Intersection of composition factors

Question

For $G=HK$, and $K= K_1\cap K_2$, both normal in $G$, if $G/K_1$ and $G/K_2$ are solvable, show that $G/K$ is solvable.

By the Third Isomorphism Theorem, $$\frac{G/K}{K_i/K}\cong \frac{G}{K_i}$$ Then $G/K$ is solvable if and only if $K/K_i$ and $G/K_i$ are solvable. We know $G/K_i$ is soluble for both, but how do I find $K_i/K$ to be solvable? Or am I just on the wrong path?

Answer 1

By correspondence theorem :

$K_{i} \unlhd G $ implies $ K_{i}/K \unlhd G/K$

Now if $G/K_{i} \cong \frac{G/K}{K_{i}/K}$ and $G/K_{i}$ is solvable then $\frac{G/K}{K_{i}/K}$ is solvable too.

That means there is a normal series:

$(1)$ $\frac{G/K}{K_{i}/K} = \frac{G_{0}^{i}/K}{K_{i}/K} \supset \frac{G_{1}^{i}/K}{K_{i}/K} \supset ... \supset \frac{G_{m_{i}}^{i}/K}{K_{i}/K} = {1}$. Where $G_{m}^{i} = K_{i}$ and $\frac{G_{j}^{i}/K}{K_{i}/K}/ \frac{G_{j+1}^{i}/K}{K_{i}/K}$ is abelian.

Now notice that by second isomorphism theorem:

$K_{1}/K_{1} \cap K_{2} \cong K_{1}K_{2}/K_{2}$

As $K_{1},K_{2} \unlhd G$ then $K_{1}K_{2}$ is a subgroup of $G$ then by correspondence theorem $K_{1}K_{2}/K_{2}$ is a subgroup of $G/K_{2}$ which is solvable therefore $K_{1}K_{2}/K_{2}$ is solvable. Now as $K_{1}K_{2}/K_{2} \cong K_{1}/K_{1} \cap K_{2}$ then there is a series:

$ (2)$ $K_{1}/K = H_{0}/K \supset H_{1}/K \supset ... \supset H_{z}/K$ With $H_{z} = K$ and $\frac{H_{k}/K}{H_{k+1}/K}$ abelian.

Now applying the correspondence theorem to $(1)$: The series $\frac{G/K}{K_{1}/K} = \frac{G_{0}^{1}/K}{K_{1}/K} \supset \frac{G_{1}^{1}/K}{K_{1}/K} \supset ... \supset \frac{G_{m_{1}}^{1}/K}{K_{1}/K} = {1}$. Where $G_{m_{1}}^{1} = K_{1}$ and $\frac{G_{j}^{1}/K}{K_{1}/K}/ \frac{G_{j+1}^{1}/K}{K_{1}/K}$ is abelian. Gives us another series: $G/K = G_{0}^{1}/K \supset G_{1}^{1}/K \supset G_{2}^{1}/K \supset ... \supset G_{m_{1}}^{1}/K$ and $\frac{G_{j}^{1}/K}{G_{j+1}^{1}/K}$ is abelian with $G_{m_{1}}^{1} = K_{1}$

Now composing the last series with $(2)$, gives us:

$G/K = G_{0}^{1}/K = \Gamma_{0}/K \supset G_{1}^{1}/K = \Gamma_{1}/K \supset G_{2}^{1}/K = \Gamma_{2}/K \supset ... \supset G_{m_{1}}^{1}/K = K_{1}/K = H_{0}/K_{2} = \Gamma_{m_{1}}/K \supset H_{1}/K = \Gamma_{m_{1} + 1}/K \supset ... \supset H_{z}/K = \Gamma_{m_{1}+z}/K$ With $\Gamma_{m_{1}+z} = K$ and $\frac{\Gamma_{i}/K}{\Gamma_{i+1}/K}$ abelian.

:) I think this works, maybe a little too long.

Answer 2

Consider the homomorphism from $G$ into $G/K_1\times G/K_2$ given by $x\in G$ maps to $(xK_1,xK_2)$. The kernel is clearly $K_1\cap K_2$. Since direct products of solvable groups are solvable and subgroups of solvable groups are solvable, you're done. Exactly the same argument can be applied to supersolvable, nilpotent, etc.