# The Lengths of Certain Real Conjugacy Classes and the Related Prime Graph
## 5 The Prime Graph
Since \(P\) lies in the kernel of the action of \(N\) on \(Z\), we have the decomposition \(Z=\operatorname{C}_{Z}(N)\times[Z,N]\), with \(\operatorname{C}_{Z}(N)\), \([Z,N]\lhd G\). Let \(x\in\operatorname{Re}^{*}(G)\) with \(\Delta_{G}(x)=\{q\}\). Then, \(x\in Z\cap\operatorname{C}_{G}(N)=\operatorname{C}_{Z}(N)\). Observe that \([Z,N]\
eq 1\), as otherwise every real class of elements of \(\operatorname{Re}^{*}(G)\) that has an odd length would have \(q\)-power length and hence \(n(\Gamma_{r}^{*}(G))\leqslant 2\). Let \(y\in[Z,N]\) be an involution. Then, from \(N\subseteq\mathrm{C}_{G}(x)\) and \(N\
subseteq\mathrm{C}_{G}(y)\), we obtain \((|x^{G}|,|y^{G}|)=1\), so, by Lemma 3(a), it follows \(xy\in\mathrm{Re}(G)\). Of course, \(xy\in\mathrm{Re}^{*}(G)\). Since \(x\) and \(y\) lie in normal subgroups of \(G\) which intersect trivially, we get \(\mathrm{C}_{G}(xy)=\mathrm{C}_{G}(x)\cap\mathrm{C}_{G}(y)\) and therefore \(\pi(|(xy)^{G}|)\supseteq\pi(|x^{G}|)\cap\pi(|y^{G}|)\). This gives \(\Delta_{G}(x)=\Delta_{G}(xy)=\Delta_{G}(y)\), a contradiction. If \(N\unlhd G\), by the fact that \(\mathrm{Re}^{*}(N)\subseteq\mathrm{Re}^{*}(G)\), we easily see that \(\Gamma_{r}^{*}(N)\) is a subgraph of \(\Gamma_{r}^{*}(G)\). However, it may not be true that \(\Gamma_{r}^{*}(G/N)\) is a subgraph of \(\Gamma_{r}^{*}(G)\). However, this is the case if \(|N|\) is odd. Lemma 12: _Let \(N\unlhd G\) with \(|N|\) odd. Then, \(\Gamma_{r}^{*}(G/N)\) is a subgraph of \(\Gamma_{r}^{*}(G)\)._  
Proof: First, we argue that \(\mathrm{V}_{r}^{*}(G/N)\subseteq\mathrm{V}_{r}^{*}(G)\). Let \(p\in\mathrm{V}_{r}^{*}(G/N)\) and let \(xN\in\mathrm{Re}^{*}(G/N)\) such that \(p\big{|}|(xN)^{G/N}|\). By Lemma 9(b), there exists an element \(y\in\mathrm{Re}^{*}(G)\) such that \(xN=yN\). Since \(|(xN)^{G/N}|=|(yN)^{G/N}||||y^{G}|\), \(p\big{|}|y^{G}|\) and \(p\in\mathrm{V}_{r}^{*}(G)\). With a similar argument, we can prove that, if there are two different vertices \(p,q\in\mathrm{V}_{r}^{*}(G/N)\), adjacent in \(\Gamma_{r}^{*}(G/N)\), then \(p,q\) are adjacent in \(\Gamma_{r}^{*}(G)\) by using Lemma 9(b) again. Thus, \(\Gamma_{r}^{*}(G/N)\) is a subgraph of \(\Gamma_{r}^{*}(G)\). Let \(H\) be a subgroup or a quotient group of \(G\). Assume that \(\Gamma_{r}^{*}(H)\) is a subgraph of \(\Gamma_{r}^{*}(G)\), and \(\Gamma_{r}^{*}(G)\) is also disconnected having two connected components with vertex sets \(\pi_{1}\) and \(\pi_{2}\), respectively. To show that \(\Gamma_{r}^{*}(H)\) is disconnected, it suffices to show that \(\mathrm{V}_{r}^{*}(H)\cap\pi_{i}\
eq 2\) for \(i=1,2\), or, equivalently, there exists two elements \(u_{1},u_{2}\) of \(\mathrm{Re}^{*}(H)\) such that \(\pi(|u_{1}^{H}|)\cap\pi_{i}\
eq\varnothing\) for \(i=1,2\). Theorem 3.2: _Suppose that, for a group \(G\), \(\Gamma_{r}^{*}(G)\) has two connected components with vertex sets \(\pi_{1}\) and \(\pi_{2}\), where \(2\
otin\pi_{2}\). Then:_  
_(1) If \(G\) is not \(2\)-closed and \(M\unlhd G\) with \(|G:M|\) odd, then \(\Gamma_{r}^{*}(M)\) is disconnected._  
_(2) If \(N\unlhd G\) with \(|N|\) odd and assuming further that \(G=\mathrm{O}^{2^{\prime}}(G)\) is not \(2\)-nilpotent, then \(\Gamma_{r}^{*}(G/N)\) is disconnected._  
Proof: By Lemma 11, \(G\) has an involution \(i\) such that \(|i^{G}|>1\) is a \(\pi_{2}\)-number and \(\mathrm{C}_{G}(i)\) has a normal Sylow \(2\)-subgroup \(S\). For (1), let \(M\unlhd G\) with \(|G:M|\) being odd. We have \(V_{r}^{*}(M)\subseteq V_{r}^{*}(G)=\pi_{1}\cup\pi_{1}\). Notice that, if \(M\) is \(2\)-closed, then \(G\) is also \(2\)-closed, hence we may assume that \(M\) is not \(2\)-closed. By Theorem 2, \(2\) divides the class length of some element of \(\mathrm{Rp}(M)\) and hence \(2\in\mathrm{V}_{r}^{*}(M)\cap\pi_{1}\). Now, \(M\) contains every real element of \(G\) by Lemma 2(d); then, \(i\in M\). If \(M\leq\mathrm{C}_{G}(i)\), then \(S\unlhd M\) as \(|G:M|\) is odd, so \(M\) is \(2\)-closed, a contradiction. Thus, \(|i^{M}|>1\) and then \(\mathrm{V}_{r}^{*}(M)\cap\pi_{2}\
eq\varnothing\).