Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Every Lie group is the direct product of a connected Lie group and a discrete group.

I think the component of the identity could be useful.

share|cite|improve this question
What did you try? – Bruno Joyal Feb 27 '13 at 23:30
Show the connected component you described is a normal subgroup first. – Thomas Andrews Feb 27 '13 at 23:35
Someone edited my question. I want to show that every Lie group is the semidirect product of a connected Lie Group and a discrete group – Gsanm Feb 28 '13 at 1:16
Please edit your question appropriately. There was no version which said something about a semi-direct product. – Martin Brandenburg Feb 28 '13 at 5:06

Warning. Apparently the proof is not correct. I will update it later.

For every topological group $G$ with identity component $G^0$ it is well-known and easy to prove that $G^0$ is a normal, in fact characteristic, closed subgroup. If $G$ is locally connected, $G^0$ is open, which implies that the quotient group $G/G^0$ is discrete. The Lie algebra of $G$ is isomorphic to the Lie algebra of $G/G^0 \times G^0$, hence these Lie groups are locally isomorphic. It is not hard to see that the isomorphisms glue to a global isomorphism, since the decomposition into a product of a connected and a discrete Lie group is essentially unique.

By the way, the corresponding statement for topological groups is wrong.

share|cite|improve this answer
What about, for example, $O(2n+1)$? As a manifold, it's diffeomorphic to $SO(2n+1) \times \mathbb{Z}/2$, but not as a group (since $O(2n+1)$ has trivial center but $SO(2n+1)\times \mathbb{Z}/2$ doesn't). Or am I missing something stupid? – Jason DeVito Feb 28 '13 at 0:13
Why does $O(2n+1)$ has trivial center? If your counterexample is correct, you should add it as an answer (or "anti-answer" ;)). – Martin Brandenburg Feb 28 '13 at 1:04
Ha! I got it backwards. $-I \in O(2n+1)$, and in fact $O(2n+1)\cong SO(2n+1)\times\mathbb{Z}/2$ (as Lie groups). On the other hand, $O(2n)$ is not Lie isomorphic to $SO(2n)\times\mathbb{Z}/2$. For example, when $n=1$, $SO(2)\times\mathbb{Z}/2$ is abelian, but generally reflections and rotations don't commute. (I'm not going to add this as an aswer because the OP has now added a correction asking for a semi-direct product instead of direct product. I'm not sure how to prove it.) – Jason DeVito Feb 28 '13 at 3:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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