I'm aware this post is related to many other posts on proving that subgroups of solvable groups are solvable. However, there's a certain claim that is necessary to show based on the definition of solvability that I'm using. In the other posts, this part of the proof isn't demonstrated because of the difference in this definition.
In this post, a composition series $\{H_i\}$ of a group $G$ means the $\{H_i\}$ form a subnormal series with each $H_i/H_{i-1}$ simple. Also, for a group $G$ to be solvable in this post, it means there exists a composition series $\{H_i\}$ of $G$ such that all the factor groups $H_i/H_{i-1}$ are abelian.
My goal is to show that if $G$ is solvable and $K$ is a subgroup of $G$, then $K$ is solvable. I got this exercise from Fraleigh's A First Course In Abstract Algebra 7th edition on p.320 (exercise 26). I used the hint that suggested I consider the subnormal series $\{H_i \cap K\}$ of $K$. I'm having difficulty showing that the factor groups are all simple.
Another part of the hint suggests to use that $H_{i-1}(H_i \cap K) \leq H_i$. Furthermore, by the second isomorphism theorem, $(H_i \cap K)/(H_{i-1} \cap K) \cong H_{i-1}(H_i \cap K)/H_{i-1}$. It's tempting to assert that since $H_{i-1}(H_i \cap K)/H_{i-1} \leq H_i/H_{i-1}$, and $H_i/H_{i-1}$ is simple, then $H_{i-1}(H_i \cap K)/H_{i-1}$ is simple and therefore $(H_i \cap K)/(H_{i-1} \cap K)$ is simple by their being isomorphic. But, this is flawed since a subgroup of a simple group isn't generally simple.
With that said, how does one show that the series $\{H_i \cap K\}$ has all simple factor groups, i.e. that all $(H_i \cap K)/(H_{i-1} \cap K)$ are simple? Any help is appreciated.