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Is it true that a compact group always has a faithful, finite-dimensional unitary representation? If not, are there any reasonably simple counter-examples?

I've done some research and know that every group has some faithful representation, all irreducible reps of a compact group are finite, and that the irreducible reps separate the points of the group. However, that doesn't quite answer the question!

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    $\begingroup$ The additive group of $p$-adic integers is an example, any finite dimensional continuous representation has some $p^n$ in its kernel. $\endgroup$
    – Torsten Ekedahl
    May 5, 2011 at 14:00

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A famous theorem is that this is true if and only if $G$ is a Lie group.

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    $\begingroup$ Also, the adjective "unitary" is redundant, as every finite-dimensional representation of a compact group is unitarizable. $\endgroup$
    – Mark
    May 5, 2011 at 11:47
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    $\begingroup$ Does this theorem have a name? Even better, are there any good references for it? $\endgroup$
    – Sabri
    May 5, 2011 at 14:01
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    $\begingroup$ @Sabri: one direction follows from the fact that a closed subgroup of a Lie group is a Lie group (I don't know if this theorem has a name). The other follows from Peter-Weyl. $\endgroup$
    – Qiaochu Yuan
    May 5, 2011 at 16:15
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    $\begingroup$ Sabri: you can find the proof in Folland's text ("A Course in Abstract Harmonic Analysis", theorem 5.13). I don't know of a name for it. The theorem which Qiaochu mentioned, though, is often called the "closed subgroup theorem" or Cartan's theorem. $\endgroup$
    – Mark
    May 5, 2011 at 16:33
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No, it is false! Take a product of alef-2011 copies of $C_2$. It is a bit too big to fit into $GL_n (C)$...

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    $\begingroup$ And I fear it will be falser next year... $\endgroup$
    – Georges Elencwajg
    May 5, 2011 at 11:30
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    $\begingroup$ More specifically, it isn't a second-countable group, while every subspace (let alone subgroup) of $GL_n (\mathbb{C}$ is second-countable. $\endgroup$
    – Mark
    May 5, 2011 at 11:50
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    $\begingroup$ Or use Burnsides's theorem that any linear group of finite exponent is finite. $\endgroup$
    – Benjamin Steinberg
    Oct 12, 2011 at 2:47
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Another famous and interesting NASC for a compact (or even locally compact) group to have a faithful finite dimensional representation is that it "not have arbitrarily small subgroups", i.e., that there exist a neighborhood of the identity with no non-trivial subgroup. This was the way that von Neumann solved the Hilbert 5th Problem in the compact case, and is explained (starting on page 1243) in:

http://www.ams.org/notices/200910/rtx091001236p.pdf

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See On closed totally disconnected subgroups of connected real Lie groups : a non-discrete totally diisconnected group has no faithful representation in $GL_n(\mathbb{C})$.

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