Definition: We say that $H$ is a minimal normal group of $G$ if $H\lhd{G}$ and $\forall K\lneq{H}$ , $K\neq{\{e}\}$ it holds that $ K\ntriangleleft{G}$.
Let $G$ be a finite solvable group, and let $H$ be a minimal normal group of $G$. Prove that $H$ is abelian.
I wrote a solution but it feels too easy and it proves a stronger claim- that $Z(G)=H$ , so I think it might be wrong:
Since every $g\in{Z(G)}$ commutes with every $h\in{H}$ , it holds that $Z(G)\leq{Z(H)}\leq{H}$. Since $Z(G)\lhd{G}$ , and $H$ is a minimal normal group of $G$, we get that $Z(G)=H$ or $Z(G)={\{e}\}$ . But $G$ is solvable, so we get $Z(G)\neq{\{e}\}$ and therefore $Z(G)=H$ and $H$ is abelian.
I'd like to know if this proof is correct, and if it's wrong, some guidence would be much appreciated.
Your argument commits several errors.
For one thing, solvable groups may have trivial centers (for example, $S_3$).
For another, It is false that $Z(G)\leq Z(H)$ for $H\leq G$. What is true is that $Z(G)\cap H\leq Z(H)$, but again, in general, there is absolutely no reason to assume that a central element of $G$ must be in $H$, and so there is no reason to think it will lie in the center of $H$. When you argue that elements of $Z(G)$ commute with every element of $H$, the conclusion is not that $Z(G)\leq Z(H)$, but rather that $Z(G)\leq C_G(H)$, the centralizer of $H$. The centralizer also contains $Z(H)$, but that doesn’t tell you anything about whether $Z(G)$ and $Z(H)$ have any containment relationship with each other.
As for an argument, here are few things you may want to prove:
Subgroups of solvable groups are solvable.
If $K$ is solvable, then $[K,K]\neq K$.
If $N\triangleleft G$, then $[N,N]\triangleleft G$.
Use those facts to show the desired result.