# The Lengths of Certain Real Conjugacy Classes and the Related Prime Graph
## 2 Preliminaries  
We will need some results on conjugacy classes, and in particular, on real classes. **Lemma 1**. : _Let \(N\trianglelefteq G\). Then, \(|x^{N}|\big{|}|x^{G}|\) for all \(x\in N\), and \(|(xN)^{G/N}|\big{|}|x^{G}|\) for all \(xN\in G/N\)._  
Lemma 1 is well known, and we will use it without further reference. **Lemma 2** ([9], Lemma 2.1). : _(a) If \(x\in\operatorname{Re}(G)\) and \(|x^{G}|\) is odd, then \(x^{2}=1\)._  
_(b) If \(x\in\operatorname{Re}(G)\), then every power of \(x\) is a real element of \(G\)._  
_(c) If \(x\in\operatorname{Re}(G)\), then \(x^{t}=x^{-1}\) for some \(2\)-element \(t\in G\)._  
_(d) If \(N\lhd G\) and \(|G/N|\) is odd, then \(\operatorname{Re}(G)=\operatorname{Re}(N)\)._  
**Lemma 3** ([10], Lemma 2.5). : _Let \(x,y\in G\) be such that \(xy=yx\)._  
_(a) Assume \(G=\operatorname{C}_{G}(x)\operatorname{C}_{G}(y)\). Then, \((xy)^{G}=x^{G}\cdot y^{G}\). If in addition \(x\) and \(y\) are in \(\operatorname{Re}(G)\), then \(xy\in\operatorname{Re}(G)\)._  
_(b) If \((|x|,|y|)=1\), or if \(\langle x\rangle\) and \(\langle y\rangle\) are normal in \(G\) and have trivial intersection, then \(\operatorname{C}_{G}(xy)=\operatorname{C}_{G}(x)\cap\operatorname{C}_{G}(y)\), and therefore \(\pi(|x^{G}|)\cup\pi(|y^{G}|)\subseteq\pi(|xy^{G}|)\)._  
**Lemma 4** ([10], Proposition 2.8). : _Let \(A\) be a group of odd order that acts faithfully on a nontrivial elementary abelian \(2\)-group \(V\) of order \(2^{n}\). Assume that \(\operatorname{C}_{V}(A)=1\). Then, the following conditions are equivalent:_  
_(a) For every nontrivial_ \(x\in V\)_,_ \(|A:\operatorname{C}_{A}(x)|\) _is a prime number._  
_(b) There exists a prime_ \(p\) _such that, for every nontrivial_ \(x\in V\)_,_ \(|A:\operatorname{C}_{A}(x)|=p\)_._  
**Lemma 5** ([2], Proposition 6.4). : _The following are equivalent:_  
_(a) Every nontrivial element in \(\operatorname{Re}(G)\) has even order;_  
_(b) Every element in \(\operatorname{Re}(G)\) is a \(2\)-element;_  
_(c) \(G\) has a normal Sylow \(2\)-subgroup._  
**Lemma 6** ([10], Lemma 3.3). : _Let \(x\) be a nontrivial real odd order element of \(G\). If \(|x^{G}|\) has prime length \(q\), then \(q=2\) and \(\langle x\rangle\trianglelefteq G\). Furthermore, for every nontrivial \(y\in\langle x\rangle\), we have \(\operatorname{C}_{G}(y)=\operatorname{C}_{G}(x)\)._  
The next Lemma gives a neat description of the structure of \(G\) when \(\operatorname{cs}_{r}(G)=\{1,2,p\}\). **Lemma 7** ([10], Theorem 3.4). : _Let \(p\) be an odd prime. Then, \(\mathsf{cs}_{r}(G)=\{1,2,p\}\) if and only if \(G=H\times O\), where \(H\) is a dihedral group of order \(2p\) and \(\mathsf{cs}_{r}(O)=\{1\}\)._  
Lemma 8 and 9 are for inductive purposes. **Lemma 8** ([7], Lemma 2.2). : _Suppose that \(N\) is a normal subgroup of \(G\) and \(xN\in\mathrm{Re}(G/N)\). If \(|N|\) or the order of \(xN\) in \(G/N\) is odd, then \(xN=yN\) for some real element \(y\) of \(G\) (of odd order if the order of \(xN\) is odd)._  
**Lemma 9**. : _Let \(\pi\) be a set of some primes and \(N\) a normal subgroup of \(G\)._  
_(a) ([11], Lemma 2.3) If \(\bar{x}=xN\) is a \(\pi\)-element, then there exists a \(\pi\)-element \(x_{1}\) of \(G\) such that \(\bar{x}=\bar{x}_{1}\)._  
_(b) If \(|N|\) or the order of \(xN\) in \(G/N\) is odd, and \(\bar{x}=xN\) is a real \(\pi\)-element, then there exists a real \(\pi\)-element \(x_{1}\) of \(G\) such that \(\bar{x}=\bar{x}_{1}\)._  
Proof. : (b) By Lemma 8, there exists a real element \(y\) of \(G\) such that \(xN=yN\). Clearly, we can write \(y=y_{1}y_{2}\), where \(y_{1}\) is a \(\pi\)-element and \(y_{2}\) is a \(\pi^{\prime}\)-element. As the order of \(yN\) is the product of the orders of \(y_{1}N\) and \(y_{2}N\), we have \(y_{2}\in N\); therefore, \(yN=y_{1}N\). Since \(y_{1}\) is some power of \(y\), by Lemma 2(b), \(y_{1}\) is real in \(G\). Choose \(y_{1}=x_{1}\) will meet the conclusion. Lemma 10 and Lemma 11 are used in Sections 4 and 5, respectively. **Lemma 10**. : _Let \(p\) be a prime and \(K\unlhd G\) such that \(|G|_{p}=|K|_{p}\).