I would like to find how many group extensions of $T$ by $\mathbb{Z}_{2}$, where $T$ is the tetrahedral group (in the sense of the definition in https://en.wikipedia.org/wiki/Group_extension).
I was told that there exists two such group extensions and I am trying to use a fact that: If there is an extension of a group $A$ by $G$, then $H^{2}(G; Z(A))$ classifies the extesions of $A$ by $G$. Hence, I would like to prove that $H^{2}(\mathbb{Z}_{2} ; Z(T)) \cong \mathbb{Z}_{2}$, but I am not getting any success. Does anyone have a hint?
Thank you.
Note that the document you linked to in the comments does not use the same terminology regarding extensions as Wikipedia does. Suppose there are groups $A$, $E$ and $G$ which fit into the following short exact sequence
$$0 \to A \to E \to G \to 0.$$
Wikipedia calls $E$ an extension of $G$ by $A$, whereas in the linked document, the group $E$ is called an extension of $A$ by $G$. You are using the same terminology as Wikipedia, so the in the statement of Proposition $4$, we need to swap $A$ and $G$.
As $\operatorname{Out}(\mathbb{Z}_2) = 0$ and $Z(\mathbb{Z}_2) = \mathbb{Z}_2$, Proposition $4$ tells us that $H^2(T; \mathbb{Z}_2)$ classifies the extensions of $T$ by $\mathbb{Z}_2$. By the Universal Coefficient Theorem, we have
$$H^2(T; \mathbb{Z}_2) \cong \operatorname{Hom}(H_2(T; \mathbb{Z}), \mathbb{Z}_2)\oplus\operatorname{Ext}(H_1(T; \mathbb{Z}), \mathbb{Z}_2).$$
As $H_1(T; \mathbb{Z}) \cong T^{\text{ab}} \cong A_4^{\text{ab}} \cong \mathbb{Z}_3$ and $\operatorname{Ext}(\mathbb{Z}_3, \mathbb{Z}_2) = 0$, we just need to determine $H_2(T; \mathbb{Z})$ which is called the Schur multiplier of $T$. I personally don't know how to compute this, but it has been computed to be $\mathbb{Z}_2$, see here for example. Therefore $H^2(T; \mathbb{Z}_2) \cong \mathbb{Z}_2$.
So there are only two extensions of $T$ by $\mathbb{Z}_2$. Namely
$$0 \to \mathbb{Z}_2 \to \mathbb{Z}_2\times T \to T \to 0$$
and
$$0 \to \mathbb{Z}_2 \to 2T \to T \to 0$$
where $2T$ is the binary tetrahedral group.