Double cover of symplectic groups

Question

What is the normal definition of double cover of Symplectic group? I couldn't find a simple and understandable definition

Answer 1

If $G,H$ are groups then $H$ is a double cover of $G$ iff $H$ has a normal subgroup $K$ of order 2 such that $K \leq [H,H] \cap Z(H)$ and $H/K \cong G$.

In other words, $H$ has an element $z$ that (1) commutes with every element of $H$, (2) can be written as a product of commutators $z=\left(h_1^{-1} h_2^{-1} h_1 h_2\right)\cdots\left(h_{2n-1}^{-1} h_{2n}^{-1} h_{2n-1}h_{2n}\right)$, (3) $z$ has order 2, and (4) if we ignore $z$, then we just get $G \cong H/\langle z\rangle$.

A triple cover is defined similarly, except $K$ has order 3.

A double cover has the neat property that every element of $G$ has two ghosts in $H$, one light and one dark. If Scott is an element of $H$, then there is always Nega-Scott in $H$ as well. In representations of $G$, both Scott and Nega-Scott have the same action and character, but the representations of $H$ (called projective representations of $G$) reveal the true character, with the value of Nega-Scott being the negative of the value of Scott.

Strictly, I think I may need that $Z(G)$ is finite of odd order for this dichotomy to hold, and I usually consider finite groups, so my apologies if an infinite evil representation interferes.