Locally compact vs. compactly generated in group theory I'm studying the Creutz-Shalom paper on the normal subgroup theorem for dense commensurators of lattices. The authors state their results for locally compact and compactly generated groups. I do not have a good intuitive idea of these objects. My naive thinking would be that in compactly generated groups there is a compact neighbourhood of the identity. Are there easy examples that help distinguish the two concepts? Thank you very much for your time.  
 A: Functional analysis  abounds with infinite dimensional (and so not locally compact) locally convex spaces which (as topological groups under addition) are compactly generated.  For example the dual of an infinite dimensional Banach space with the weak $\ast$ topology.  The unit ball is compact.
Edit in view of the comment below.  It seems that there is some ambiguity in the phrase "compactly generated" in the OP.  I assumed from the context that it was used in the sense of topological group theory, i. e., that there is a compact subset which generates the group algebraically.  If it is intended to mean compactly generated as a topological space, then the same example works---we just take the topology of uniform convergence on the compact subsets of the Banach space rather than the weak $\ast$ topology. This is compactly generated à la Kelley (Banach-Dieudonné theorem).
A: Edit: $(\mathbb{Q},+)$ with the order topology is compactly generated (by the set $\{1/n \mid n \in \mathbb{N}\} \cup \{0\}$) but not locally compact.  I would say this is the 'easiest' example that answers the OP's question.  (It's also a good example of an infinite totally disconnected group with no proper open subgroups, which is something that cannot occur for locally compact groups.)  The groups mentioned by YCor in his comment have similar topological properties.
Old answer: If 'compactly generated' is used in the loose sense that there is a compact subset that generates a dense subgroup, then all bets are off.  However, if there is a generating set in the strict sense that is compact, then we can at least say 'completely metrisable $\Rightarrow$ locally compact':
We may assume that the compact generating set $X$ contains inverses (since the set of inverses of $X$ is also compact), and we also note that $X^n$ is compact for all $n$.  Thus the group $G$ is a union of a countable ascending chain of compact sets.  One of these has non-empty interior, by the Baire category theorem.
