Lie groups having a circle subgroup

Question

in an answer to this question it is said the following:

Why Lie groups are null-cobordant (are the boundary of a compact manifold)?

I assume it refers to Lie groups of dimension (as manifolds) grater than zero.

Proof:

"Take a circle subgroup in the Lie group. Take all the cosets. This gives a fiber bundle with total space G (the Lie group) and fiber $S^1$. Take the associated disc bundle. Its boundary is $G$." Q.E.D.

I assume for the context that it refers to compact Lie groups. So, my questions are:

1. Could you point to a reference (or prove if the proof is easy and not too long) where it is proved that every compact Lie group has a circle subgroup.
2. Are there any general conditions on a Lie group $G$ that guarantee that $G$ has a circle subgroup?

Thanks in advance!

EDIT: 1 is explained here.

Answer 1

Consider a compact Lie group. A one parameter subgroup is abelian , so its closure is compact connected and abelian. But a compact connected abelian Lie group is a torus, and a torus is a product of circle. To prove that a compact abelian connected Lie group is a torus, note that its Lie algebra is abelian, so it must be a quotient of Rn (the Lie algebra) by a lattice, the kernel of the exponential map (if the Lie algebra is commutative the exponential map is a homomorphism)

Answer 2

Hint: If the group is not semi-simple, then it is a semi-direct product of a semi-simple group and a non trivial torus since it is compact, done.

If the group is semi-simple take the root system, you find a Lie subalgebra isomorphic to $sl(2)$ this means that the universal cover has a Lie subgroup isomorphic to the circle which projection on $G$ by the covering map induces a subgroup isomorphic to $S^1$.