I am reading "Algebra, a graduate course" of Martin Isaacs. While looking at exercises of chapter 8 ("Solvable and Nilpotent Groups"), I couldn't solve the part (d) of the following question.
8.2 Let $G$ be finite and solvable and suppose $M<G$ is a maximal subgroup and $core_G(M)=1$. Let $N$ be minimal normal subgroup of $G$. Show the following:
a. $ NM=G$ and $N\cap M=1$.
b. $ C_M(N)=1$.
c. $ N=C_G(N)$.
d. $ N$ is the unique minimal normal subgroup of $G$.
Firstly, $NM$ is a subgroup of $G$, because $N$ is normal, and $N\cap M < N$ because $core_G(M)=1$ (So $M$ cannot contain any normal subgroup of $G$).
$\Rightarrow NM > M \Rightarrow NM=G=MN$.
$N$ is solvable because $G$ is solvable $\Rightarrow N'<N \Rightarrow N'=1 $ because $N$ is minimal normal subgroup. So N is abelian. If $t\in N\cap M$, then for any $n\in N$ $n^{-1}tn=t \Rightarrow t\in core_G(M) \Rightarrow t=1$ (Because $core_G(M)=\bigcap_{g\in G} M^g=\bigcap_{mn\in G} M^{mn}=\bigcap_{n\in N} M^n$). So $N\cap M=1$.
If $s\in C_M(N)$, then for any $n\in N$ $n^{-1}sn=s \Rightarrow s\in core_G(M) \Rightarrow s=1$. So $C_M(N)=1$ and this implies $C_G(N)=N$ because $N$ is abelian and $G=MN$.
But I couldn't prove the uniqueness of minimal normal subgroup.
If $K$ is another minimal normal subrgoup, then $K \cap N=1$, whence $K$ and $N$ commute ($[K,N]=1$). So $K \subseteq N$ by c. And by c. also $N \subseteq K$ (replace in the entire argument $N$ by $K$, or use that $N$ is minimal normal). So $K=N$.