When every minimal subgroup is contained in the center

I am not sure that I can achieve elegance.

Let's deal with (b); then (a) is a corollary. So we have a normal subgroup $N$ of $G$.

Step 1: Let $K$ be a minimal normal subgroup of $N$ with $K\not\leqslant Z(N)$. Then $K\cap Z(N)=1$ or we are done.

Step 2: Consider $K^G$. It is not hard to prove that this is a normal subgroup of $G$, and indeed that $K^G=K_1\times K_2\times\dots\times K_s$, where $K_1=K$ and each $K_i$ is a $G$-conjugate of $K$, and each is normal in $N$.

Step 3: Let $L$ be a minimal normal subgroup of $G$ lying in $K^G$. Then by hypothesis $L\leqslant Z(G)$, and so $L\leqslant Z(N)$.

Step 4: Let $1\not=(x_1,x_2,\dots,x_s)\in L\cap Z(N)$, where $x_i\in K_i$. Now $N$ fixes each $K_i$, so we have that $N$ centralises each $x_i$.

Step 5: Some $x_i\not=1$, so we get $K_i\cap Z(N)\not=1$. Conjugating by an appropriate element of $G$ we get that $K_1\cap Z(N)\not=1$; this is a contradiction, so $K\leqslant Z(N)$.

[I think Step 2 is standard stuff.]