Through a previous question, I understood how we can lift the action of a group $G$ on a topological space $X$ to an action of a covering group $G'$ of $G$ on the universal cover $\tilde{X}$ in such a way that $X/G\cong \tilde{X}/G'$.
Now I am trying to work out an explicit example and I am stuck -
Let $\mathbb Z_2$ act on $S^1$ by conjugation. That is $-1\cdot(x,y)=(x,-y)$. The universal cover of $S^1$ is $\mathbb R$ with covering map $p:\mathbb R\to S^1$ given by $r\mapsto e^{2\pi ir}$. But I can't figure out a right candidate for $G'$. Can someone help?
Thank you.
Hint:
Consider the group $G'$ which is the semi product of $\mathbb Z_2$ and $\mathbb Z$ with the product defined by $((-1)^i,n).((-1)^j,m)=((-1)^{ij},n+(-1)^im)$.
$G'$ acts on $\mathbb R$ by $(-1,n).x =-x+n$, you have $e^{2i\pi (-x+n)}=e^{-2i\pi x}$.