This is the exercise 23.10 p. 135 of Groups and symmetry of Armstrong :
Let $G$ be an abelian group and write $G \tilde{\times}\mathbb{Z}_2 $ for the semidirect product $G\rtimes_\phi\mathbb{Z}_2$, where $\phi : \mathbb{Z}_2 \rightarrow \text{Aut}(G)$ sends $1$ to $x\rightarrow x^{-1}$.
[...]
Prove that $O_2 \cong SO_2 \tilde{\times} \mathbb{Z}_2 $.
My idea was to take the morphism : $$ \Phi(A, k) = (-1)^k A$$ But it seem to be wrong. Any hints ?
A useful tool in questions about semi-direct product is the splitting lemma.
Consider an exact sequence $$1\to SO(n) \hookrightarrow O(n)\xrightarrow{\det} O(1) \to 1.$$
There is a homomorphism $p\colon O(1)\to O(n)$ given by $p(-1)=R$, where $R\in O(n)$ is any reflection. Obviously $\det \circ p = \mathrm{id}_{O(1)}$, so the splitting lemma guarantees us that $O(n) \simeq O(1) \rtimes SO(n)$.
Let's be more explicit for $n=2$. Take $$R = \begin{pmatrix} 0 & 1\\1 & 0 \end{pmatrix} \in O(2)$$ to be the reflection. We need an action $\phi\colon O(1) \to \mathrm{Aut}(SO(2))$ (which can be easily reconstructed from the splitting lemma):
$$\phi(1)(B) := A^{-1}BA = B^{-1}.$$
Alternatively, you can start from this map $\phi$, construct the semi-direct product as usual, and then explicitly show that you get $O(2)$.