# Which countable discrete groups have a metrisable group compactification?

Let $$G$$ be a countable discrete group. A group compactification of $$G$$ is a compact Hausdorff topological group $$H$$ such that there is a group homomorphism $$\iota\colon G\to H$$ with dense image. For example, for every group $$G$$ there is the Bohr compactification, which is in some sense the largest one. But for some reasons, I am looking for metrisable compactifications, equivalently, for second countable compactifications. Which discrete countable groups have non-trivial largest metrisable compactifications? And a follow up question: For which groups we can also demand that $$\iota$$ is injective? I found here on MO that a finitely generated group admits an injective homomorphism to a compact group if, and only if, it is residually finite. It is not hard to see that all residually finite groups embbed injectively in a metrisable compactification. So are there any other examples of such countable discrete groups embedded injectively into compact metrisable groups?

• For every group $G$, the homomorphism $G\to\{1\}$ is a (non-injective) metrizable compactification. What are the missing requirements? – YCor Nov 12 '18 at 22:29
• Anyway, (discrete) groups with a group (injective) compactification are called maximally almost periodic (MAP); for countable groups such a compactification can be chosen to be metrizable. For f.g. groups, MAP is equivalent to residually finite. For more general groups, residually finite implies MAP, which implies locally residually finite. This gives a wealth of examples. What else do you want to know? – YCor Nov 12 '18 at 22:31
• You changed to "nontrivial": then these are groups with at least one nontrivial finite-dimensional representation. For f.g. groups, this is equivalent to the existence of a proper subgroup of finite index. – YCor Nov 12 '18 at 22:47
• Dear @YCor, your comments provide the answers I was looking for! Many thanks! I will be happy to accept it if you submit them as an answer. Or delete the question if it was too trivial for MO (it certainly was't trivial for me). – Dominik Kwietniak Nov 12 '18 at 22:50