Galois group of a cyclotomic extension of $\mathbb Q$ isomorphic to a direct product of its subgroups.

Question

Let $n,m$ be coprime. Let $\zeta_k$ be primitive k$^{th}$ root of unity.

I know Group of $x^{nm} -1$ over $\mathbb Q$ is isomorphic to the unit group $\mathbb Z_{nm}^\times$ i.e. $G(\mathbb Q(\zeta_{nm})/\mathbb Q) \cong \mathbb Z_{nm}^\times$.

$\mathbb Z_{nm} \cong \mathbb Z_n \times \mathbb Z_m$ as rings implies $\mathbb Z_{nm}^\times \cong \mathbb Z_n^\times \times \mathbb Z_m^\times$ as groups.

Then is it true that $\mathbb Z_{nm}^\times \cong \mathbb Z_n^\times \times \mathbb Z_m^\times$ solely implies $$G(\mathbb Q(\zeta_{nm})/\mathbb Q) \cong G(\mathbb Q(\zeta_{n})/\mathbb Q) \times G(\mathbb Q(\zeta_{m})/\mathbb Q)$$ ?

Answer 1

a) $\cong$ is an equivalence relation, so it is transitive.

b) $A\cong B\wedge C\cong D\Rightarrow A\times C\cong B\times D$

So it seems that you know

c) $G(\mathbb Q(\zeta_{nm})/\mathbb Q) \cong \mathbb Z_{nm}^\times$

d) $\mathbb Z_{nm}^\times \cong \mathbb Z_n^\times \times \mathbb Z_m^\times$

With those $$G(\mathbb Q(\zeta_{nm})/\mathbb Q) \cong G(\mathbb Q(\zeta_{n})/\mathbb Q) \times G(\mathbb Q(\zeta_{m})/\mathbb Q)$$

follows.