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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, 2018 at 8:40
  • $\begingroup$ Can you give details please? $\endgroup$
    – user372565
    Feb 8, 2018 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, 2018 at 10:12
  • $\begingroup$ Your surmise is correct. A confusing aspect is that in various contexts the notion of "representation" implicitly includes "finite-dimensional"... so we don't know whether infinite-dimensional repns are excluded "by def'n", or because there aren't any... $\endgroup$
    – paul garrett
    Jan 24 at 23:28

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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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    $\begingroup$ Interestingly, for a finite group over a field with characteristic dividing the order such that the representation type is infinite (and this happens iff the Sylow subgroup is not cyclic) there are infinite dimensional indecomposable representations. This is a theorem of Auslander. $\endgroup$
    – Mariano Suárez-Álvarez
    Jan 25 at 0:10
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You can prove stronger statement: every complex irreducible representation of a compact group is finite dimensional. Since finite group is compact, the statement for finite group follows. Here is a good reference for the subject: https://en.wikipedia.org/wiki/Compact_group

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