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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.

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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.

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