Is the group generated by a finite group and a compact Lie group still a compact Lie group? Suppose we are given two matrix groups comprised of unitary matrices: a finite group $G_1$ and a compact group $G_2$. Let $G$ be the group generated by $G_1$ and $G_2$, that is, every element of $G$ is the matrix product of elements of $G_1$ and $G_2$, i.e., $U_1U_2U_3\cdots$,  where $U_i$ is either an element of $G_1$ or that of $G_2$. My question is:
Is $G$ still a compact Lie group?
If not, what kind of group could $G$ be?
Thanks for your help.
 A: "The group generated by $G_1$ and $G_2$" is quite ambiguous; if you have in mind an embedding of $G_1$ and $G_2$ as subgroups in some larger group, the generated group will generally depend a lot on the embedding.
The most general possibility is the free product, which will almost never be a compact Lie group. Already $G_2$ could be another finite group and the free product $G_1 \ast G_2$ is countable (with the discrete topology, hence noncompact) unless either $G_1$ or $G_2$ is trivial. For example, the free product $C_2 \ast C_2$ is the infinite dihedral group, and the free product $C_2 \ast C_3$ is the modular group $PSL_2(\mathbb{Z})$.
Edit: The answer to the revised question is still no although this is less obvious and I'm not sure how to give a complete proof off the top of my head; free products as above can embed into unitary groups (although the subspace topology will no longer be discrete, it is still noncompact). This is related to the embeddings of free groups used to prove the Banach-Tarski paradox. For example, if we pick two elements of order $2$ in $U(2)$ uniformly at random then they generate the infinite dihedral group $C_2 \ast C_2$ with probability $1$.
However, these subgroups won't generally be closed, and for physical applications you may find it more natural to take the closure. The closure of any subgroup of a compact Lie group is again a compact Lie group, by the closed subgroup theorem.
