Consider a polynomial $f(x) = \prod_{i=1}^{m} g_{i}(x) \in\mathbb{Z}[x]$ of degree $n$, where $g_{i}(x)$ are irreducible factors of degree $m_{i} > 1$. For every $1 \leq i \leq m$, let $\alpha_{i1}, \alpha_{i2}, \ldots, \alpha_{im_{i}}$ be the roots of $g_{i}(x)$.
Let $L_{1}$ be splitting field of $f$ and $G_{1} :=$ Gal $(L/\mathbb{Q})$. Now, for every $1 \leq i \leq m$ and every $1 \leq j \leq m_{i}$, let $\beta_{ij}^{(1)}, \ldots, \beta_{ij}^{(n)}$ be the roots of $f(x) - \alpha_{ij}$. In other words, $\beta_{ij}^{(\nu)}$ are roots of $f(f(x))$. For ease, I was assuming that each $f(x)-\alpha_{ij}$ are separable over $\mathbb{Q}(\alpha_{ij})$.
Let $L_{2}^{(ij)}$ be the splitting field of $f(x) - \alpha_{ij}$ as a polynomial in $\mathbb{Q}(\alpha_{ij})$, let $L_{2}$ be the composite of all $L_{2}^{(ij)}$. Let $G_{2} :=$ Gal $(L_{2}/L)$ and $G_{3} :=$ Gal $(L_{2}/\mathbb{Q})$. Clearly, $G_{3}$ contains $G_{2}$ but I am trying to prove something stronger.
I am trying to see whether $G_{1} \times G_{2} \hookrightarrow G_{3}$, where $\hookrightarrow$ is meant to be an injective homomorphism.
Try $f=x^2-2$, $L=\Bbb{Q}(\sqrt2),L_2=\Bbb{Q}(\sqrt{2+\sqrt2},\sqrt{2-\sqrt2})= \Bbb{Q}(\sqrt{2+\sqrt2})$.
$Gal(L/\Bbb{Q})\times Gal(L_2/L)\cong C_2\times C_2$ doesn't embed into $Gal(L_2/\Bbb{Q}) \cong C_4$.