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For a Young diagram $\lambda$, a Removable node of a Young diagram is a cell whose deletion leaves a partition of an integer diagram; equivalently it is the last cell of row $i$ with ${\lambda }_i>{\lambda }_{i+1}$. Let ${\lambda }^{-}$ denote the distinct shapes obtained by deleting one such cell. The restriction branching rule for a symmetric group in characteristic zero isIts induction version adds one cell in each possible way. Over an arbitrary field, restriction instead has a Specht filtration with these successive quotients, each once; it need not split.

To prove restriction, list the removable nodes from top to bottom. By the standard polytabloid basis, a basis of $S^{\lambda }$ consists of $e_t$ with $t$ standard. Its entry $n$ is necessarily at a removable node. Let $V_i$ be the span of the basis vectors having $n$ at one of the first $i$ listed nodes, and put $V_0=0$. These are $S_{n-1}$-invariant. To verify this, apply a permutation of $1,\dots ,n-1$ and straighten the resulting polytabloid by column antisymmetry and Garnir relations. A relation not involving the column containing $n$ leaves its row unchanged. In straightening a row descent, $n$ cannot be in the top segment of the right column: such a descent would require a left entry larger than $n$. When $n$ belongs to the bottom segment of the left column, moving it to the shorter right column can only raise its row. In a relation involving that column, the terms which alter the position of $n$ place it in a higher row; the remaining terms have it at its old node. Thus straightening never moves it past the current top-to-bottom cutoff.

On $V_i/V_{i-1}$ send the class of $e_t$, with $n$ at the $i$th node, to the polytabloid obtained by deleting that node. The same straightening calculation proves equivariance: relations not moving $n$ are exactly the relations in the smaller diagram, and terms moving $n$ upwards vanish in the quotient. Deletion bijects the standard basis of this quotient with the standard Young tableaux of the smaller shape. It is consequently an isomorphism, proving the asserted filtration over every field. In characteristic zero, Maschke's theorem splits the filtration and gives the direct sum. Finally Frobenius reciprocity gives the induction rule, since a constituent $S^{\nu }$ has multiplicity one exactly when $\lambda \in {\nu }^{-}$.

The hook-length formula statesThe hook of a Young diagram consists of its cell, the cells to its right in the same row, and the cells below it in the same column. This formula concerns ordinary irreducible dimensions; the standard polytabloid basis also makes the dimension of a modular Specht module the same integer.

All transpositions are conjugate, have order two, and generate $S_n$. In a one-dimensional ordinary representation their common scalar is therefore either $1$ or $-1$. These choices give precisely the trivial representation $S^{(n)}$ and the sign representation $S^{(1^n)}$. At $n=1$ they coincide. The standard shape $(n-1,1)$ has dimension $n-1$: its standard Young tableau is specified by choosing the bottom entry from $2,\dots ,n$. Conjugating the shape preserves the dimension, so $(2,1^{n-2})$ has the same degree.

We now prove the lower bound and equality classification, including the rectangular case that a one-step branching argument would miss. Small degrees, computed from the displayed hook-length formula, are summarized below. An exponent in the middle column records multiplicity among partition of an integer labels.

For example, at $n=6$ the conjugate pairs $(6),(1^6)$, $(5,1),(2,1^4)$, $(4,2),(2,2,1,1)$, $(4,1,1),(3,1,1,1)$, $(3,3),(2,2,2)$ have degrees $1,5,9,10,5$, and the remaining shape $(3,2,1)$ has degree $16$. These checks also identify every small equality case.

Induct on $n\ge 9$, assuming the lower bound for all smaller degrees. If a non-linear, nonstandard shape has at least two removable nodes, none of its predecessors is a single row or single column: adding one cell to a row or column could produce only a linear or standard shape. Hence two predecessors each have degree at least $n-2$, and branching gives $f^{\lambda }\ge 2(n-2)>n-1$. If it has only one removable node, it is a rectangle $(a^b)$ with $a,b\ge 2$. Its sole predecessor is $(a^{b-1},a-1)$, which has two removable nodes. Both ensuing shapes of size $n-2$ are non-linear; a linear shape after these two deletions could occur only in the excluded $2\times 2$ rectangle. Therefore two-step branching gives $f^{\lambda }\ge 2(n-3)>n-1$. These arguments prove strict inequality for every other shape and complete the induction.

Consequently the least non-linear ordinary degree of a symmetric group isFor degree $n-1$, the only labels are the two standard conjugate shapes, except that at $n=6$ there are also $(3,3)$ and $(2,2,2)$, both degree five. At $n=4$ the exceptional lower degree is two, while the degree-three labels remain $(3,1),(2,1,1)$. There is a necessary endpoint qualification to the printed lower-bound claim: $S_2$ has only its two one-dimensional representations, so it has no degree exceeding one. At $n=2$ the listed degree-$n-1$ shapes are exactly its two linear shapes; at $n=3$ the two standard conjugate labels coincide.

The polytabloids $e_t$ indexed by standard Young tableaux of shape $\lambda$ form a basis of $S^{\lambda }$ over every field, and a basis of its integral lattice over $\mathbb{Z}$. Column antisymmetry and Garnir relations straighten any polytabloid into this set. Ordering leading tabloids by the rows containing successive entries makes the corresponding coefficient matrix triangular with diagonal entries one, proving independence. In particular the dimension is the standard-tableau count in every characteristic.