# The Lengths of Certain Real Conjugacy Classes and the Related Prime Graph
## 4 The Class Lengths of Primary Real Elements
Since \(\mathrm{O}_{2^{\prime}}(G/\mathrm{O}_{2^{\prime}}(G))=1\), by ([16], the main theorem), the last term of the derived series of \(G/\mathrm{O}_{2^{\prime}}(G)\) is isomorphic to \(G_{1}\times G_{2}\times\cdots\times G_{s}\), where each \(G_{i}\) is isomorphic to either \(\mathrm{SL}_{2}(q)\) with odd number \(q>5\), or \(2\cdot\mathrm{A}_{7}\), the perfect double cover of \(\mathrm{A}_{7}\). We claim that, for some \(i\), there exists a real element \(a\) of \(G_{i}\), and therefore of \(G/\mathrm{O}_{2^{\prime}}(G)\), such that \(|a|=2^{m}\geqslant 4\). First, assume that \(G_{i}\cong 2\cdot\mathrm{A}_{7}\) for some \(i\). Using [17], \(G_{i}\) has a real element of order 4. Next, assume that \(G_{i}\cong\mathrm{SL}_{2}(q)\) with odd number \(q>5\). As we know (see ([18], Theorem 2.8.3)), the Sylow 2-subgroup \(S\) of \(\mathrm{SL}_{2}(q)\) is a generalized quaternion group \(Q_{2^{k+1}}\) of order \(2^{k+1}\geqslant 8\). From its presentation \(\langle a,b:a^{2^{k-1}}=b^{2},a^{2k}=1,b^{-1}ab=a^{-1}\rangle\), we obtain a real element \(a\) of \(\mathrm{SL}_{2}(q)\) of order \(2^{k}\geqslant 4\). Of course, \(a\) is also real in \(G/\mathrm{O}_{2^{\prime}}(G)\) as \(G_{i}\) is a subgroup of \(G/\mathrm{O}_{2^{\prime}}(G)\). The claim follows. \(a\) must lie in the center of \(G/\mathrm{O}_{2^{\prime}}(G)\), which yields \(a\in Z(G_{i})\). Notice that \(|Z(G_{i}|=2\) for all \(i\geqslant 1\), and we get a contradiction. Therefore, \(G\) is solvable. Suppose that \(G>\mathcal{O}^{2^{\prime}}(G)\). Using Lemma 10 where we let \(p=2\) and \(K=\mathrm{O}^{2^{\prime}}(G)\), we see that \(|x^{K}|_{2}=2^{n}\) for all non-central \(x\in\mathrm{Rp}(K)\) and by induction \(\mathrm{O}^{2^{\prime}}(G)\) is 2-nilpotent. Hence, we may assume that \(G=\mathcal{O}^{2^{\prime}}(G)\). Let \(\overline{G}=G/\mathrm{O}_{2^{\prime}}(G)\) and use bar as the customary. Let \(\overline{P}=\mathrm{O}_{2}(\overline{G})\). \(\overline{G}\) has a Hall \(2^{\prime}\)-subgroup \(\overline{H}\) since it is solvable. By the second paragraph, we see that every nontrivial real 2-element of \(\overline{G}\) lies in \(Z(\overline{G})\). This means that the conjugation action of \(\overline{H}\) on \(\overline{P}\) fixes all real elements of order 2 or 4, and then \(\overline{H}\) centralizes \(\overline{P}\) by ([19],Theorem B). By ([12], Theorem 3.21), we get \(\overline{H}\leq\mathrm{C}_{\overline{G}}(\overline{P})\leq\overline{P}\), which yields \(\overline{H}=1\). Hence, \(\overline{P}=\overline{G}\) is a 2-group and we conclude that \(G\) is 2-nilpotent. Theorem 4.1: _Let \(p\) be a prime, \(|x^{G}|\) a power of \(p\) for every \(2\)-element, and every \(p\)-element \(x\) of \(G\), and also for every real primary \(p^{\prime}\)-element \(x\) of \(G\). Then, \(G\) has a normal \(p\)-complement._  
Proof: First, by ([8], Theorem 4.2), we see that \(G\) is a solvable group. Next, we show that \(G\) has a normal \(p\)-complement by induction on \(|G|\). Let \(P\in\mathrm{Syl}_{p}(G)\) and \(M\) be a normal maximal subgroup of \(G\). Then, \(|G/M|=q\) a prime. Note that \(\mathrm{Rp}(M)\subseteq\mathrm{Rp}(G)\) and \(|x^{M}|||x^{G}|\) for all \(x\in M\), By induction, we know that \(M\) has a normal \(p\)-complement \(H\). Hence, we may assume that \(q\
eq p\). Let \(\overline{G}=G/H\). We claim that \(\overline{G}\) inherits the hypothesis on \(G\). Let \(\tilde{x}\) be a \(p\)-element (2-element, resp. ) of \(\overline{G}\). By Lemma 9(a), there exists a \(p\)-element (2-element, resp. ) \(x_{1}\) of \(G\) such that \(\tilde{x}=\tilde{x}_{1}\). Therefore, \(|\tilde{x}^{\overline{G}}|=|\tilde{x_{1}}^{\overline{G}}|||x_{1}^{G}|\), which yields that \(|\tilde{x}^{\overline{G}}|\) is a \(p\)-power. Let \(\tilde{y}\) be a real primary \(p^{\prime}\)-element of \(\overline{G}\), and we can assume that \(|\tilde{y}|=r^{m}\) for some prime \(r\
eq 2\), since otherwise, by hypothesis and Lemma 9(a), we easily get that \(|\tilde{y}^{\overline{G}}|\) is a power of \(p\).