We know that the alternating group of degree $n>7$ has an irreducible character of degree $n-1$. The latter number is the smallest nontrivial one for each the alternating group has an irreducible character of that degree.


Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree $n-1$?



Does the Alternating group of degree $n>7$ have exactly one irreducible character of degree $n-1$?
**Old answer:** You know already that the answer is ``yes.'' For a reference, see result 2 of



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> Rasala, Richard *On the minimal degrees of characters of $S\_n$*. J. Algebra 45 (1977), no. 1, 132–181.
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This gives the answer for $n\geq 9$. The *Atlas of Finite Groups* then gives the result for $n=7,8,9$. (Note that the statement is also true for $n=7$, but not for $n=6$.)


I can email you copies of these sources should you need them.


**New answer:** I was unhappy with the previous answer, because one should be able to answer this using basic facts about the representation theory of the symmetric group. Here is such an answer: my reference is Fulton and Harris' *Representation theory*.


We know that irreducible reps of $S\_n$ are associated to partitions $\lambda$ of $n$. Given such a partition $\lambda$, there are several formulae for the dimension of the associated irreducible rep - see (4.10), (4.11) and (4.12) of F&H. The latter is the hook length formula, which reads:


$$ \dim(V\_\lambda)=\frac{d!}{\prod (\textrm{ Hook lengths})}.$$


Using this formula it is easy to work out that, for $n\geq 7$, there are only four reps of $S\_n$ of degree less than $n$ - two of dim $1$ and two of dim $n-1$ (this is Exercise 4.14 of F&H). It's pretty obvious what they all are - each pair correspond to partitions that are conjugate to each other.


Now one uses Proposition 5.1 of F&H to see which of these remain as irreducibles when one restricts to the alternating group $A\_n$ - the answer is that they remain irreducible so long as the associated partition is not self-conjugate. None of the partitions in question are self-conjugate so they remain irreducible; what is more representations corresponding to partitions that are conjugate yield isomorphic representations when one restricts to $A\_n$ - this yields our irreducible reps for $A\_n$ of dimension $1$ and $n-1$.


Now to complete the proof one needs to check that there are no self-conjugate partitions of $n$ for which the associated irreducible rep has dimension $\leq 2(n-1)$. (If there were these would split in half to yield irreducible representations of $A\_n$.) This is easy.


