# Why does a one-to-one homomorphism from $G$ to $G/H\times G/K$ imply $G$ is solvable?

Suppose $$H,K\triangleleft G$$ and $$H\cap K=1$$. Prove that if $$G/H$$ and $$G/K$$ are soluble, then $$G$$ is also soluble.

I know we can define a homomorphism, $$\theta:G\rightarrow G/H\times G/K$$ by $$\theta(g)=(gH,gK)$$, and its kernel will be $$\{g\in G| (gH,gK)=(H,K)\}=H\cap K=1$$, so $$\theta$$ is one-to-one. We are supposed to "immediately" deduce from this that $$G$$ is soluble also, but I just can't see it.

We know a fact from class about commutators of direct products $$G=A\times B$$ then $$G'=A'\times B'$$, $$G^{n}=A^n\times B^n$$ etc. if that helps.

If $$G/H$$ and $$G/K$$ are solvable, then so is their direct product. If $$\Gamma$$ is a solvable group, with subgroup $$G$$, then $$G$$ is solvable. Your one-to-one homomorphism exhibits $$G$$ as a subgroup of a solvable group, so $$G$$ is solvable.