TITLE: Are locally compact, Hausdorff, locally path-connected topological groups locally Euclidean?
QUESTION [9 upvotes]: Is every locally compact, Hausdorff, locally path-connected topological group $G$ locally Euclidean? (That would imply of course also being a Lie group.) Is it true when countable basis is assumed? I wasn't able to find a discussion of this question in the literature on topological groups and the Hilbert 5th problem.
EDIT: As YCor rightly pointed out, let's assume that $G$ is of finite topological dimension. The general theme of my question is to identify a minimal set of conditions on a topological group making it into a Lie group.

REPLY [2 votes]: This is an addendum to Pedro's answer:
Theorem 1. Suppose that $G$ is a locally compact finite dimensional (Hausdorff) topological group. Then $G$ is locally homeomorphic to the product of a totally disconnected space and ${\mathbb R}^n$.
A reference for this result (that should be better known) is
Montgomery, Deane; Zippin, Leo, Topological transformation groups, Mineola, NY: Dover Publications (ISBN 978-0-486-82449-9). xi, 289 p. (2018). ZBL1418.57024.
section 4.9.3. (The actual theorem says a bit more, namely that locally, at $e$, $G$ is a product of a local Lie group and a totally disconnected group.
In the same book you will also find Theorem 4.10.1 answering your original question:
Theorem 2. If $G$ is a locally compact, locally connected and finite dimensional topological group, then $G$ is a Lie group (possibly non-2nd countable).
Of course, this is a consequence of Theorem 1.