Universal central extension

Remarks

A group $G$ has a universal central extension if and only if $G$ is perfect, that is, $G = [G, G]$. Then the universal central extension is unique up to a unique isomorphism over $G$, and the group $U$ is itself perfect. The kernel of $\rho$ is isomorphic to $H_2(G, \Z)$, the Schur multiplier of $G$.

Example: for a ring $R$, the map $St(R) \to E(R)$ from the Steinberg group to the group generated by elementary matrices is a universal central extension, with kernel $K_2(R)$. Example: the map $SL_2(\F_5) \to PSL_2(\F_5) \cong A_5$ is a universal central extension, with kernel $\set{\pm I} \cong \Z/2\Z$.

Non-example: $G = \Z/2\Z$ has no universal central extension. Let $(U, \rho)$ be any central extension and let $\pi \colon G \times G \to G$ be the first projection, a central extension since $G \times G$ is abelian. The maps $u \mapsto (\rho(u), 0)$ and $u \mapsto (\rho(u), \rho(u))$ are two different homomorphisms $f$ with $\pi \circ f = \rho$, because $\rho$ is onto and so not zero. So the uniqueness clause fails.