= Solution
For $n\ge2$, a <cycle type> $\alpha\vdash n$ lies in $A_n$ exactly when $n-\ell(\alpha)$ is even. Its $S_n$ <conjugacy class> is either one $A_n$ class or two equal-sized classes. Splitting occurs exactly when its <centralizer> in $S_n$ contains no odd <permutation>. An even-length cycle is itself odd; two equal odd-length cycles can be interchanged by an odd <permutation>. Conversely, for distinct odd cycle lengths the <centralizer> is a product of cyclic groups of odd order, all contained in $A_n$. Thus \b[the classes that split are exactly the <partitions of an integer> into distinct odd parts]. Repeated fixed points count as repeated parts of length one.
For $n=1$ the group is trivial, with its single <trivial representation>. Assume $n\ge2$ for the index-two argument that follows. For ordinary <representations> work over $\mathbb C$. Tensoring $S^\lambda$ by the <sign representation> gives $S^{\lambda'}$. The index-two restriction identity, obtained by <Frobenius reciprocity>, is
$$
\left\langle\operatorname{Res}_{A_n}\chi^\lambda,\operatorname{Res}_{A_n}\chi^\mu\right\rangle
=\delta_{\lambda\mu}+\delta_{\lambda'\mu}.
$$
If $\lambda\ne\lambda'$, restriction is <irreducible>, and the conjugate pair gives the same <irreducible>. If $\lambda=\lambda'$, the norm is two, so restriction is the sum of two distinct <irreducibles>. An odd <permutation> interchanges them, so both have degree $f^\lambda/2$. Every <irreducible> of $A_n$ occurs in some such restriction: induce it to $S_n$ and choose an <irreducible> constituent, then apply reciprocity. The displayed <inner product> distinguishes all the listed constituents. Hence
$$
\boxed{\operatorname{Irr}(A_n)=\{V^{\{\lambda,\lambda'\}}:\lambda\ne\lambda'\}\ \cup\ \{V^{\lambda,+},V^{\lambda,-}:\lambda=\lambda'\}.}
$$
The first family has degree $f^\lambda$, the second degree $f^\lambda/2$. A choice of labels for the two split classes and constituents fixes the otherwise interchangeable signs.
The one-dimensional <representations> are <characters> of the <abelianization>. For $n\ge5$, simplicity and noncommutativity of $A_n$ make its <abelianization> trivial. For $n=4$, its <commutator subgroup> is the <Klein four-group> and its <abelianization> is $C_3$, giving the three <characters> obtained by sending a quotient generator to $1,\omega,\omega^2$. For $n=3$, $A_3=C_3$ has the same three <characters>; for $n=1,2$ the group is trivial and has just one. The two additional <linear characters> at $n=4$ are the split constituents of shape $(2,2)$; at $n=3$ they come from $(2,1)$.
Here is a <dimension> argument that also proves the asserted uniqueness without assuming a classification of small <character> degrees. Write $f^\lambda$ for the number of <standard tableaux>. Removing the cell containing the largest label gives
$$
f^\lambda=\sum_{\mu\in\lambda^-}f^\mu.
$$
<Hook lengths>, or direct <Young tableau> counting, give the following small cases: for $S_5,S_6$ every non-linear degree is at least $4,5$, respectively. Among non-self-conjugate shapes excluding the row, column and standard pair, the minimum degree at $n=5,6,7$ is respectively $5,5,14$. The self-conjugate shapes at these sizes have half-degrees $3,8,10$, respectively, from $(3,1^2),(3,2,1),(4,1^3)$.
Inductively, for every $n\ge7$, a shape other than the row, column or standard pair has all its one-cell predecessors non-linear. If it has at least two removable corners, each predecessor has degree at least $n-2$, so its degree is at least $2(n-2)$. If it has only one corner, it is a nontrivial rectangle $(a^b)$. Remove that corner and then either of the two corners of $(a^{b-1},a-1)$. For $n\ge8$ these two size-$n-2$ shapes are non-linear, so its degree is at least $2(n-3)$. At $n=7$ there is no nontrivial rectangle. Together with the size-5 and size-6 checks this proves by induction that the standard pair is the only pair of shapes of degree $n-1$ for $n\ge7$, and that every other non-linear degree is larger.
We must additionally exclude a self-conjugate shape of degree $2(n-1)$, since its restriction splits. For $n\ge8$, if such a shape has at least three corners, the preceding lower bound gives $f^\lambda\ge3(n-2)>2(n-1)$. With two corners, its predecessors are a conjugate pair and are not standard shapes; the bounds just proved give $f^\lambda\ge4(n-4)>2(n-1)$. With one corner it is a square $(a^a)$. For $a\ge4$, the two size-$n-2$ predecessors obtained after two removals are nonstandard, giving $f^\lambda\ge4(n-5)>2(n-1)$. The remaining square $(3^3)$ has degree $42>16$ by the <hook-length formula>. The size-7 self-conjugate case has degree $20>12$. This excludes both smaller split degrees and equality with the standard degree.
It follows that the lowest non-linear ordinary degree is \b[$3$ for $A_4$ and $n-1$ for every $n\ge6$]. For $A_1,A_2,A_3$ no non-linear <irreducible> exists. The excluded case $A_5$ has lowest degree $3$, although its standard degree-$4$ <representation> is unique. At $A_6$ there are two degree-$5$ <irreducibles>, from the standard conjugate pair and the pair $(3,3),(2,2,2)$.
Finally, for every $n\ge7$ the preceding proof gives a unique degree-$n-1$ <irreducible>, namely the restriction of $S^{(n-1,1)}$. The small cases $n=4,5$ have unique degree $3,4$, respectively, and $A_2$ has its unique trivial degree-$1$ <representation>. These prove exactly the requested uniqueness range; the omitted cases $n=3,6$ genuinely fail.
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