From a Young diagram we can build an irreducible complex representation for the symmetric group. It turns out that they are all dinite dimensional and that every other representation of $S_n$ is isomorphic to one constructed via a Young diagram.

So in particular, there are really no infinite dimensional irreducible complex representations of $S_n$?

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    $\begingroup$ There are no infinite dimensional irreducible representations of any finite group $G$ over any field. It is easy to see that such representations have dimension at most $|G|$. $\endgroup$ – Derek Holt Feb 8 '18 at 8:40
  • $\begingroup$ Can you give details please? $\endgroup$ – user372565 Feb 8 '18 at 9:47
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    $\begingroup$ Let the representation act on vector space $V$, and let $0 \ne v \in V$. Then the subspace of $V$ spanned by $\{ gv : g \in G \}$ is invariant under $G$. So if the representation is irreducible then this subspace must equal $V$, and hence $\dim V \le |G|$. $\endgroup$ – Derek Holt Feb 8 '18 at 10:12

Perhaps I should make my comment into an answer.

Let $\rho$ be any representation of any finite group $G$ over any field $K$, and suppose that the representation is acting on the (nonzero) vector space $V$.

Choose any nonzero vector $v \in V$. Then it is straightforward to check that the subspace of $G$ spanned by the vectors $\{ \rho(g)(v) : g \in G \}$ is invariant under the action of $G$ (that's because $G$ permutes the elements of the spanning set). So if the representation is irreducible then it has degree at most $|G|$.

You can go a little further, because if this space has dimension exactly $|G|$, then the spanning set forms a basis, and it has a proper $G$-invariant subspace of codimension $1$, consisting of those vectors for which the coefficient sum is $0$. So any irreducible representation of $G$ has degree at most $|G|-1$.

This bound is attained for example for cyclic groups of prime order $p$ over finite fields of order $q$ such that the multiplicative order of $q$ mod $p$ is $p-1$.


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