Does $O(n)$ have a proper normal subgroup that contains a reflection?

Question

Here is what I have conjectured.

Let $H$ be a normal subgroup of $O(n)$ (the orthogonal group of an n-dimensional vector space). If there $\exists h\in H,\det h=-1$, then $H=O(n)$.

The cases $n=1,2$ are extremely easy to prove - only some simple algebraic calculations are required. However, I find that this approach doesn't seem to work. I am trying to do an induction on $n$, but this yields no results.

I could not find a counterexample either.

Question: Is the above statement true or false?

Answer 1

It fails for $n=3$. Just take $H=\{\operatorname{Id},-\operatorname{Id}\}$.

Answer 2

If you replace $\det h=-1$ with something better matching the word "reflection" in your question title (namely, that $h$ is negative identity in a one-dimensional subspace and the identity on its orthogonal complement), then the claim is true. First note that all such reflections are conjugate (just map the corresponding invariant subspaces to one other), then note that you can map any ON basis to any other ON basis one by one using reflections .