the identity, the representation dimension is $n-1$: a standard tableau of shape $(2,1^{n-2})$ is determined by the entry in its top-right box, any one of $2,\dots ,n$.

For $2\le k\le n-2$, the only removable border strip of length $k$ is the vertical bottom strip. Its height is $k-1$, and the remaining hook has size $n-k$ and dimension $n-k-1$. A strip of length $n-1$ would leave the top-left box but disconnect the removed top-right box from the removed column, so none exists and the value is zero. For $k=n$, the whole hook is a border strip of height $n-2$ and contributes $(-1{)}^{n-2}$. These cases combine asIn particular the formula includes zero at $k=n-1$ and the full-cycle sign at $k=n$.

If the cycle partition contains a part $r$, the Murnaghan–Nakayama rule isA border strip is a connected skew Young diagram with no $2\times 2$ block; its height is its number of occupied rows minus one. The remaining boxes must be a partition diagram. The recursion starts with character one for the empty diagram at the empty cycle type, and an empty sum is zero. Repeated removal for all cycle lengths computes the character value.

The equality ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2=n!$ exhausts the dimension of the regular representation. There is no room for another irreducible constituent, since it would contribute its positive squared dimension. ThusEach has the standard-polytabloid basis just proved. The usual split characteristic-zero statement follows over any characteristic-zero field because these integral Specht models remain irreducible after extension to an algebraic closure and already exhaust all dimensions there. In the literal wording of the source, polytabloids themselves are vectors; it is the Specht modules with those bases that form the family of representations.

Let $f^{\lambda }$ be the number of standard tableaux of shape $\lambda$. First the associated polytabloids are linearly independent. Order tabloids lexicographically by reading each row after sorting its entries. For a column-increasing tableau $t$, the term $\{t\}$ in $e_t$ has coefficient one; every other tabloid term has larger row word. To see this, take the first row changed by a nontrivial column permutation. Earlier rows are fixed entry by entry, since each entry belongs to one original column. The changed row can only receive entries from lower positions of those increasing columns, so after sorting it is lexicographically larger. In a linear relation between standard polytabloids, choose the smallest row word with a nonzero coefficient. Its own tabloid coefficient cannot be cancelled by a larger standard polytabloid, a contradiction. Thus $\dim S^{\lambda }\ge f^{\lambda }$.

Work first over $\mathbb{C}$. The RSK bijection between permutations and pairs of standard tableaux of the same shape gives ${\sum }_{\lambda ⊢n}(f^{\lambda }{)}^2=n!$. The assumed pairwise nonisomorphic irreducible Specht modules, and the decomposition of the regular representation with irreducible multiplicities equal to their dimensions, give ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2\le n!$. Combining the two inequalities forces equality term by term: $\dim S^{\lambda }=f^{\lambda }$. Hence the independent standard polytabloids are a basis.

The same conclusion holds over every characteristic-zero field. All polytabloid coordinates in the tabloid basis are integers, so the ranks of their coordinate matrices are the same over any characteristic-zero field as over $\mathbb{C}$. In particular the dimension equality and independence transfer without requiring an embedding of the whole field into $\mathbb{C}$.

Use the convention that insertion returns the insertion tableau $P$ first and the recording tableau $Q$ second. Successive row insertions of $3,4,1,6,7,5,2$ giveFor example inserting five bumps six to the second row, and the final insertion of two bumps four from the first row and six from the second, creating the third row. The newly created boxes at these steps receive recording labels six and seven respectively.

The deletion input is this pair with the two tableaux interchanged. The symmetry property of RSK therefore gives the inverse permutation. One can also reverse-insert the boxes in descending recording-label order. The result isIndeed the displayed second row lists the positions of one through seven in the original permutation.