I have just learned from mighty Wikipedia that the symmetric group of an infinite set is not a matrix group. Why?


I can see rep-theory reasons: sizes of minimal non-trivial, non-sign representations of $S\_n$ grow as $n$ grows. But I believe that there should really be an elementary linear algebra argument for this. Is there one?



Why is symmetric group not matrix?
**Edit:** My original idea doesn't work, but unknown's does. Here are the details.


Let $k$ be a field, which is WLOG algebraically closed. Let $V$ be a finite-dimensional representation over $k$ of dimension $n$. Then $S\_{\infty}$ contains $(\mathbb{Z}/p\mathbb{Z}))^{p^n}$ (in fact any finite group) as a subgroup, where $p \neq \text{char}(k)$. Let $g\_1, ... g\_{p^n}$ be its generators. Then by "elementary linear algebra" $V$ is a direct sum of $1$-dimensional irreps of $\langle g\_1 \rangle$ which $g\_2, ... g\_{p^n}$ must preserve, while still having order $p$. But there are only $p^n - 1$ nontrivial ways to do this; hence either one of the $g\_i$ acts as the identity or two of them are the same and $V$ cannot be faithful.


Of course, whether this is "elementary linear algebra" or representation theory is debatable, and I think irrelevant. All I did was find a sequence of finite groups such that the dimension of the smallest faithful representation goes to infinity and I could have done this any number of ways, e.g. I could have chosen $\text{PSL}\_2(\mathbb{F}\_q)$.


