# What are the relation between Rep(G) and Rep(S_n)?

Let G be a finite group. We know it can be written as a subgroup of S_n. On the other hand, people sometimes say Rep(G) --- the category of all finite dimensional representations, are more interesting to study.

So my question is, if we pass to Rep(G), how can we relate Rep(G) and Rep(S_n)? I can imagine there will be some functor given by restrictions. But are there other way to connect them?

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Sure. For example, if $V$ is any representation, then $V^{\otimes n}$ is naturally a representation of $S_n$ (in fact this action commutes with that of $G$). But this question strikes me as a little broad. What do you actually want to know? – Qiaochu Yuan Mar 9 '12 at 4:38
@Qiaochu : I can't make this precise, but I want to ask, e.g. Can all Rep(G) appear by some construction and taking sub quotient from Rep(S_n)? – temp Mar 9 '12 at 17:47
Since every finite group embeds in some symmetric group, it seems very doubtful that anything meaningful can be done to compare their representations (which I assume are taken in characteristic 0). For instance, all symmetric group representations are realizable over the rational numbers, which is quite unusual for other finite groups. – Jim Humphreys Mar 10 '12 at 16:50
@temp: well, it depends on exactly how much structure you want to endow these categories of representations with. As an abelian category $\text{Rep}(G)$ (say over an algebraically closed field of characteristic $0$) is just $\text{Vect}^n$ where $n$ is the number of conjugacy classes of $n$. – Qiaochu Yuan Mar 16 '12 at 16:38
I agree that this question is too general. If $G\leq S_n$ then there's an adjoint pair of functors between $Rep(G)$ and $Rep(S_n)$ given by $Res$ and $Ind.$ It is not the case that every representation of $G$ arises as the restriction of a representation of $S_n$ for some $n.$ Take, for example, either of the non-trivial representations of $\mathbb{Z}/3\mathbb{Z}.$ It has character values which are not rational, thus it cannot arise as the restriction of any $S_n$ rep. However, inducing the regular representation of $G$ gives the regular rep of $S_n,$ hence all of its representations. – Grant Rotskoff Apr 5 '12 at 17:40