Is multiplicative group of real numbers a Lie group? Is group $\mathbb{R} \setminus \lbrace 0\rbrace$ together with multiplication operation a Lie group?
I'm just learning group theory and I would appreciate it if you could explain in greater detail or give a few pointers. Also, if I posted the question incorrectly, please, let me know. Thank you!
 A: Since you are new to Lie groups, probably the part of the question that you find intimidating is proving that the multiplication map is smooth. If I give you a random manifold and a map on it, proving that the map is smooth straight from the definition is typically a little painful.
However, this manifold (call it $X$) has the practical advantage of already being an open submanifold of $\mathbb{R}^n$ (and in fact $n$ has the practical advantage of being $1$). So when we look at the multiplication map
$X \times X \to X$
we may check that it is smooth by extending that map to the map
$$
\mathbb{R}^2 \to \mathbb{R}
$$
given by $(x, y) \mapsto xy$. This map is certainly smooth, because you can just write down all the partials (more generally, and usefully, polynomials are smooth).
You need a similar argument for the inversion map (and there you will take advantage of the fact that $0 \not\in X$, a fact we did not need in checking this first step).
You will not need this yet, but this same way of thinking works for matrix groups: an invertible matrix is in particular a matrix; you can think of an $n$ by $n$ matrix as an element of $\mathbb{R}^{n^2}$, and the entries in a product of matrices under matrix multiplication is a polynomial in the entries of the original matrices. 
