#
Quasirecognition by Prime Graph of the Groups ^{2}D_{2n}(q) Where q < 10^{5}

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

**Main**

**Theorem.**

## 2. Preliminary Results

**Lemma**

**1**

- 1.
- there exists a finite nonabelian simple group S such that$$S\le \overline{G}=G/K\le Aut\left(S\right)$$
- 2.
- for every independent subset ρ of $\pi \left(G\right)$ with $\left|\rho \right|\ge 3$ at most one prime in ρ divides the product $\left|K\right|\xb7|\overline{G}/S|$. In particular, $t\left(S\right)\ge t\left(G\right)-1$.
- 3.
- one of the following holds:
- (a)
- every prime $r\in \pi \left(G\right)$ nonadjacent to 2 in $\mathsf{\Gamma}\left(G\right)$ does not divide the product $\left|K\right|\xb7|\overline{G}/S|$; in particular, $t(2,S)\ge t(2,G)$;
- (b)
- there exists a prime $r\in \pi \left(K\right)$ nonadjacent to 2 in $\mathsf{\Gamma}\left(G\right)$; in this case $t\left(G\right)=3$, $t(2,G)=2$, and $S\cong {A}_{7}$ or ${L}_{2}\left(q\right)$ for some odd q.

**Remark**

**1.**

**Lemma**

**2**

**Lemma**

**3**

**Lemma**

**4**

**Lemma**

**5**

**Lemma**

**6.**

- 1.
- there is a primitive prime ${p}^{\prime}$ for ${p}^{n}-1$, that is , ${p}^{\prime}\mid ({p}^{n}-1)$ but ${p}^{\prime}\vee ({p}^{m}-1)$, for every $1\le m<n$,
- 2.
- $p=2$, $n=1$ or 6,
- 3.
- p is a Mersenne prime and $n=2$.

## 3. Proof of the Main Theorem

**Lemma**

**7.**

**Proof.**

**Case 1.**Let $\{47,59\}\u228a\pi ({q}^{2}-1)$, where $q={p}^{\alpha}<{10}^{5}$. Therefore, we get that $e(47,q)\ge 23$ or $e(59,q)\ge 29$. $\{47,59\}\subseteq \pi \left(S\right)$, so G contains an element $a\in {A}_{m}$, such that $e(a,q)\ge 23$, which implies that $n\ge 24$. Now we have $min\left\{t\right(47,G),t(59,G\left)\right\}\ge 19$. Hence, according to Remark 1, $min\left\{t\right(47,S),t(59,S\left)\right\}\ge 18$ in S. On the other hand, 47 is not connected to the prime numbers in the interval $[m-46,m]$ in the prime graph of ${A}_{m}$ and similarly, 59 is not connected to the prime numbers in the interval $[m-58,m]$, in the prime graph of ${A}_{m}$. However, these intervals contain at most 16 prime numbers, and this implies that $t\left(S\right)<t\left(G\right)-1$, is a contradiction.

**Case 2.**Let $\{47,59\}\subseteq \pi ({q}^{2}-1)$, where $q={p}^{\alpha}<{10}^{5}$. Using GAP, we get that:

**Lemma**

**8.**

**Proof.**

**Case 1.**Let ${r}_{2n-2}\in \pi \left(S\right)$. In addition, let ${p}_{1}$ and ${p}_{2}$ be two primitive prime divisors of ${p}^{(2n-2)\alpha}-1$ and ${p}^{2n\alpha}-1$, respectively. So we may assume that ${p}_{1}$ and ${p}_{2}$ are ${r}_{2n-2}$ and ${r}_{2n}$, respectively. This implies that $\{{r}_{2n-2},{r}_{2n}\}\subseteq \pi \left(S\right)$. Thus ${r}_{2n-2}$ is a primitive prime divisor of ${q}^{\prime s}-1$ and ${r}_{2n}$ is a primitive prime divisor of ${q}^{\prime t}-1$, where $s=e({r}_{2n-2},{p}^{\beta})$ and $t=e({r}_{2n},{p}^{\beta})$. It follows that $(2n-2)\alpha \mid s\beta $ and $2n\alpha \mid t\beta $. On the other hand, using Zsigmondy’s theorem, we conclude that $t\beta \le 2n\alpha $ and so $t\beta =2n\alpha $. Furthermore, since $2n<2(2n-2)$, we have $s\beta =(2n-2)\alpha $ and $s<t$.

**Subcase 1.1.**Let $S\cong {L}_{m}\left({q}^{\prime}\right)$. By [[12] Proposition 2.6], we see that each prime divisor of $|{L}_{m}\left({q}^{\prime}\right)|$ is adjacent to p, except ${r}_{m}^{\prime}$ and ${r}_{m-1}^{\prime}$. Hence $\rho (p,S)=$$\{p,{r}_{m-1}^{\prime},{r}_{m}^{\prime}\}$. Therefore, ${p}_{1}$ and ${p}_{2}$ are some primitive prime divisors of ${q}^{\prime m}-1$ and ${q}^{\prime m-1}-1$. Since $s<t$, we conclude that $m=t$ and $m-1=s$. Hence $2n\alpha =m\beta $ and $(2n-2)\alpha =(m-1)\beta $. Consequently, we get that $\beta =2\alpha $ and $m=n$, that is $S\cong {L}_{n}\left({p}^{2\alpha}\right)$. Then S has a maximal torus of order $({p}^{2n\alpha}-1)/\left(({p}^{2\alpha}-1)(n,{p}^{2\alpha}-1)\right)$, say T. Obviously, ${r}_{n},{r}_{2n}\in \pi \left(T\right)$. Therefore, ${r}_{n}\sim {r}_{2n}$ in $\mathsf{\Gamma}\left({L}_{n}\left({p}^{2\alpha}\right)\right)$, whereas ${r}_{n}\nsim {r}_{2n}$ in $\mathsf{\Gamma}\left(G\right)$, by Lemma 2, which is a contradiction.

**Subcase 1.2.**Let $S\cong {U}_{m}\left({q}^{\prime}\right)$. If $m=3$, then $\rho (p,S)=\{p,{r}_{1}^{\prime}\ne 2,{r}_{6}^{\prime}\}$ and so $s=1$ and $t=6$. Hence $(2n-2)\alpha =\beta $ and $2n\alpha =6\beta $. Therefore $n=6/5$, is a contradiction.

**Subcase 1.3.**Let $S\cong {B}_{m}\left({q}^{\prime}\right)$ or ${C}_{m}\left({q}^{\prime}\right)$. Since $t(p,S)\ge 3$, using [[12] Table 4], we get that m is odd. In this case, $\rho (p,S)=\{p,{r}_{m}^{\prime},{r}_{2m}^{\prime}\}$. Hence $s=m$ and $t=2m$ and so $(2n-2)\alpha =m\beta $ and $2n\alpha =2m\beta $, which implies that $n=2$, which is a contradiction. Let $S\cong {}^{2}{D}_{m}\left({q}^{\prime}\right)$, where m is odd. Since $\rho (p,S)=\{p,{r}_{2m-2}^{\prime},{r}_{2m}^{\prime}\}$, we conclude that $(2n-2)\alpha =(2m-2)\beta $ and $2n\alpha =2m\beta $ and so $m=n$, which is impossible, since n is even.

**Case 2.**Let ${r}_{2n-2}\notin \pi \left(S\right)$. Hence ${r}_{n-1}\in \pi \left(S\right)$. Let ${p}_{1}$ and ${p}_{2}$ be as ${r}_{n-1}$ and ${r}_{2n}$, respectively. Therefore ${r}_{n-1}$ and ${r}_{2n}$ are primitive prime divisors of ${q}^{\prime s}-1$ and ${q}^{\prime t}-1$, respectively, where $s=e({r}_{n-1},{p}^{\beta})$ and $t=e({r}_{2n},{p}^{\beta})$. Now we conclude that $(n-1)\alpha \mid s\beta $ and $2n\alpha \mid t\beta $. On the other hand, using Zsigmondy’s theorem, we conclude that $t\beta \le 2n\alpha $ and so $t\beta =2n\alpha $. If $s\beta >(n-1)\alpha $, then using Zsigmondy’s theorem, we conclude that $s\beta =(2n-2)\alpha $, which implies that ${r}_{2n-2}\in \pi \left(S\right)$, which is a contradiction. Hence we suppose that $s\beta =(n-1)\alpha $.

**Subcase 2.1.**Let $S\cong {L}_{m}\left({q}^{\prime}\right)$, where ${q}^{\prime}={p}^{\beta}$. We know that $\rho (p,S)=$ $\{p,{r}_{m-1}^{\prime},{r}_{m}^{\prime}\}$. Hence $t=m$, $s=m-1$, $2n\alpha =m\beta $, and $(n-1)\alpha =(m-1)\beta $. These equations imply that $m=2-2/(n+1)$, which is impossible.

**Subcase 2.2.**Let $S\cong {}^{2}{D}_{m}\left({q}^{\prime}\right)$, where m is odd. We note that $\rho (p,S)=\{p,{r}_{2m-2}^{\prime},{r}_{2m}^{\prime}\}$ and so $(n-1)\alpha =(2m-2)\beta $ and $2n\alpha =2m\beta $ and so $m=2-2/(n+1)$, which is impossible.

**Subcase 2.3.**Let $S\cong {}^{2}{D}_{m}\left({q}^{\prime}\right)$, where m is even. Since $\rho (p,S)=$ $\{p,{r}_{m-1}^{\prime},{r}_{2m-2}^{\prime},{r}_{2m}^{\prime}\}$, we get that $2n\alpha =2m\beta $ and $(n-1)\alpha =(2m-2)\beta $ or $(n-1)\alpha =(m-1)\beta $. If $(n-1)\alpha =(2m-2)\beta $, then we get that $m=2-2/(n+1)$, which is impossible. Hence $(n-1)\alpha =(m-1)\beta $, which implies that $m=n$ and $\alpha =\beta $, and so $S\cong D$, which is a contradiction, since ${r}_{2n-2}\notin \pi \left(S\right)$.

**Lemma**

**9.**

**Proof.**

## Author Contributions

## Conflicts of Interest

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**MDPI and ACS Style**

Moradi, H.; Darafsheh, M.R.; Iranmanesh, A.
Quasirecognition by Prime Graph of the Groups ^{2}*D*_{2n}(*q*) Where *q* < 10^{5}. *Mathematics* **2018**, *6*, 57.
https://doi.org/10.3390/math6040057

**AMA Style**

Moradi H, Darafsheh MR, Iranmanesh A.
Quasirecognition by Prime Graph of the Groups ^{2}*D*_{2n}(*q*) Where *q* < 10^{5}. *Mathematics*. 2018; 6(4):57.
https://doi.org/10.3390/math6040057

**Chicago/Turabian Style**

Moradi, Hossein, Mohammad Reza Darafsheh, and Ali Iranmanesh.
2018. "Quasirecognition by Prime Graph of the Groups ^{2}*D*_{2n}(*q*) Where *q* < 10^{5}" *Mathematics* 6, no. 4: 57.
https://doi.org/10.3390/math6040057