Every minimal normal subgroup is contained in the center G is a finite group in the following questions:
(X):Every minimal normal subgroup is contained in the center.
(1) Let $N$ and $M$ be normal subgroups of $G$, both of which satisfy (X), then prove:
$NM$ satisfies (X).
(2) If $G$ satisfies (X), then prove: every normal subgroup of $G$ satisfies (X).
(Resource:  Hans Kurzweil Bernd Stellmacher
The Theory of Finite Groups
An Introduction pp39)
I solve question(1):
Suppose $U$ is minimal normal subgroup of MN and $1\neq[U,MN]=U$, without loss of any generality,  we suppose $[U,M]=U\leq U\cap M$, which means $U\unlhd M$.
Considering the fact (X), if K is the minimal normal subgroup of M contained in $U$, we get $K\leq Z(M)\Rightarrow |K|=p$, where $p$ is a prime number. $K$ is also the minimal normal subgroup of $U$, $I=\{K^{mn}|mn\in MN\}$,$\exists K_1,K_2,...,K_l \in I,U=\times_i K_i$.
$[U,M]=\prod_i [K_i,M]=1$,contradiction. $[U,M]=[U,N]=1\Rightarrow [U,MN]=1\Rightarrow U\leq Z(MN)$
But I cannot solve question (2), could you help me? Plz let the topic on.(I'm trying to improve my question.)
 A: (2)Proof: Suppose $U$ is a minimal normal subgroup of $N\unlhd G$, $U^{g}\unlhd N$, so $U^{g}$ is also a minimal normal subgroup of $N$, let E be a minimal normal subgroup of $U$: $I_1=\{E^{n}|n\in N\}$, $\exists E_1,E_2,...,E_m\in I, U=\times_i E_i$.
Define $M=\prod_{g\in G}U^{g}$ and $I_2=\{U^{g}|g\in G\}$,
$\exists U_1,U_2,...,U_k\in I, M=\times_j U_j=\times_j (\times_i E_i)^{g_j}\unlhd G$, Suppose $C$ is the minimal subgroup of $G$ contained in $M$, $C\leq Z(G)\Rightarrow |C|=p$, where $p$ is a prime number.
For $1\neq c\in C$, $c=\times_j (\times_i e_i)^{g_j}$, where $e_i \in E_i$. Apparently, not all $e_i\in E_i$ is identity element $1$ (or contradicts the fact $c\neq 1$), suppose $e_1\neq 1$.
$U$ is an elementary Abelian p-group , (or $U$ is semi-simple, $M$ is semi-simple too. $C\unlhd M\Rightarrow E_l=C\cong C_p$, contradiction),
$\forall n\in N, 1\neq c=c^{n}=(e_1^{n}\times e_2^n\times ...)\times (e_{21}\times ...)^{g_jn}\times...\Rightarrow e_1^n=e_1\Rightarrow [E_1,N]=1\Rightarrow [U,N]=\prod_{i} [E_i,N]=1\Rightarrow U\leq Z(N)$.
