# Normal Subgroups of $SU(n)$

I was wondering if there is any classification for normal subgroups of $SU(n)$? In particular, I think that the answer is no for $n = 2$ by looking at the covering map onto $SO(3)$, but I was curious if there were any results for arbitrary $n$?

I'm actually looking for finite normal subgroups of the unitary group (a lot to ask for, I know).

Here are two general facts.

1. If $G$ is a connected topological group, then any discrete normal subgroup $N$ of $G$ is in fact contained in its center. This is a standard exercise and the proof is straightforward: by hypothesis, $N$ is setwise fixed by conjugation. But since $N$ is discrete, a connected topological group can't act nontrivially on it, and so $N$ is in fact pointwise fixed by conjugation. This already implies that any finite normal subgroup of $SU(n)$ is contained in its center.
2. If $G$ is a connected Lie group and $N$ is a closed normal subgroup, then $N$ is a Lie subgroup, and so its Lie algebra $\mathfrak{n}$ is an ideal of the Lie algebra $\mathfrak{g}$. If $\mathfrak{g}$ is simple (which is the case when $G = SU(n)$), then $\mathfrak{n} = 0$, and so $N$ must in fact be discrete, and then by the previous point we know that $N$ must in fact be central. This implies that any nontrivial closed normal subgroup of $SU(n)$ is contained in its center.
• Thanks! These types of standard results in Lie groups/algebras always seem to answer my questions. Do you have a recommendation for a good textbook to learn Lie groups and Lie algebras side by side? Let's say at an early graduate level.
– MGN
Mar 22, 2015 at 16:02

Qiaochi's answer deals with closed normal subgroups. However, a priori, there could have been non-closed proper normal subgroups in $SU(n)$ (not contained in its center). One can show that this cannot happen "with bare hands" (this is done for instance in Berger's "Geometry" book in the case of $SO(3)$ merely using elementary geometry), or use the following general theorem:

Theorem. Suppose that $G$ is a compact Hausdorff topological group, which is topologically simple, i.e. contains no proper closed normal subgroups. Then $G$ is simple as an abstract group.

See Theorem 9.90 of

Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.

Addendum. There exist Hausdorff topologically simple groups which are not simple as abstract groups, see section 3 in

George A. Willis, Compact open subgroups in simple totally disconnected groups, Journal of Algebra, Volume 312, Issue 1 (2007) pages 405-417.

I doubt there are any finite normal subgroups besides subgroups contained in the center $\lbrace\exp(\frac{2ik\pi}n)\mathrm{id}_V\mid k=1,\dots,n\rbrace$, as any element $u\in SU(n)$ that is not of this kind should have a non discrete conjugacy class. Indeed, its eigenspaces form an orthogonal decomposition of the underlying vector space into at least two subspaces, $$V=\bigoplus_{i=1,\dots, r}^{\perp} E_{\lambda_i}(u)$$ and by conjugation by an element in $SU(n)$ amounts to rotating the eigenspaces, which can be done in a continuous fashion, and hence the conjugacy classes are nondenumerable.

• Good point! In fact, now that I think about it, $SU(n)$ likely has no normal subgroups at all! Other than those in the center of course.
– MGN
Mar 22, 2015 at 5:04