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\begin{document}
\begin{lemma}\label{lemma1}
Let  $x \in  G$  such that      $\ \vert E_{\frac{G}{\mathbb{R}_2}}^2 ( x \mathbb{R}_2 ) \vert  =\mathcall{P} $,  P a prime   where  $ \mathbb{R}_2 =\mathbb{R}_2 (G  )$.Then  for  all $ y \in   \bigtextl{G} $ with $ E_{\frac{G}{\mathbb{R}_2}}^2 ( x \mathbb{R}_2 )=E_{\frac{G}{\mathbb{R}_2}}^2 ( y \mathbb{R}_2 ) $ ,   we have $E_{G}^2 (x)=E_{G}^2 (y)$
 
\end{lemma}
\begin{proof}
We have that $\frac{E_{G}^2(x)}{\mathbb{R}_2} \leq E_{\frac{G}{\mathbb{R}_2}}^2(x \mathbb{R}_2)$. Suppose that  $\frac{E_{G}^2(x)}{\mathbb{R}_2} < E_{\frac{G}{\mathbb{R}_2}}^2(x \mathbb{R}_2)$ since $\vert E_{\frac{G}{\mathbb{R}_2}}^2(x \mathbb{R}_2) \vert =P $ and $\vert    \frac{E_{G}^2(x)}{\mathbb{R}_2} \vert $ divides  $\vert  E_{\frac{G}{\mathbb{R}_2}}^2 (x\mathbb{R}_2) \vert$ so  
$ \vert   \frac{E_{G}^2(x)}{\mathbb{R}_2}\vert =1 \Longrightarrow   E_{G}^2(x)= \mathbb{R}_2  \Longrightarrow x \in \mathbb{R}_2$ , a contradiction clearly , $\frac{E_{G}^2 (y)}{\mathbb{R}_2} \leq E_{\frac{G}{\mathbb{R}_2}}^2(y\mathbb{R}_2)= E_{\frac{G}{\mathbb{R}_2}}^2(x\mathbb{R}_2)$. Therefore  $ \vert  E_{\frac{G}{\mathbb{R}_2}}^2(x\mathbb{R}_2) \vert = \vert \frac{E_{G}^2 (y)}{\mathbb{R}_2}} \vert$ and so $\frac{E_{G}^2 (y)}{\mathbb{R}_2}}=\frac{E_{G}^2 (x)}{\mathbb{R}_2}}$. Thus $$\frac{E_{G}^2 (x)}{\mathbb{R}_2}}=\frac{E_{G}^2 (y)}{\mathbb{R}_2}} = \{ \mathbb{R}_2}, t_1 \mathbb{R}_2}, t_2 \mathbb{R}_2}, \dots , t_{p-1}\mathbb{R}_2}  \}$$
where$ \{t_1,\dots,t_{p-1} \} \in E_{G}^2(x) \bigcap E_{G}^2(y) -\mathbb{R}_2 $. So $ E_{G}^2 (x)=E_{G}^2 (y)$. Hence , the lemma follows.
\end{proof}
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\begin{theorem}
Let G be a group and P be a prime. If $\frac{G}{\mathbb{R}_2(G)} \cong C_P \times C_P$ then $\vert  E_{G}^2 (x) \vert  =P+2$
\end{theorem}
\begin{proof}
Suppose, first that $\frac{G}{\mathbb{R}_2(G)} \cong C_P \times C_P$ then 
\begin{align*}
\frac{G}{\mathbb{R}_2(G)} =& \left\langle    \mathbb{R}_2 x,\mathbb{R}_2 y \arrowvert (    \mathbb{R}_2 x)^P = (    \mathbb{R}_2 y)^P = \mathbb{R}_2 \right\rangle , (    \mathbb{R}_2 x) (    \mathbb{R}_2 y) = (   \mathbb{R}_2 y)(    \mathbb{R}_2 x)\\
=& \left\langle    \mathbb{R}_2 x,\mathbb{R}_2 y \arrowvert  x^P, y^P , x^{-1} y^{-1}xy \in \mathbb{R}_2 \right\rangle
\end{align*}
If $\frac{H}{\mathbb{R}_2} < \frac{G}{\mathbb{R}_2}$ , then $\left\vert   \frac{\frac{G}{\mathbb{R}_2}}{\frac{H}{\mathbb{R}_2}} \right\vert =P \Longrightarrow \left\vert \frac{H}{\mathbb{R}_2} \right\vert = P$. There for $H=\mathbb{R}_2 \bigcup \mathbb{R}_2 t_2 \bigcup \dots \bigcup \mathbb{R}_2 t_{p-1}$ where $ t_i \in H-\mathbb{R}_2$ and $i \in \{ 1,2,\dots , p-1 \}$ so the proper subgroups of G properly containing  $\mathbb{R}_2$ are 
\begin{align*}
H_1 =& \mathbb{R}_2 \bigcup \mathbb{R}_2 x^2 \bigcup \mathbb{R}_2 x^3 \bigcup \dots \bigcup \mathbb{R}_2 x^{P-1}\\
H_2=& \mathbb{R}_2 \bigcup \mathbb{R}_2 y \bigcup \mathbb{R}_2 y^2 \bigcup \dots \mathbb{R}_2 y^{P-1}\\
H_3=& \mathbb{R}_2 \bigcup \mathbb{R}_2 x^2 y \bigcup \mathbb{R}_2 x^2 y^2 \bigcup \mathbb{R}_2 x^2 y^{P-1}\\
\vdots &\\
H_{P+1}=&  \mathbb{R}_2 \bigcup \mathbb{R}_2 x^{P-1}y \bigcup \mathbb{R}_2 x^{P-1} y^2 \bigcup \dots  \mathbb{R}_2 x^{P-1} y^{P-1}
\end{align*}
No we will show that $H_1 , H_2 ,\dots H_{P+1} $ are only proper Engelizer of G. Let $a \in G-\mathbb{R}_2$ , $a \in G-\mathbb{R}_2$ then $\mathbb{R}_2 a = \mathbb{R}_2 k$ such that  $$k \in \left\{ x,\dots , x^{P-1},y ,\dots , y^{P-1}, xy ,xy^2, \dots, xy^{P-1}, \dots x^{P-1}y, \dots, x^{P-1}y^{P-1}  \right\}.$$\\
There for $E_{\frac{G}{\mathbb{R}_2}}^2 (\mathbb{R}_2 a) = E_{\frac{G}{\mathbb{R}_2}}^2 (\mathbb{R}_2 k)$ from lemma \ref{lemma1}
  we have $E_{G}^2(a)=E_{G}^2(k)  $. Again let  $  k \in H_i - \mathbb{R}_2}(G)$ then $ E_{G}^2(k) \in \bigcup_{j=1}^{P+1} H_j$ as $H-1 ,H_2, \dots H_{P+1}$ are the only proper subgroups of G.
 Also $k \in E_{G}^2(k) $, therefor $E_{G}^2(k) \neq H_j , 1\leq j \leq P+1$ and $i \neq j$.\\
therefore $E_{G}^2(k)=H_i$ . Hence $H_1, H_2, \dots H_{P+1}$ are the only  proper Engelizer of G. thus $\left\vert E_{G}^2 \right\vert =P+2$. This complete the proof. To  the reverse of theorem we check for $p=2,3,5,7$.
\end{proof}
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\begin{theorem}
Let G be agroup. Then $\vert Cent(G) \vert =4 $ if only if $\frac{G}{R_2(G)} \cong C_2 \times C_2$
\end{theorem}
\begin{proof}
به طور مشابه قضیه 1
\end{proof}
\end{document}