# Elements of a Galois group: can one make this claim?

Say $E = \mathbb{Q}(\alpha)$, $F = \mathbb{Q}(\beta)$, and $EF = \mathbb{Q}(\alpha, \beta)$, all of which are Galois extensions of $\mathbb{Q}$, where $E$ and $F$ have distinct prime degrees $p_1$, $p_2$ over $\mathbb{Q}$.

If you know that $\mathrm{Gal}(EF/\mathbb{Q})$ is isomorphic to $\mathbb{Z}_{p_1}\times \mathbb{Z}_{p_2}$, can you say that its elements must be $\sigma_{ij}(\alpha \beta) = \alpha_i \beta_j$, where $i \in [1, p_1]$, $j \in [1, p_2]$ and $\alpha = \alpha_1$, $\beta = \beta_1$?

Your assumptions imply that $$\operatorname{Gal}(EF/\mathbb Q) \to \operatorname{Gal}(E/\mathbb Q) \times \operatorname{Gal}(E/\mathbb Q) \cong \mathbb Z_{p_1} \times \mathbb Z_{p_2}, \sigma \mapsto (\sigma_{|E},\sigma_{|F})$$ is an isomorphim, hence the answer to your question is affirmative.