## 3 Proofs of Theorems 1 and 2  
Proof of Theorem 1.: (1) By Proposition 1 (2), \(D_{p}(G)\) is soluble. Then \(D_{p}(G)\) has a Hall \(p^{\prime}\)-subgroup, denoted by \(A\). Let \(P\) be a Sylow \(p\)-subgroup of \(D_{p}(G)\). Then \(D_{p}(G)=PA\). Firstly, \(A\) is a Dedekind group. Case 1. \(A\) is a \(q\)-group. For a subgroup \(H\) of \(A\). Since \(A\leq D_{p}(G)\), we have \(A\) normalizes \(H^{\prime}(p)\). It follows from \(H^{\prime}(p)=H\) that \(H\) is normal in \(A\), that is, \(A=N(A)\) is a Dedekind group. Case 2. \(A\) is not a \(q\)-group. Let \(A_{q}\) and \(A_{r}\) be any Sylow \(q\)-subgroup and Sylow \(r\)-subgroup of \(A\), respectively, \(q\
eq r\). Since \(A_{q}\) and \(A_{r}\) are subgroups of \(D_{p}(G)\), \(A_{q}\) normalizes \(A_{r}^{\prime}(p)\) and \(A_{r}\) normalizes \(A_{q}^{\prime}(p)\). Then it follows from \(A_{r}^{\prime}(p)=A_{r}\) and \(A_{q}=A_{q}^{\prime}(p)\) that \([A_{q},A_{r}]=1\), that is, \(A\) is nilpotent. For a subgroup \(H\) of \(A_{q}\), by the same argument above, \(A_{q}\) is Dedekind group, hence \(A\) is Dedekind group. Secondly, \(P=D(P)\) is a D-group. For a subgroup \(K\) of \(P\). Since \(P\leq D_{p}(G)\), we have \(P\) normalizes \(K^{\prime}(p)\). It follows from \(K^{\prime}(p)=K^{\prime}\) that \(K^{\prime}\) is normal in \(P\), that is, \(P=D(P)\) is a D-group. Finally, \(D_{p}(G)=P[A]\). Since \(P\) normalizes \(A^{\prime}(p)\) and \(A^{\prime}(p)=A\), we have \(D_{p}(G)=P[A]\). (2) By (1), the Hall \(p^{\prime}\)-subgroup of \(D_{p}(G)\) is Dedekind group for any prime \(p\in\pi(G)\), then \(\cap_{p||G|}D_{p}(G)\leq N(G)\). Hence \(\cap_{p||G|}D_{p}(G)=N(G)\) by Proportion \(1\) (1). Suppose that a group \(H\) acts on a group \(G\). We say that \(H\) acts hypercentrally on \(N\) if \(N\) has a subnormal series \(1=N_{0}\leq N_{1}\leq\cdots\leq N_{s}=N\) such that \([H,N_{i}]\leq N_{i-1}\), for all \(i=1,2,\cdots,s\) (cf. [11]). Clearly, if \(N\) is a normal subgroup of \(H\) then \(H\) acts hypercentrally on \(N\) if and only if \(N\leq Z_{\infty}(H)\). Lemma 2: _Let \(G\) be a \(\{p,q\}\)-group. Assume that \(N\) is a normal \(q\)-subgroup of \(G\) and \(H\) is a subgroup of \(G\) with \(H=O^{p}(H)\). If \(N\leq D_{p}(G)\), then \(H\) acts hypercentrally on \(N\)._  
**Proof. ** Suppose that the lemma is not true. Let \(G\) be a counterexample of minimal order. Then  
(1) \(G=NH\). If \(NH<G\), then \(NH\) satisfies the condition of the lemma by Proposition 1 (3) and the choice of \(G\) shows that \(H\) acts hypercentrally on \(N\), a contradiction. (2) _Let \(T\) be a minimal supplement of \(C_{G}(N)\) in \(G\), then \(O^{p}(T)=T\)._  
Since \(H=O^{p}(H)\) and \(N\) is a normal \(q\)-subgroup, we have \(G=HN=O^{p}(H)N=O^{p}(G)\). Let \(T\) be a minimal supplement of \(C_{G}(N)\) in \(G\). Then \(G=C_{G}(N)T\). Assume that \(O^{p}(T)<T\). Then \(C_{G}(N)O^{p}(T)<G\) by the minimality of \(T\). It is easy to see that \(G/C_{G}(N)O^{p}(T)=C_{G}(N)T/C_{G}(N)O^{p}(T)\cong T/C_{T}(N)O^{p}(T)\) is a \(p\)-group, and then \(O^{p}(G)\leq C_{G}(N)O^{p}(T)<G\), a contradiction. (3) \(G=NT\), _and \(T\trianglelefteq G\)._  
If \(NT<G\), then \(NT\) satisfies the condition of the lemma by Proposition 1 (3). By the choice of \(G\), \(T\) acts hypercentrally on \(N\). Let \(T_{p}\) and \(C_{G}(N)_{p}\) be Sylow \(p\)-subgroup of \(T\) and \(C_{G}(N)\), respectively. Then \(T_{p}\) acts trivially on \(N\), and then \(G_{p}=C_{G}(N)_{p}T_{p}\) acts trivially on \(N\). Since \(G\) is a \(\{p,q\}\)-group, \(G/C_{G}(N_{i}/N_{i-1})\) is a \(q\)-group for each \(G\)-chief factor \(N_{i}/N_{i-1}\) of \(N\). However, \(O_{q}(G/C_{G}(N_{i}/N_{i-1}))=1\) by ([12], Lemma 1.7.11). It follows that \(G/C_{G}(N_{i}/N_{i-1})=1\). This shows that \(G\) acts hypercentrally on \(N\), and so does \(H\), a contradiction. Thus, \(G=NT\)  
Since \(N\) normalizes \(T^{\prime}(p)\) and \(T=T^{\prime}(p)\), we have \(T\trianglelefteq G=NT\).