## 3 Proofs of Theorems 1 and 2
(4) \(G=RT\), _where \(R\) is a nontrivial normal subgroup in \(G\) with \(R\leq N\)._  
If \(RT<G\), then one can see that \(RT\) satisfies the condition by Proposition 1 (3). Hence \(T\) acts hypercentrally on \(R\) by the choice of \(G\). Since \(N/R\leq D_{p}(RT/R)\) and \(O^{p}(RT/R)=RT/R\), then, by the choice of \(G\), \(RT/R\) acts hypercentrally on \(N/R\). Then \(T\) acts hypercentrally on \(N\), that is, \(G=NT\) acts hypercentrally on \(N\) by ([12], Lemma 1.7.11), so does \(H\), a contradiction. (5) _Final contradiction._  
Since \(G/R=TR/R\) acts hypercentrally on \(N/R\), without generality, we can assume \(R=N\) is minimal normal in \(G\). Then, by the minimality of \(N\) and the normality of \(T\), we have that \(G=N\times T\) or \(G=T\). If \(G=N\times T\), then \(N\leq Z(G)\), a contradiction. Let \(G=T\). Since \(T\) is the minimal supplement of \(C_{G}(N)\) in \(G\), we have that \(T\cap C_{G}(N)\leq\Phi(T)\) by ([12], Lemma 2.3.4). Thus, \(C_{G}(N)\leq\Phi(G)\). By the minimality of \(N\) and \(N\), \(O_{q}(G)\leq C_{G}(N)\leq\Phi(T)=\Phi(G)\). It follows that \(O_{q,p}(G)\) is \(p\)-closed. Choose \(P\) to be a Sylow \(p\)-subgroup in \(O_{q,p}(G)\). Then \(P\trianglelefteq G\) and so, \(P\leq C_{G}(N)\leq\Phi(G)\). Therefore \(O_{q,p}(G)\leq\Phi(G)\), a contradiction. **Proof of Theorem 2. ** (1) Since \(C_{G}(G^{\prime}(p))\leq D_{p}(G)\), the necessity is clear. Conversely, assume that \(C_{G}(G^{\prime}(p))=1\) and \(D_{p}(G)>1\). It implies that \(G^{\prime}(p)\cap D_{p}(G)>1\). Otherwise, \(D_{p}(G)\leq C_{G}(G^{\prime}(p))\) and \(C_{G}(G^{\prime}(p))\
eq 1\). By Proposition 1 (2), \(D_{p}(G)\) is soluble. So \(G\) has a minimal normal subgroup \(N\) such that \(N\leq G^{\prime}(p)\cap D_{p}(G)\). Then \(N\) is elementary abelian. \(N\leq Z(G^{\prime})\). Assume \(G^{\prime}\cap D_{p}(G)=1\). Since \([G,D_{p}(G)]\leq[G,G]=G^{\prime}\) and \([G,D_{p}(G)]\leq D_{p}(G)\), \([G,D_{p}(G)]\leq G^{\prime}\cap D_{p}(G)=1\). It follows that \(D_{p}(G)\leq Z(G)\), a contradiction and thus \(G^{\prime}\cap D_{p}(G)\
eq 1\). Since \(G^{\prime}(p)\cap D_{p}(G)\geq G^{\prime}\cap D_{p}(G)\), we can assume that \(N\leq G^{\prime}\cap D_{p}(G)\). Now, by the ([13], Theorem 2.3 (1)), we have \(N\leq G^{\prime}\cap D(G)\leq Z_{\infty}(G^{\prime})\). It follows from the minimality of \(N\) that \(N\leq Z(G^{\prime})\). \(N\leq C_{G}(O^{p}(G))\). Let \(N\) be \(q\)-group for some prime \(q\) and \(r\) a prime divisor of \(|G|\) different to \(p\) and \(q\). If \(R\) is a \(r\)-group. Then \(N\leq N_{G}(R)\) by \(N\leq D_{p}(G)\) and hence \([N,R]\leq N\cap R=1\). Thus, \(R\leq C_{G}(N)\) and it follows from the choice of \(r\) that \(G/C_{G}(N)\) is a \(\{p,q\}\)-group. Therefore, without generality, we can assume that \(G\) is a \(\{p,q\}\)-group. If \(q\
eq p\), then, by Lemma 2, \(N\leq Z_{\infty}(O^{p}(G))\). It follows from the minimality of \(N\) that \(N\leq Z(O^{p}(G))\). If \(N\) is a \(p\)-group, then \([N,Q]=[N,Q^{\prime}O^{p}(Q)]=1\) for any Sylow \(r\)-subgroup of \(G\) with \(r\
eq p\). Then \([N,O^{p}(G)]=1\), and \(N\leq C_{G}(O^{p}(G))\). Hence, one can see that \(N\leq C_{G}(G^{\prime}(p))\), a contradiction. (2) If \(O_{p^{\prime}}(D_{p}(G))=1\), the result is clear. If \(O_{p^{\prime}}(D_{p}(G))\
eq 1\), then \(G\) has a minimal normal subgroup \(N\) with \(N\leq O_{p^{\prime}}(D_{p}(G))\). For any Sylow \(r\)-subgroup \(R\) of \(G\), we have \([N,R]=1\). Then \(G/C_{G}(N)\) is a \(\{p,q\}\)-group, hence, without loss of generality, we assume that \(G\) is a \(\{p,q\}\)-group. If \(N\) is a \(q\)-group, then, by Lemma 2, \(N\leq Z_{\infty}(O^{p}(G))\). It follows from the minimality of \(N\) that \(N\leq Z(O^{p}(G))\). If \(N\) is a \(p\)-group, then \([N,Q]=[N,Q^{\prime}O^{p}(Q)]=1\) for any Sylow \(q\)-subgroup of \(G\). Then \([N,O^{p}(G)]=1\), and \(N\leq Z(O^{p}(G))\). By induction, \(O_{p^{\prime}}(D_{p}(G)/N)\leq Z_{\infty}(O^{p}(G)/N)\), then \(O_{p^{\prime}}(D_{p}(G))\leq Z_{\infty}(O^{p}(G))\).