Semidirect products as (amalgamated) free product

Question

It is well known that $\mathbb{Z}\rtimes \mathbb{Z}_2$ is free product $\mathbb{Z}_2 \star \mathbb{Z}_2$. Are there more examples of these kind?

Answer 1

Let $R,A,B$ be groups. Take the semidirect products $G_1 = A \rtimes R$ and $G_1 = B \rtimes R$. Then you can amalgamate $G_1$ and $G_2$ along $R$ and the resulting amalgamated product will split as semidirect product $(A\ast B)\rtimes R$.

Also, if $G = A \ast B$ is a free product then $G$ splits as a semidirect product:
Let $\beta \colon A \to B$ be a group homomorphism. Then by universal property of free products $\beta$ extends to a homomorphism $\gamma \colon A\ast B$ given by $\gamma(a) = \beta(a)$ for all $a\in A$ and $\gamma(b) = b$ for all $b \in B$. The map $\gamma$ is a retraction of $A * B$ onto $B$ and the group $A \ast B$ splits as a redirect product $\ker(\gamma) \rtimes B$.

In your specific case you have group given by presentation $\langle a, b \|a^2, b^2 \rangle$ and the map $\beta$ is given by $\beta(a) = b$.