Let Z be a standard normal R.V.. Show that Sgn(Z) and $\lvert Z\rvert$ are independent.

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This seems intuitively obvious, but I'm struggling with showing it explicitly. My attempt is to shatter $\lvert Z\rvert \in B$ over the possible outcomes of $\mathrm{Sgn}(Z)$. It goes something like:

Let $B$ be a Borel set.

$$P(\lvert Z\rvert \in B)= P(\lvert Z\rvert \in B\mid \mathrm{Sgn}(Z)=1)P(\mathrm{Sgn}(Z)=1) + P(\lvert Z\rvert \in B\mid \mathrm{Sgn}(Z)=-1)P(\mathrm{Sgn}(Z)=-1)$$

I know $P(\mathrm{sgn}(Z)=1)=P(\mathrm{sgn}(Z)=-1) = 1/2$ and that combined with the symmetry of $Z$ leads to the result I want, but only for $\mathrm{Sgn}(Z)= \pm 1$. Am I even close with this approach? Thanks in advance.