# The Lengths of Certain Real Conjugacy Classes and the Related Prime Graph
## 4 The Class Lengths of Primary Real Elements  
Recognizing \(p\)-nilpotency (i.e., has a normal \(p\)-complement or not) is clearly an important problem. In this section, we provide some evidence for a conjecture proposed in [8],which said that, if the lengths of all the nontrivial real classes in \(G\) have the same 2-part, then \(\mathrm{O}^{2^{\prime}}(G)\) is 2-nilpotent; in particular, \(G\) is solvable. In Theorem 4, we prove that, with only real classes of primary elements being considered, this conjecture is held under some condition on the Sylow 2-subgroups. In ([13], Theorem 3.4), [PERSON] et al. showed that all real class lengths of \(G\) are powers of 2 if and only if \(G\) has a normal 2-complement \(K\) and \(\mathrm{Re}(G)\subseteq\mathrm{C}_{G}(K)\). However, when \(p\) is an arbitrary prime, the situation for the existence of a normal \(p\)-complement is different. We also remark that, in ([14], Lemma 2.2), [PERSON] and [PERSON] proved that, if \(|x^{G}|\) is a \(p\)-number for every primary \(p^{\prime}\)-element \(x\) of \(G\), then \(G\) is a solvable group with abelian \(p\)-complements. Here, we obtain Theorem 5 as an analogous result. At the end of this section, a particular class of groups is concerned, namely, the \(\mathbf{P}\)-groups. For every subgroup \(H\) of \(G\), if \(H<P_{G}(H)\) where \(P_{G}(H)=\langle x|x\in G,\langle x\rangle H=H\langle x\rangle\rangle\), then \(G\) is said to satisfy the permutizer condition \(\mathbf{P}\), or to be a \(\mathbf{P}\)-group. In [15], the authors proved that \(\mathbf{P}\)-groups are solvable by making use of the Classification Theorem of Finite Simple Groups (CFSG), they also proved that every maximal subgroup of a \(\mathbf{P}\)-group has index 4 or a prime. It is meaningful to consider how close such groups come to being supersolvable. Theorem 4: _Suppose that \(|x^{G}|_{2}=2^{n}\) for all non-central \(x\in\mathrm{Rp}(G)\). Assume further that \(G\) has a Sylow 2-subgroup \(S\) with \(\mathrm{Re}(S)\subseteq Z(S)\). Then, \(G\) is solvable and \(\mathcal{O}^{2^{\prime}}(G)\) is 2-nilpotent._  
Proof. : We first prove that \(G\) is solvable. Let \(G\) be a counterexample of minimal order. Then, \(G\) is non-solvable and has no normal Sylow 2-subgroup. By Corollary 1, there exists a nontrivial real primary \(2^{\prime}\)-element \(z\in G\). Clearly, \(z\
otin Z(G)\) and so \(|z^{G}|\) is even. Thus, \(|z^{G}|_{2}=2^{n}\geqslant 2\). That is, \(|x^{G}|_{2}=2^{n}\geqslant 2\) for all non-central \(x\in\mathrm{Rp}(G)\). Now we claim that every nontrivial real 2-element of \(G/\mathrm{O}_{2^{\prime}}(G)\) lies in the center of \(G/\mathrm{O}_{2^{\prime}}(G)\). Let \(x\mathrm{O}_{2^{\prime}}(G)\) be a nontrivial real 2-element of \(G/\mathrm{O}_{2^{\prime}}(G)\). By Lemma 9(b), there exists a real 2-element \(y\) of \(G\) such that \(x\mathrm{O}_{2^{\prime}}(G)=y\mathrm{O}_{2^{\prime}}(G)\). Considering \(y\), we have \(y^{t}=y^{-1}\) for some 2-element \(t\in G\) by Lemma 2(c). Write \(U=\langle y,t\rangle\). Since \(t\) normalizes \(\langle y\rangle\), \(U\) is a 2-subgroup of \(G\). We have \(U^{S}\leq S\) for some \(g\in G\) by the [PERSON] theorem. We want to show that \(y\in U\) is a central involution of \(G\). Notice that, if \(y^{8}\) is a central involution of \(G\), then so is \(y\). Therefore, we may assume that \(U\leq S\). Since \(y\in\mathrm{Re}^{*}(U)\), then \(y\in\mathrm{Re}(S)\subseteq Z(S)\), so it follows easily that \(y\) is an involution and \(|y^{G}|\) is odd. However, by the first paragraph, \(|y^{G}|\) is even if \(y\) is not central, and this forces \(y\in Z(G)\). We conclude that \(y\) is a central involution of \(G\), hence \(x\mathrm{O}_{2^{\prime}}(G)=y\mathrm{O}_{2^{\prime}}(G)\) is also a central involution of \(G/\mathrm{O}_{2^{\prime}}(G)\), and the claim follows.