I have the following definition of double cover ($Z_2$ is the multiplicative group of order $2$):
Suppose that we have a short exact sequence
$$1 \longrightarrow Z_2 \stackrel{j}{\longrightarrow} G_1 \stackrel{\theta}{\longrightarrow} G_2 \longrightarrow 1$$
Then $j(Z_2)$ is a normal subgroup of $G_1$, from which it follows that $j(Z_2)$ is contained in the centre of $G_1$. In this case, we say that $G_1$ is a double cover of $G_2$.
Could you give me a proof for $j(Z_2)$ to be in the centre of $G_1$?
$j(Z_2)$ is in the center because every automorphism od $Z_2$ is the identity, take $x$ in $G_1$, since $Z_2$ is normal, you have $ad_x:Z_2\rightarrow Z_2$ defined by $ad_x(y)=xyx^{-1}$ well defined $Z_2$ has two element $1,u$, $x1x^{-1}=1$ and $xux^{-1}=u$ since $ad_x$ is bijective, $xux^{-1}=u$ is equivalent to $xu=ux$