9
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • 17
    $\begingroup$ Assume that your infinite symmetric group $G$ has an $n$ dimensional representation over a field $k$. You will be certainly able to find infinitely many elements of $G$ of order two that commute with each other. If $char(k)\neq 2$ then you can simultaneously diagonalized the matrices corresponding to these elements of order two. But then you see that you have only 2^n-1 choices for them. Contradiction. Does it work? $\endgroup$
    – unknown
    Oct 28, 2010 at 21:02

3 Answers 3

Reset to default
13
$\begingroup$

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 $\text{PSL}_2(\mathbb{F}_q)$.

Improve this answer
$\endgroup$
1
  • 12
    $\begingroup$ So can we agree that the main point is the following: If $S_\infty$ had a finite-dimensional faithful representation over some field $k$, then the dimension of minimal faithful representations of finite groups over $k$ was uniformly bounded, which is "clearly" wrong, and there are many ways to make that clearly really clear. $\endgroup$
    – Tobias Hartnick
    Oct 28, 2010 at 22:37
3
$\begingroup$

This might not be really elementary, but ... We know that $M_n(k)$ satisfies polynomial identities. For instance, the Amitsur-Levitski Theorem tells us that $$\sum_{\sigma\in S_{2n}}\epsilon(\sigma)A_{\sigma(1)}\cdots A_{\sigma(2n)}=0_n, \qquad\forall A_1,\ldots,A_{2n}\in M_n(k).$$ On the contrary, the symmetric group of an infinite set does not satisfy polynomial identities.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ @Denis: What is a polynomial identity of a group? If you mean the usual identity (law), then your answer is wrong. The free group does not satisfy any identity, but is inside $M_2(k)$. Perhaps you meant that $S_\infty$ does not embed into the multiplicative semigroup of any PI ring. That is true, but requires a proof which is certainly more complicated than the original question. $\endgroup$
    – user6976
    Oct 29, 2010 at 9:37
  • $\begingroup$ @Mark. Right. It is more involved than I thought initially. $\endgroup$
    – Denis Serre
    Oct 29, 2010 at 10:45
  • $\begingroup$ @Denis Take $A=kS_\infty /I$ where $I$ is the ideal generated by the values of the standard polynomial. You are saying that $S_\infty \rightarrow kS_\infty /I$ is not injective. I am not sure how difficult it is to find $1-g$ in $I$. $\endgroup$
    – Bugs Bunny
    Oct 29, 2010 at 11:13
  • $\begingroup$ @Mark Are there nonmatrix groups embeddable into PI-algebras at all? $\endgroup$
    – Bugs Bunny
    Oct 29, 2010 at 11:14
  • $\begingroup$ @Bugs Bunny: Of course there are. For example, $({\mathbb Z}/2{\mathbb Z})^\infty$ embeds into a commutative ring (its group algebra). In the finitely generated case, though, the situation is different because of Braun's theorem that the radical of PI rings is nilpotent. $\endgroup$
    – user6976
    Oct 29, 2010 at 13:23
1
$\begingroup$

The symmetric group of an enumerable set contains every enumerable group as a subgroup. Its linearity would thus imply the linearity of every enumerable group which is false for example by the Tits's alternative.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Roland, the group I'm thinking of is the group of symmetries which fix all but finitely many elements. This group is way bigger, although it might be the group that is actually meant by the question. $\endgroup$
    – Qiaochu Yuan
    Oct 29, 2010 at 11:27
  • $\begingroup$ His name is Tits, not Tit. $\endgroup$
    – S. Carnahan ♦
    Oct 29, 2010 at 11:42
  • $\begingroup$ Considering the group of all permutations with compact support (fixing all but a finite number of elements), the above argument applies to all finite groups and implies the existence of an universal bound $M$ such that every finite group has a faithful representation of dimension at most $M$. $\endgroup$
    – Roland Bacher
    Oct 29, 2010 at 11:55
  • $\begingroup$ @Roland: Is there an elementary reason why that is false? $\endgroup$
    – Igor Rivin
    Feb 1, 2011 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.