If $A \times B$ is a Lie group, are both $A$ and $B$ Lie groups?

Question

Suppose $A,B$ are smooth manifolds and there exists a binary operation on the product manifold $A \times B$ making it into a Lie group.

1. Does this guarantee that there exist binary operations on both $A$ and $B$ making them each into Lie groups?
2. If the answer is yes, can it be done in such a way that the product group $A \times B$ is equal to the original Lie group?
3. If the answer to that is yes, is it necessary?

I tried briefly to find a counterexample browsing a table of Lie groups but was unsuccessful.

Answer 1

Question 2 has well-known counterexamples. Generally, it's known that every connected Lie group $G$ is diffeomorphic to a product $K \times \mathbb{R}^n$ where $K$ is its maximal compact subgroup, but $G$ is generally not isomorphic to a product of Lie groups diffeomorphic to $K$ and $\mathbb{R}^n$. A nice small example is $G = SL_2(\mathbb{R})$, whose maximal compact is $SO(2)$, and hence which is diffeomorphic to a product

$$SL_2(\mathbb{R}) \cong S^1 \times \mathbb{R}^2.$$

(this can be established quite explicitly, e.g. using Iwasawa decomposition). However, $SL_2(\mathbb{R})$ has a simple Lie algebra $\mathfrak{sl}_2(\mathbb{R})$, so it isn't isomorphic to any nontrivial product of positive-dimensional Lie groups.

Answer 2

1. First, the example mentioned by Jack Lee: The group $SO(8)$ acts transitively on the unit sphere $S^7$ with point-stabilizers isomorphic to $SO(7)$. This gives $SO(8)$ structure of a principal $SO(7)$-bundle over $S^7$. This bundle is nothing but the orthonormal frame bundle of $S^7$. Since $S^7$ is parallelizable, its orthonormal frame bundle is trivial, hence, $SO(8)$ is diffeomorphic to $S^7\times SO(7)$. However, $S^7$ (as any sphere apart from $S^1$ and $S^3$) is not homeomorphic to any Lie group.

2. A more difficult example is an exotic $R^4$: It is a smooth 4-dimensional manifold $W$ homeomorphic to $R^4$ but not diffeomorphic to it (there is actually continuum of diffeomorphism classes of exotic $R^4$'s). On the other hand, $W\times R$ is homeomorphic to $R^5$, hence, diffeomorphic to $R^5$ since there are no exotic $R^n$'s for $n\ne 4$. Now, $R^5$, of course, has structure of a Lie group. But if an $n$-dimensional Lie group is contractible, it has to be diffeomorphic to $R^n$. I am quite sure the same works in one dimension lower, when one uses $W$ equal to the Whitehead manifold.