Linear Space Data Structures for Finite Groups with Constant Query-Time

Abstract

A finite group of order n can be represented by its Cayley table. In the word-RAM model the Cayley table of a group of order n can be stored using $O(n^2)$ words and can be used to answer a multiplication query in constant time. It is interesting to ask if we can design a data structure to store a group of order n that uses $o(n^2)$ space but can still answer a multiplication query in constant time. Das et al. (J Comput Syst Sci 114:137–146, 2020) showed that for any finite group G of order n and for any $\delta \in [1/\log {n}, 1]$, a data structure can be constructed for G that uses $O(n^{1+\delta }/\delta )$ space and answers a multiplication query in time $O(1/\delta )$. Farzan and Munro (ISSAC, 2006) gave an information theoretic lower bound of $\Omega (n)$ on the number of words to store a group of order n. We design a constant query-time data structure that can store any finite group using O(n) words where n is the order of the group. Since our data structure achieves the information theoretic lower bound and answers queries in constant time, it is optimal in both space usage and query-time. A crucial step in the process is essentially to design linear space and constant query-time data structures for nonabelian simple groups. The data structures for nonabelian simple groups are designed using a lemma that we prove using the Classification Theorem for Finite Simple Groups.

Appendices

Appendix I

Here we prove the information theoretic lower bound on the space required to store groups due to Farzan and Munro [14]. For the proof we need the following concepts.

Let G be a group and $\Omega$ be a nonempty set. Let $\chi : G \times \Omega \rightarrow \Omega$ be a function. We denote $\chi (g,\alpha )$ by $\alpha ^g$ for all $g \in G$ and $\alpha \in \Omega$. Then we say that $\chi$ is an action of G on the set $\Omega$ if

• (i) $\alpha ^{e}=\alpha$ for all $\alpha \in \Omega$.

• (ii) $(\alpha ^{g_1})^{g_2}=\alpha ^{g_1 g_2}$ for all $\alpha \in \Omega$ and all $g_1,g_2 \in G$.

The set $\alpha ^G=\{\alpha ^g: g \in G \}$ is called the orbit of $\alpha$ under G. For $\alpha \in \Omega$, the subgroup $\textrm{Stab}_{G}(\alpha )=\{g \in G: \alpha ^g=\alpha \}$ is called the point stabilizer of $\alpha$ in G.

Theorem 9

(The orbit-stabilizer theorem, see e.g., [28]) For all $\alpha \in \Omega$, $\vert \alpha ^{G} \vert = [ G:\textrm{Stab}_{G}(\alpha )]$.

If $f:A\longrightarrow B$ is a function we sometime denote f(a) by $a^f$. In this notation function composition takes the following form: If $f:A\longrightarrow B$ and $g:B\longrightarrow C$ are two function, (gof)(a) is denoted as $a^{fg}$.

Let G and H be two groups. Let $\odot _G$ and $\odot _H$ be the binary operation on G and H respectively. A function $f:G\longrightarrow H$ is called a homomorphism if $(g_1\odot _G g_2)^f=g_1^f\odot _H g_2^f$ for all $g_1$ and $g_2$ in G. A bijective homomorphism $g:G\longrightarrow G$ is called an automorphism. The set of automorphism ${\textrm{Aut}(G)}$ of a groups G forms a subgroup of the group Sym(G) of permutations of G.

Theorem 3

[14] The space required to store a group from the set of all groups with n elements is $\Omega (n\log n)$ bit or $\Omega (n)$ words.

Proof

Let $G_1$ and $G_2$ be two groups on elements $1,2,\ldots ,n$. Let $\odot _{G_1}$ and $\odot _{G_2}$ be the binary operations in $G_1$ and $G_2$ respectively. Let $M_{G_1}$ and $M_{G_2}$ be the Cayley tables of $G_1$ and $G_2$, respectively. The (i, j)th entry of $M_{G_1}$ (or $M_{G_2}$) gives the product of i and j in $G_1$ ($G_2$ respectively). Let $\sigma : G_1 \longrightarrow G_2$ be a homomorphism. Since $\sigma$ is a homomorphism we have $M_{G_2}[i^{\sigma }, j^{\sigma }]=(M_{G_1}[i, j])^{\sigma }$ for all $i,j\in [n]$.

We consider an action of the group $S_n$ on the set containing all Cayley tables of all groups whose elements are $\{1,2,\ldots ,n\}$. The action is defined as follows: A $\sigma \in S_n$ maps a Cayley table M to $M^\sigma$ where $M^{\sigma }[i, j]:=(M[i^{\sigma ^{-1}}, j^{\sigma ^{-1}}])^{\sigma }$. It is easy to check that this is an action.

Let G be a group and let M be its Cayley table. Then by orbit counting lemma 9, the orbit of M under the above action is of size $\frac{\vert S_n \vert }{ \vert \textrm{Stab}_{S_n}(M) \vert }$, where $\textrm{Stab}_{S_n}(M)$ consists of all permutations of $S_n$ that maps M to M.

We claim that $\textrm{Stab}_{S_n}(M)=\mathrm{Aut(G)}$. Note that the claim will imply that the number of different Cayley tables representing groups isomorphic to G defined on set [n] is $\frac{n!}{\vert {\textrm{Aut}(G)}\vert }$. Let us now prove the claim.

Let $\sigma \in \textrm{Aut}(G)$. Then for all i and j we have,

$$\begin{aligned} M^{\sigma }[i, j]&=(M[i^{\sigma ^{-1}}, j^{\sigma ^{-1}}])^{\sigma } \\&= (i^{\sigma ^{-1}} \star _{G} j^{\sigma ^{-1}})^{\sigma } \\&= (i \star _{G} j), \quad ({\text { since } \sigma \text { is an automorphism of } G})\\&= M[i,j]. \end{aligned}$$

On the other hand if $\sigma \in \textrm{Stab}_{S_n}(M)$ then for all i and j,

$$\begin{aligned} M^{\sigma }[i, j]&=M[i, j] \\ (M[i^{\sigma ^{-1}}, j^{\sigma ^{-1}}])^{\sigma }&= M[i, j]\\ (i^{\sigma ^{-1}} \star _{G} j^{\sigma ^{-1}})^{\sigma }&= i \star _{G} j \\ (i^{\sigma ^{-1}} \star _{G} j^{\sigma ^{-1}})&= (i \star _{G} j)^{{\sigma }^{-1}}. \end{aligned}$$

Therefore, $\sigma ^{-1}$ is actually a homomorphism from G to G. Since $\sigma \in S_n$ is a bijection as well, we get $\sigma \in \textrm{Aut}(G)$.

Let ${\mathcal {G}}_n$ be the set of all non-isomorphic groups on [n]. Then the number of Cayley tables representing isomorphic copies of groups from ${\mathcal {G}}_n$ is $\sum _{G \in {\mathcal {G}}_n} \frac{n!}{\vert \textrm{Aut}(G) \vert }$.

We know that ${\mathcal {G}}_n$ contains a group isomorphic to ${\mathbb {Z}}_n$. Also, note that $\vert \textrm{Aut}({\mathbb {Z}}_n) \vert = \Phi (n)$, where $\Phi (n)$ is the Euler’s totient function (see [21, p. 135]). Then

$$\begin{aligned} \sum _{G \in {\mathcal {G}}_n} \frac{n!}{\vert \textrm{Aut}(G) \vert } \ge \frac{n!}{\Phi (n)} > \frac{n!}{n}= (n-1)!. \end{aligned}$$

Therefore, the domain whose elements are being stored by the data structure has size at least $(n-1)!$. Thus, it needs $\log ((n-1)!)$ bits to store this. But, $\log (n-1)! = \Omega (n \log n)$. $\square$

Appendix II

Proof of Lemma 1

In this section, we prove Lemma 1 in detail. Let us first recall Lemma 1 and Theorem 8.

Lemma 1

There are positive constants $b_1$ and $b_2$ such that for any nonabelian simple group H there exist subgroups $H_1$ and $H_2$ such that $1\le H_2 \le H_1 \le H$ and $\vert H_2 \vert \le \sqrt{\vert H \vert }$, $[H:H_1]\le b_1\sqrt{\vert H \vert }$, and $[H_1:H_2] \le b_2\sqrt{\vert H \vert }$.

Theorem 8

(see [23, p. 3])(The Classification Theorem of Finite Simple Group)

Every finite simple group is isomorphic to one of the following:

• (i) a cyclic group $C_p$ of prime order p;

• (ii) an alternating group $A_m,$  for $m \ge 5$;

• (iii) a classical group;

  • (a) linear:         $A_{\textrm{m}}(q) ( \text {or}\,\, { \mathrm PSL}_{\textrm{m}+1}(q) ), \textrm{m} \ge 1$, except $\textrm{PSL}_2(2)$ and $\textrm{PSL}_2(3)$;

  • (b) unitary:      $^2A_{\textrm{m}}(q^2) ( \text {or}\, \textrm{PSU}_{\textrm{m}+1}(q))\,, \textrm{m} \ge 2,$ except ${ \mathrm PSU}_{3}(2)$;

  • (c) symplectic: $C_{\textrm{m}}(q)) ( \text {or~}\, \textrm{PSp}{_{2\textrm{m}}}(q)), \textrm{m} \ge 2$ except ${\textrm{PS}_{p}}_{4}(2)$;

  • (d) orthogonal: $B_{\textrm{m}}(q) ( \text {or~}\,\textrm{P}\Omega _{2\textrm{m}+1}(q)), \textrm{m} \ge 3, q$ odd;          $D_{\textrm{m}}(q) ( \text {or~}\, \textrm{P} \Omega _{2\textrm{m}}^{+}(q)), \textrm{m} \ge 4$;          $^2D_{\textrm{m}}(q^2) ( \text {or~}\, \textrm{P}\Omega _{2\textrm{m}}^{-}(q))\,, \textrm{m} \ge 4$

  where q is a power $p^a$ of some prime;

• (iv) an exceptional group of Lie type:

  $$\begin{aligned} \mathrm {G_2(q), q \ge 3; F_4(q); E_6(q); ^2E_6(q); ^3D_4(q); E_7(q); E_8(q) \,\, \text {or} } \end{aligned}$$

  where q is a power $p^a$ of some prime;

  $$\begin{aligned} \mathrm {^2B_2(2^{2m+1})}, \textrm{m} \ge 1; ^2G_2(3^{2\textrm{m}+1}), \textrm{m} \ge 1; ^2F_4(2^{2\textrm{m}+1}), \textrm{m} \ge 1 \end{aligned}$$

  or the Tits group $^2F_4(2)^{'}$;

• (v) one of 26 sporadic simple groups:

  • (a) the five Mathieu groups $\mathrm {M_{11}, M_{12}, M_{22}, M_{23}, M_{24}}$;

  • (b) the seven Leech Lattice groups $\mathrm {Co_1, Co_2, Co_3, McL, HS, Suz, J_2}$;

  • (c) the three Fischer groups $\mathrm {Fi_{22}, Fi_{23}, Fi_{24}^{'};}$

  • (d) the Monstrous groups ${\mathbb {M}}, {\mathbb {B}},\textrm{ Th, HN, He}$;

  • (e) the six pariahs $\mathrm {J_1, J_2, J_4, O'N, Ly, Ru.}$

We now present the detailed calculations in the ordering mentioned in the Classification Theorem of Finite Simple Group, i.e., Theorem 8. As we mentioned in Sect. 2, we just need to use some known results on the order of certain subgroups of simple groups. The detailed description of these groups may be skipped for the purpose of the proof. The results that are used in the proof are on the orders of the finite simple groups, on the orders of maximal subgroups of simple groups, and on the normalizers of certain types of Sylow subgroups of simple groups. The information about the order of these simple groups can be obtained from [24] (see p. 252).

In case (ii) of Theorem 8, H is an alternating group. The stabilizer of any set is a subgroup of an alternating group under natural action. The subgroups $H_1$ and $H_2$ are suitably-picked stabilizer subgroups of the given alternating group H.

For the cases (iii) and (iv) of Theorem 8, we use Method 1 and Method 2 to get the desired subgroups as required in Lemma 1. In this cases the finite simple group H is of Lie-type and is defined over a finite field ${\mathbb {F}}_q$ where q is a power of some prime p. In Method 1, we take $H_2$ to be certain Sylow p-subgroup of H. The existence of such $H_2$ follows from the well-known Sylow theorem. For the existence of $H_1$, we take the normalizer of $H_2$ or the Borel subgroup. The information about the order of normalizer has been obtained from (see [22, p. 76], [23, p. 46]).

For the groups in which we use Method 2, we consider a maximal subgroup of H as $H_1$ and $H_2$ to be some Sylow p-subgroup of $H_1$. The index of a maximal subgroup (and hence its order) can be obtained from [25, p. 175], and [23, p. 156].

For the simple groups in case (v), we use Method 2, and the information about the order of maximal subgroup ($H_1$) can be obtained from [29]. Also, for the choice of $H_2$, we choose a certain Sylow subgroup of $H_1$.

The inequalities in the following two remarks are used in the calculation multiple times. For the sake of completeness we provide proofs of the inequalities.

Remark 4

For all integer $q>2,$ we have $\frac{q}{(q-1)^{2}}< 1.$

Proof

$$\begin{aligned} 1-\frac{q}{(q-1)^{2}}=&\frac{(q-1)^2-q}{(q-1)^{2}}=\frac{q^2-3q+1}{(q-1)^{2}}\\ =&\frac{(q-\frac{3+\sqrt{5}}{2})(q-\frac{3-\sqrt{5}}{2})}{(q-1)^{2}} > 0. \end{aligned}$$

Hence, $\frac{q}{(q-1)^{2}}< 1$, for $q >2$ $\square$

Remark 5

$$\begin{aligned} \prod _{i=1}^{m}(q^{i+1}-(-1)^{i+1})<q^{\sum _{i=1}^{m}(i+1)} \end{aligned}$$

.

Proof

We can observe that the sign of 1 changes alternatively. When i is even then sign of 1 is negative and when i is odd then sign of 1 is positive.

Now,

$$\begin{aligned} (q^{2j-1}-(-1)^{2j-1})(q^{2j}-(-1)^{2j})=(q^{4j-1}-1-q^{2j}+q^{2j-1}) < q^{4j-1} \,( j \ge 1). \end{aligned}$$

If m is even, then it is easy to see that we can pair two consecutive odd and even term, the product of these terms is less than the sum of powers of q. If m is odd, then the term which is not paired is $q^m-1$. Notice that $(q^m-1)< q^m$. Thus, in this case also the product of all the terms is less than the sum of powers of q. $\square$

We now prove Lemma 1 for each of the cases of Theorem 8 in the following few subsections.

Organization of the proof of Lemma 1. Before we go into the proof of Lemma 1, we first describe the subsections below that address specific subcase of Theorem 8. To begin, Sect. 6.1 contains case (ii) of Theorem 8. Section 6.2 contains case (iii) of Theorem 8. In which, Sect. 6.2.1 deals with the group $A_{m}(q)$, Sect. 6.2.2 contains the case for the group $~^2A_m(q^2)$, Sect. 6.2.3 deals with the group $C_{m}(q)$, and Sect. 6.2.4 contains the cases for the groups $B_{m}(q), D_{m}(q), ~^2D_m(q^2)$. The cases for all the exceptional simple groups of Lie types described in case (iv) of Theorem 8 are proved in Sect. 6.3.

1.1 Alternating group

The Alternating group $A_m$ is a group of all even permutations of a finite set. It is well known that $A_m$ is a simple group when $m \ge 5$. The set

$$\begin{aligned} \{\sigma \in A_m\vert \sigma (\ell )=\ell \text { for } \ell =i+1,\ldots ,m\} \end{aligned}$$

forms a subgroup of $A_m$ which is isomorphic to $A_i.$

Thus, there exists a subgroup $K_j$ of $A_m$ such that $K_j \cong A_j,$ $\forall$ j and $|K_j |= \frac{j!}{2}.$ Let $k \in {\mathbb {Z}}$ such that $\frac{k!}{2} \le \sqrt{\frac{m!}{2}} < \frac{(k+1)!}{2},$ and

$$\begin{aligned} H_2 \cong A_k \text { amd } H_1 \cong A_{k+1}. \end{aligned}$$

Thus, $A_k$ is subgroup of $A_{k+1}$ as $A_k$ is stabilizer which fixes the point $k+1$ and $H_2 \le H_1.$ Notice that

$$\begin{aligned} |H_2 |^2 = \Big (\frac{k!}{2}\Big )^2 \le \frac{m!}{2}. \end{aligned}$$

Now, consider

$$\begin{aligned} {\frac{m!}{2}}&< \Big (\frac{(k+1)!}{2}\Big )^2\\ \frac{(k+2) (k+3) \cdots m}{2}&< \frac{1\cdot 2 \cdots k+1}{4}\\ (k+2) (k+3) \cdots m&< 3 \cdot 4 \cdots k+1. \end{aligned}$$

Notice that each term on the right-hand side is less than that of each term on the left-hand side. Thus, the number of terms on the left-hand side must be strictly less than the number of terms on the right-hand side, which implies that

$$\begin{aligned} m-(k+1)&< (k+1)-2\\ m-2k&< 0\\ \frac{m}{2}&< k. \end{aligned}$$

Thus, $k > \frac{m}{2}$ and it is easy to find such k.

Consider,

$$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |}\Big )^2 = \Big (\frac{\frac{(k+1)!}{2}}{\frac{k!}{2}}\Big )^2 = (k+1)^2. \end{aligned}$$

Now, when $m=5,k=3$ and $m=6,k=4$, we can see that the following inequality holds:

$$\begin{aligned} (k+1)^2< \frac{m!}{2}. \end{aligned}$$

Consider, $m\ge 7$ and $k\ge 4,$ then,

$$\begin{aligned} \frac{k!}{2}&\le \sqrt{\frac{m!}{2}}\\ \Big (\frac{k!}{2}\Big )^2&\le \frac{m!}{2}\\ \frac{k^2 (k-1)^2 (k-2)^2 }{4}&\le \Big (\frac{k!}{2}\Big )^2\le \frac{m!}{2}. \end{aligned}$$

Let,

$$\begin{aligned} \frac{4 (k+1)^2}{ k^2 (k-1)^2 (k-2)^2 }&= \frac{4 (1+\frac{1}{k})^2}{(k-1)^2 (k-2)^2 }\\&< \frac{4}{4 \cdot 9} \frac{25}{16}\\&<1\\ (k+1)^2 < \frac{ k^2 (k-1)^2 (k-2)^2 }{ 4 }&\le \Big (\frac{k!}{2}\Big )^2\le \frac{m!}{2}. \end{aligned}$$

Thus, it implies that,

$$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |}\Big )^2 = (k+1)^2 \le \frac{m!}{2}. \end{aligned}$$

Clearly,

$$\begin{aligned} \Big (\frac{|H |}{|H_1 |}\Big )^2 = \Big (\frac{\frac{(m)!}{2}}{\frac{(k+1)!}{2}}\Big )^2 < \Big (\frac{\frac{m!}{2}}{\sqrt{\frac{m!}{2}}}\Big )^2 = \frac{m!}{2}. \end{aligned}$$

1.2 The Classical Groups

In this section, we consider H to be a classical simple group described in case (iii) of Theorem 8. Let q be a power of some prime p. As described earlier, we use Method 1 and Method 2 to show the existence of subgroups $H_2$ and $H_1$ of the simple group H.

1.2.1 Classical Groups of Linear Type:

• 1.1 $H=A_{m}(q)$; $m \ge 1$, $q>2$ (Method 1) The finite simple group $A_{m}(q)$ is isomorphic to the projective special linear group $\textrm{PSL}_{\textrm{m}+1}(q)$, where $\textrm{PSL}_{m+1}(q)$ is the group obtained by taking special linear group $\textrm{SL}_{m+1}(q)$ and quotienting out by its center, i.e. $A_{m}(q) \cong \frac{\textrm{SL}_{m+1}(q)}{Z(\textrm{SL}_{m+1}(q))}$ (see [23, p. 44]). It is known that (see [24, p. 252]) its order is, $|H |= \frac{q^{\frac{m(m+1)}{2} } \prod _{i=1}^{m} (q^{i+1}-1) }{\textrm{gcd } (q-1,m+1)}.$ Let $H_2$ be the Sylow p-subgroup of $A_{m}(q)$, then $|H_2 |= q^{\frac{m(m+1)}{2}}.$  Notice that $|H_2 |^2 \le |H |.$ Consider the Borel subgroup $H_1$ of H, its order is (see [23, p. 46]),

  $$\begin{aligned} \vert H_1 \vert =\frac{q^{\frac{m(m+1)}{2}}}{\textrm{gcd }(q-1,m+1)} (q-1)^m. \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ \frac{q^{\frac{m(m+1)}{2}}}{\textrm{gcd }(q-1,m+1)} (q-1)^m }{q^{m(m+1)} }\\&= \frac{1}{\textrm{gcd }(q-1,m+1)} \frac{(q-1)^m }{q^{\frac{m(m+1)}{2}} }\\&< \frac{1}{\textrm{gcd }(q-1,m+1)} \frac{q^m }{q^{\frac{m(m+1)}{2}}}\\&< \frac{q^m }{q^{1+2+3+ \ldots +m}}\\&<1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{\vert H_1 \vert }{|H_2 |}\Big )^2< \vert H_1 \vert < \vert H\vert . \end{aligned}$$

  Also,

  $$\begin{aligned} \frac{|H |}{\vert H_1 \vert ^2}&=\frac{\frac{q^{\frac{m(m+1)}{2} } \prod _{i=1}^{m} (q^{i+1}-1) }{\textrm{gcd }(q-1,m+1)}}{\left( \frac{q^{\frac{m(m+1)}{2}}}{\textrm{gcd }(q-1,m+1)} (q-1)^m\right) ^2 }\\&= \frac{ \textrm{gcd }(q-1,m+1) \prod _{i=1}^{m} (q^{i+1}-1) }{ q^{\frac{m(m+1)}{2}} (q-1)^{2m}}\\&< \frac{ q^{1+2+ \ldots + m} q^{m+1}}{q^{\frac{m(m+1)}{2}}(q-1)^{2m} }\\&= \frac{ q^{m+1}}{(q-1)^{m+1} (q-1)^{m-1} }\\&= \frac{1}{(1-\frac{1}{q})^{m+1} (q-1)^{m-1} }\\&= \frac{1}{ (1-\frac{1}{q})^{2} (1-\frac{1}{q})^{m-1} (q-1)^{m-1} }\\&= \frac{1}{(1-\frac{1}{q})^{2} (q-2+ \frac{1}{q})^{m-1} }\\&< 3. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2< 3\, |H |\implies \frac{|H |}{|H_1 |} < 2\, \sqrt{ |H |}. \end{aligned}$$

• 1.2 $A_m(q)$; $m \ge 1$, $q=2$ (Method 2) The finite simple group $A_{m}(2)$ is of order $2^{\frac{m(m+1)}{2} } \prod _{i=1}^{m} (2^{i+1}-1)/\textrm{gcd }(q-1,m+1)$ and is isomorphic to projective special linear group $\textrm{PSL}_{m+1}(2)$ or $\textrm{L}_{m+1}(2).$ It has a maximal subgroup of index $(2^{m+1}-1)$ (see [25, p. 175]). Let $H_1$ be one such maximal subgroup of $A_m(2)$. Then,

  $$\begin{aligned} |H_{1} |&= \frac{ \vert A_m(2) \vert }{(2^{m+1}-1)}\\&= \frac{ 2^{\frac{m(m+1)}{2} } \prod _{i=1}^{m} (2^{i+1}-1) }{ (2^{m+1}-1) }. \end{aligned}$$

  Let $H_2$ be the Sylow 2-subgroup of $H_1.$ Then, $H_2$ has order $2^{\frac{m(m+1)}{2}}$ and $\vert H_{2} \vert ^2 < \vert A_m(2) \vert .$ This implies that

  $$\begin{aligned} \frac{ |H_1 |}{|H_2 |}= \prod _{i=1}^{m-1} (2^{i+1}-1). \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{\Big (\frac{ |H_1 |}{|H_2 |} \Big )^2}{ \vert A_m(2) \vert }&= \frac{\big (\prod _{i=1}^{m-1} (2^{i+1}-1) \big )^2}{2^{\frac{m(m+1)}{2} } \prod _{i=1}^{m} (2^{i+1}-1) }\\&= \frac{\prod _{i=1}^{m-1} (2^{i+1}-1)}{ 2^{\frac{m(m+1)}{2} (2^{m+1}-1)}}\\&< \frac{2^{\frac{m(m+1)}{2}-1}}{ 2^{\frac{m(m+1)}{2} (2^{m+1}-1)}}\\&< 1. \end{aligned}$$

  Thus, we get

  $$\begin{aligned} \Big (\frac{ |H_1 |}{|H_2 |} \Big )^2 < |A_m(2) |. \end{aligned}$$

  Now,

  $$\begin{aligned} \Big (\frac{ \vert A_m(2) \vert }{|H_1 |} \Big )^2&= (2^{m+1}-1)^2\\&< 2^{\frac{m(m+1)}{2} } \prod _{i=1}^{m} (2^{i+1}-1)\\&= |A_m(q) |. \end{aligned}$$

1.2.2 A Classical Groups of Unitary Type:

• 1.1 $H= ~^2A_m(q^2)$; $m \ge 2$, $q>2$ (Method 1) The finite simple group $^2A_m(q^2)$ is isomorphic to the projective special unitary group $\textrm{PSU}_{m+1}(q)$. The group $\textrm{PSU}_{m+1}(q)$ is the group obtain by taking special unitary group $\textrm{SU}_{m+1}(q)$ and quotienting it by its center, i.e. $^2A_m(q^2) \cong \frac{\textrm{SU}_{m+1}(q)}{Z(\textrm{SU}_{m+1}(q))}$ (see [23, p. 66]). It is known that (see [24, p. 252]) the order of $^2A_m(q^2)$ is,

  $$\begin{aligned} |H |= \frac{q^{\frac{m(m+1)}{2} } }{{\textrm{gcd} (q+1,m+1)}} \prod _{i=1}^{m} (q^{i+1}-(-1)^{i+1}). \end{aligned}$$

  Let $H_2$ be the Sylow p-subgroup of $^2A_m(q^2)$, then $|H_2 |= q^{\frac{m(m+1)}{2}}$ and $|H_2 |^2 \le |H |.$ Let $H_1$ be the Borel subgroup of H of order (see [22, 23]), $\vert H_1 \vert =\frac{q^{\frac{m(m+1)}{2} }}{{\textrm{gcd} (q+1,m+1)}}(q-1)^{\lfloor m/2 \rfloor } (q+1)^{\lceil \frac{m-1}{2} \rceil }$. Consider,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2&= \Big (\frac{(q-1)^{\lfloor m/2 \rfloor } (q+1)^{\lceil \frac{m-1}{2} \rceil } }{{\textrm{gcd} (q+1,m+1)}}\Big )^2\\&\le \frac{(q-1)^{m} (q+1)^{m}}{{\textrm{gcd} (q+1,m+1)}^2}\\&= \frac{(q^{2}-1)^{m}}{{\textrm{gcd} (q+1,m+1)}^2} \\&\le |H_1 |. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2 \le |H_1 |< |H |. \end{aligned}$$

  Also,

  $$\begin{aligned} \frac{|H |}{|H_1 |^2}&=\frac{ \frac{q^{\frac{m(m+1)}{2} } }{{\textrm{gcd} (q+1,m+1)}} \prod _{i=1}^{m} (q^{i+1}-(-1)^{i+1}) }{\Big (\frac{ (q-1)^{\lfloor m/2 \rfloor } (q+1)^{\lceil \frac{m-1}{2} \rceil } q^{\frac{m(m+1)}{2} }}{{\textrm{gcd} (q+1,m+1)} } \Big )^2 }\\&= \frac{{\textrm{gcd} (q+1,m+1)} \prod _{i=1}^{m} (q^{i+1}-(-1)^{i+1}) }{ (q-1)^{m} (q+1)^{m-1} q^{\frac{m(m+1)}{2} } }\\&< \frac{ {\textrm{gcd} (q+1,m+1)} q^{\frac{(m+1)(m+2)}{2}-1}}{ (q-1)^{m} (q+1)^{m-1} q^{{\frac{m^{2}+m}{2}}} } \quad \quad \quad ( \text {by Remark}\, 5) \\&= \frac{{\textrm{gcd} (q+1,m+1)} q^{\frac{(m+1)(m+2)}{2}-1}}{ (q-1) (q^{2}-1)^{m-1} q^{\frac{m^{2}+m}{2}}} \\&= \frac{{\textrm{gcd} (q+1,m+1)}q^{m}}{(q-1)(q^{2}-1)^{m-1}}\\&< 2\, \frac{q}{\big (q-\frac{1}{q}\big )^{m-1}}\\&< 4. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 \le 4 |H |\implies \frac{|H |}{|H_1 |} \le 2 \sqrt{|H |} \end{aligned}$$

• 1.2 $H=~^2A_m(q^2)$; $m \ge 2$, $q=2$ (Method 2) The finite simple group $^2A_m(2^2)$ is of order $2^{\frac{m(m+1)}{2} } \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})/{{\textrm{gcd}(3,m+1)}}$ and is isomorphic to projective special unitary group $\textrm{PSU}_{m+1}(2)$ or $\textrm{U}_{m+1}(q)$. The group $\textrm{U}_{m+1}(q)$ has a maximal subgroup of index ${\frac{ (2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m})}{3} }$ when $6 \not \mid (m-1)$ and of index $\frac{2^{m}(2^{m+1}-1)}{3},$ when $6 \mid (m-1)$ (see [25, p. 175]).

  • (Case 1) $6 \not \mid (m-1)$ Let $H_1$ be corresponding maximal subgroup of $^2A_m(2^2)$ whose index is

    $$\begin{aligned} {\frac{ (2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m})}{3} } \end{aligned}$$

    in $^2A_m(2^2).$ Then, the order of $H_1$ is,

    $$\begin{aligned} |H_1 |&= \frac{ \vert ~^2A_m(2^2) \vert }{{\frac{ (2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m})}{3} }}\\&= \frac{3}{{\textrm{gcd}(3,m+1)}} \frac{ {2^{\frac{m(m+1)}{2}}} \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})}{(2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m})}. \end{aligned}$$

    Let $H_2$ be the Sylow 2-subgroup of $H_1.$ Then, $|H_2 |= 2^{\frac{m(m+1)}{2}}$ and $|H_{2} |^2 < |~^2A_m(2^2) |.$ Also,

    $$\begin{aligned} \frac{ |H_1 |}{|H_2 |}= \frac{3}{{\textrm{gcd}(3,m+1)}} \frac{ \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})}{(2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m})}. \end{aligned}$$

    Consider,

    $$\begin{aligned} \frac{\Big (\frac{ |H_1 |}{|H_2 |} \Big )^2}{ \vert ~^2A_m(2^2) \vert }&= \frac{ \Big (\frac{3}{{\textrm{gcd}(3,m+1)}} \frac{ \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})}{(2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m})}\Big )^2}{\frac{2^{\frac{m(m+1)}{2} } }{{\textrm{gcd}(3,m+1)}} \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})}\\&= \frac{9}{{\textrm{gcd}(3,m+1)}} \frac{\prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})}{2^{\frac{m(m+1)}{2} } ((2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m}))^2 }\\&< \frac{9}{{\textrm{gcd}(3,m+1)}} \frac{2^{\frac{(m+1)(m+2)}{2}-1 }}{2^{\frac{m(m+1)}{2} } ((2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m}))^2 }( \text {by Remark}\, 5)\\&= \frac{9}{{\textrm{gcd}(3,m+1)}} \frac{2^{m}}{ ((2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m}))^2}\\&< \frac{9}{{\textrm{gcd}(3,m+1)}} \frac{1}{ \Big (\big (2^{\lfloor \frac{m+1}{2}\rfloor }- \frac{(-1)^{m+1}}{2^{\lfloor \frac{m}{2} +1 \rfloor }}\big )(2^{m}-(-1)^{m})\Big )^2}\\&< 1. \end{aligned}$$

    Therefore,

    $$\begin{aligned} \Big (\frac{ |H_1 |}{|H_2 |} \Big )^2 < |~^2A_m(2^2)|\end{aligned}$$

    Now,

    $$\begin{aligned} \Big (\frac{ \vert ~^2A_m(2^2) \vert }{|H_1 |}\Big )^2&= \Big ({\frac{ (2^{m+1}- (-1)^{m+1})(2^{m}-(-1)^{m})}{3} }\Big )^2\\&< \frac{2^{\frac{m(m+1)}{2} } }{{\textrm{gcd}(3,m+1)}} \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1}) \\&= |~^2A_m(2^2) |. \end{aligned}$$

  • (Case 2) $6 \vert (m-1)$ (i.e. $m \ge 7$) In this case, as we know that the group $^2A_m(q^2)$ has a maximal subgroup of index $\frac{2^{m}(2^{m+1}-1)}{3}.$ Let $H_1$ be one such maximal subgroup. Then,

    $$\begin{aligned} |H_1 |&= \frac{ \vert ~^2A_m(2^2) \vert }{\frac{2^{m}(2^{m+1}-1)}{3}}\\&= \frac{\frac{2^{\frac{m(m+1)}{2} } }{{\textrm{gcd}(3,m+1)}} \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})}{\frac{2^{m}(2^{m+1}-1)}{3}}\\&= \frac{3}{{\textrm{gcd}(3,m+1)}} {2^{\frac{m(m-1)}{2}}} \prod _{i=1}^{m-1} (2^{i+1}-(-1)^{i+1}). \end{aligned}$$

    Let $H_2$ be the Sylow 2-subgroup of $H_1,$ then $H_2$ has order $2^{\frac{m(m-1)}{2}}$ and $|H_{2} |^2 < |~^2A_m(2^2) |.$ Also,

    $$\begin{aligned} \frac{ |H_1 |}{|H_2 |}= \frac{3}{{\textrm{gcd}(3,m+1)}} \prod _{i=1}^{m-1} (2^{i+1}-(-1)^{i+1}). \end{aligned}$$

    Consider,

    $$\begin{aligned} \frac{\Big (\frac{ |H_1 |}{|H_2 |}\Big )^2}{ \vert ~^2A_m(2^2) \vert }&= \frac{\Big ( \frac{3}{{\textrm{gcd}(3,m+1)}} \prod _{i=1}^{m-1} (2^{i+1}-(-1)^{i+1})\Big )^2}{\frac{2^{\frac{m(m+1)}{2} } }{{\textrm{gcd}(3,m+1)}} \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1})}\\&= \frac{9}{{\textrm{gcd}(3,m+1)}} \frac{\prod _{i=1}^{m-1} (2^{i+1}-(-1)^{i+1})}{2^{\frac{m(m+1)}{2} } (2^{m+1}- (-1)^{m+1}) }\\&< \frac{9}{{\textrm{gcd}(3,m+1)}} \frac{2^{\frac{m^2+m-2}{2}}}{2^{\frac{m^{2}+m}{2} } (2^{m+1}- (-1)^{m+1})} \quad \quad \quad ( \text {by Remark}\, 5)\\&= \frac{9}{{\textrm{gcd}(3,m+1)}} \frac{1}{2\,(2^{m+1}- (-1)^{m+1}) }\\&< 1. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (\text {since} \, {m \ge 7}) \end{aligned}$$

    This implies that,

    $$\begin{aligned} \Big (\frac{ |H_1 |}{|H_2 |}\Big )^2 < |~^2A_m(2^2) |. \end{aligned}$$

    Now,

    $$\begin{aligned} \Big (\frac{ |~^2A_m(2^2) |}{|H_1 |}\Big )^2&= \Big (\frac{2^{m}(2^{m+1}-1)}{3}\Big )^2\\&= \frac{ 2^{2m}(2^{m+1}-1)^2}{9}\\&< \frac{2^{\frac{m(m+1)}{2} } }{{\textrm{gcd}(3,m+1)}} \prod _{i=1}^{m} (2^{i+1}-(-1)^{i+1}) \\&= |~^2A_m(2^2) |. \end{aligned}$$

1.2.3 Classical Groups of Symplectic Linear Type:

• 1.1 $H= C_{m}(q)$; $m \ge 2$, $q>2$ (Method 1) The order of the finite simple group $C_{m}(q)$ could be found in (see [24, p. 252]) and it is, $|H |= \frac{q^{m^2 } \prod _{i=1}^{m} (q^{2i}-1) }{{\textrm{gcd}(2,q-1)}}.$ Let $H_2$ be the Sylow p-subgroup of H, then $|H_2 |= q^{m^2}$ and $|H_2 |^2 \le |H |.$ Let $H_1$ be the normalizer of $H_2$ in H then the order of $H_1$ (see [30, p. 3]) is,

  $$\begin{aligned} |H_1 |=\frac{q^{m^2}}{{\textrm{gcd}(2,q-1)}} (q-1)^m. \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{\frac{q^{m^2}}{{\textrm{gcd}(2,q-1)}} (q-1)^m }{q^{2m^2} }\\&= \frac{1}{{\textrm{gcd}(2,q-1)}} \frac{(q-1)^m }{q^{m^2} }\\&< \frac{1}{{\textrm{gcd}(2,q-1)}} \frac{q^m }{q^{m^2} }\\&< 1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2< \vert H_1 \vert < |H |. \end{aligned}$$

  Also,

  $$\begin{aligned} \frac{|H |}{|H_1 |^{2}}&=\frac{\frac{q^{m^2 } \prod _{i=1}^{m} (q^{2i}-1) }{{\textrm{gcd}(2,q-1)}}}{\left( \frac{q^{m^2}}{{\textrm{gcd}(2,q-1)}} (q-1)^m\right) ^2 }\\&= \frac{ {\textrm{gcd}(2,q-1)} \prod _{i=1}^{m} (q^{2i}-1) }{ q^{m^2} (q-1)^{2m}}\\&< \frac{{\textrm{gcd}(2,q-1)}}{q^{m^2}} \frac{\prod _{i=1}^{m} q^{2i}}{(q-1)^{2m} } \\&= {\textrm{gcd}(2,q-1)} \frac{q^{m+m^2}}{q^{m^2} (q-1)^{2m}} \\&= {\textrm{gcd}(2,q-1)} \left( \frac{q}{(q-1)^2}\right) ^{m}\\&\le 2 \quad \quad \quad ( \text {by Remark}\, 4) \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H |}{\vert H_1 \vert } \Big )^2 \le 2 \, |H |\implies \frac{|H |}{\vert H_1 \vert } < 2 \, \sqrt{|H |}. \end{aligned}$$

• 1.2 $H= C_{m}(q)$; $m \ge 2$, $q=2$ (Method 2) The simple group $C_m(q)$ (or $\textrm{PSp}_{2m}(q)$) has order $2^{m^2 } \prod _{i=1}^{m} (2^{2i}-1)$. It is known that the group ${ \mathrm PSp}_{2\,m}(q)$ has a maximal subgroup of index $2^{m-1} (2^{m}-1)$ (see [25, p. 175]). Let $H_1$ be one such subgroup, then the order of $H_1$ is,

  $$\begin{aligned} |H_1 |&= \frac{ \vert C_m(2) \vert }{2^{m-1} (2^{m}-1)}\\&= \frac{ 2^{m^2 } \prod _{i=1}^{m} (2^{2i}-1)}{ 2^{m-1} (2^{m}-1) }\\&= \frac{ 2^{m^{2} -m+1} (2^{2m} -1) \prod _{i=1}^{m-1} (2^{2i}-1)}{ (2^{m}-1) }\\&= 2^{m^{2} -m+1} (2^{m} +1) \prod _{i=1}^{m-1} (2^{2i}-1). \end{aligned}$$

  Let $H_2$ be the Sylow 2-subgroup of $H_1.$ Then, the order of $H_2$ is $2^{m^{2} -m+1}.$ Notice that $|H_{2} |^2 < |C_m(2) |.$ Thus,

  $$\begin{aligned} \frac{ |H_1 |}{|H_2 |}= (2^{m} +1) \prod _{i=1}^{m-1} (2^{2i}-1). \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{\Big (\frac{ |H_1 |}{|H_2 |}\Big )^2}{ |C_m(2) |}&= \frac{((2^{m} +1) \prod _{i=1}^{m-1} (2^{2i}-1))^2}{2^{m^2 } \prod _{i=1}^{m} (2^{2i}-1) }\\&=\frac{(2^{m} +1)^2 (\prod _{i=1}^{m-1} (2^{2i}-1))^2}{2^{m^2 } (2^{2m}-1) \prod _{i=1}^{m-1} (2^{2i}-1) }\\&=\frac{(2^{m} +1) \prod _{i=1}^{m-1} (2^{2i}-1)}{2^{m^2 } (2^{m}-1) }\\&< \frac{(2^{m} +1) \prod _{i=1}^{m-1} 2^{2i}}{2^{m^2 } (2^{m}-1) }\\&= \frac{(2^{m} +1) 2^{m(m-1)}}{2^{m^2 } (2^{m}-1) }\\&= \frac{(1+2^{-m})}{ (2^{m}-1) }\\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{ |H_1 |}{|H_2 |}\Big )^2 < |C_m(2) |. \end{aligned}$$

  Now,

  $$\begin{aligned} \Big (\frac{ \vert C_m(2) \vert }{|H_1 |}\Big )^2&= (2^{m-1} (2^{m}-1))^2\\&= 2^{2m-2} (2^{m}-1)^2\\&< 2^{m^{2}} (2^{m}-1) (2^{m}+1)\\&< 2^{m^2} \prod _{i=1}^{m} (2^{2i}-1) \\&= |C_m(2) |. \end{aligned}$$

1.2.4 Classical Groups of Orthogonal Type:

• 1.1 $H=B_{m}(q);$ $m \ge 3$ and q odd (Method 1) The finite simple group $B_{m}(q)$ has order (see [24, p. 252]), $|H |= \frac{q^{m^2 } \prod _{i=1}^{m} (q^{2i}-1) }{{\textrm{gcd}(2,q-1)}}.$ Let $H_2$ be the Sylow p-subgroup of H, then $|H_2 |= q^{m^2}$ and $|H_2 |^2 \le |H |.$ Let $H_1$ be the normalizer of $H_1$ in H then its order is (see [30, p. 3]),

  $$\begin{aligned} |H_1|=\frac{q^{m^2}}{{\textrm{gcd}(2,q-1)}} (q-1)^m. \end{aligned}$$

  Since, $|B_{m}(q) |= |C_{m}(q) |$ and the order of the normalizer of $H_1$ in both the groups are also equal and thus, all the calculations will also work for $B_{m}(q).$

• 2.1 $H= D_{m}(q)$; $m \ge 4$, $q>2$ (Method 1) The order of the finite simple group $D_{m}(q)$ could be found in (see [24, p. 252]) and it is, $\vert H \vert = \frac{q^{{m(m-1)}} (q^{m}-1) \prod _{i=1}^{m-1} (q^{2i}-1) }{{\textrm{gcd} (4, q^{m}-1)}}.$ Let $H_2$ be the Sylow p-subgroup of $D_{m}(q)$, then $|H_2 |= q^{{m(m-1)}}$ and $|H_2 |^2 \le |H |.$ Let $H_1$ be the Borel subgroup of H. It has order (see [22, 23]),

  $$\begin{aligned} |H_1 |=\frac{q^{{m(m-1)}} }{ {\textrm{gcd} (4, q^{m}-1)}} (q-1)^m. \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ \frac{q^{{m(m-1)}} }{ {\textrm{gcd} (4, q^{m}-1)}} (q-1)^m }{q^{2m(m-1)} }\\&= \frac{1}{{\textrm{gcd} (4, q^{m}-1)}} \frac{(q-1)^m }{q^{m(m-1)} }\\&< \frac{1}{{\textrm{gcd} (4, q^{m}-1)}} \frac{q^m }{q^{m(m-1)} }\\&< 1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big ( \frac{|H_1 |}{|H_2 |} \Big )^2< \vert H_1 \vert < |H |\end{aligned}$$

  Also,

  $$\begin{aligned} \frac{|H |}{|H_1 |^{2}}&=\frac{\frac{q^{{m(m-1)}} (q^{m}-1) \prod _{i=1}^{m-1} (q^{2i}-1) }{{\textrm{gcd} (4, q^{m}-1)}}}{\left( \frac{q^{{m(m-1)}} }{ {\textrm{gcd} (4, q^{m}-1)}} (q-1)^m \right) ^2 }\\&= \frac{ {\textrm{gcd} (4, q^{m}-1)} (q^{m}-1) \prod _{i=1}^{m} (q^{2i}-1) }{ q^{{m(m-1)}} (q-1)^{2m}}\\&< \frac{{\textrm{gcd} (4, q^{m}-1)} (q^{m}-1) }{ q^{{m(m-1)}}} \frac{\prod _{i=1}^{m-1} q^{2i}}{(q-1)^{2m} } \\&< \frac{{\textrm{gcd} (4, q^{m}-1)} q^{m} }{ q^{{m(m-1)}}} \frac{ q^{{m(m-1)}} }{(q-1)^{2m} } \\&={\textrm{gcd} (4, q^{m}-1)} \Big (\frac{q}{(q-1)^2}\Big )^{m} \\&\le 4. \quad \quad \quad ( \text {by Remark}\, 4) \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{|H |}{\vert H_1 \vert } \Big )^2 \le 4\, |H |\implies \frac{|H |}{\vert H_1 \vert } \le 2 \sqrt{ |H |}. \end{aligned}$$

• 2.2 $H= D_{m}(q)$; $m \ge 4$, $q=2$ (Method 2) The simple group $D_m(q)$ (or $\mathrm{P\Omega }_{2\,m}^{+}(q)$) is of order $2^{{m(m-1)}} (2^{m}-1) \prod _{i=1}^{m-1} (2^{2i}-1)$. The group $\mathrm{P\Omega }_{2\,m}^{+}(2)$ has a maximal subgroup of index $2^{m-1} (2^{m}-1)$ (see [25, p. 175]). Let $H_1$ be a corresponding maximal subgroup of $D_m(2)$ whose index is $2^{m-1} (2^{m}-1).$ Then the order of $H_1$ is,

  $$\begin{aligned} |H_1 |&= \frac{ \vert D_m(2) \vert }{2^{m-1} (2^{m}-1)}\\&= \frac{2^{{m(m-1)}} (2^{m}-1) \prod _{i=1}^{m-1} (2^{2i}-1)}{ 2^{m-1} (2^{m}-1) }\\&= 2^{m^{2}-2m+1} \prod _{i=1}^{m-1} (2^{2i}-1). \end{aligned}$$

  Let $H_2$ be the Sylow 2-subgroup of $H_1,$ then $H_2$ has order $2^{m^{2}-2\,m+1}$ and $|H_{2} |^2 < |D_m(2) |.$ Also,

  $$\begin{aligned} \frac{ |H_1 |}{|H_2 |}= \prod _{i=1}^{m-1} (2^{2i}-1). \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{\Big (\frac{ |H_1 |}{|H_2 |}\Big )^2}{ \vert D_m(2) \vert }&= \frac{(\prod _{i=1}^{m-1} (2^{2i}-1))^2}{2^{{m(m-1)}} (2^{m}-1) \prod _{i=1}^{m-1} (2^{2i}-1) }\\&= \frac{\prod _{i=1}^{m-1} (2^{2i}-1)}{2^{{m(m-1)}} (2^{m}-1) }\\&< \frac{\prod _{i=1}^{m-1} 2^{2i}}{2^{{m(m-1)}} (2^{m}-1) }\\&= \frac{2^{m(m-1)} }{2^{{m(m-1)}} (2^{m}-1) }\\&= \frac{1 }{(2^{m}-1)}\\&< 1 \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{ |H_1 |}{|H_2 |}\Big )^2 < |D_m(2) |. \end{aligned}$$

  Now,

  $$\begin{aligned} \Big (\frac{ |D_m(2) |}{|H_1 |}\Big )^2&= 2^{2(m-2)} (2^{m}-1)^2\\&<2^{{m(m-1)}} (2^{m}-1) \prod _{i=1}^{m-1} (2^{2i}-1) \\&= |D_m(2) |. \end{aligned}$$

• 3.1 $H= ~^2D_m(q^2)$; $m \ge 4$, $q>2$ (Method 1) The order of the finite simple group $^2D_m(q^2)$ could be found in (see [24, p. 252]) and it is, $|H |=\frac{q^{m(m-1)}(q^{m}+1) }{{\textrm{gcd} (4, q^{m}+1))}} \prod _{i=1}^{m-1} (q^{2i}-1).$ Let $H_2$ be the Sylow p-subgroup of $^2D_m(q^2)$, then $|H_2 |= q^{m(m-1)}$ and $|H_2 |^2 \le |H |.$ Let $H_1$ to be the Borel subgroup of H. The order of $H_1$ is (see [22, 23]). Thus, $\vert B\vert =\frac{q^{{m(m-1)}} }{ \textrm{gcd}(4, q^{n}+1)} (q-1)^m$. Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ \frac{q^{{m(m-1)}} }{ \textrm{gcd}(4, q^{n}+1)} (q-1)^m }{q^{2m(m-1)} }\\&= \frac{1}{\textrm{gcd}(4, q^{n}+1)} \frac{(q-1)^m }{q^{m(m-1)} }\\&< \frac{1}{\textrm{gcd}(4, q^{n}+1)} \frac{q^m }{q^{m(m-1)} }\\&\le 1. \end{aligned}$$

  Thus, we have

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2< \vert H_1 \vert <|H |. \end{aligned}$$

  Also,

  $$\begin{aligned} \frac{|H |}{|H_1 |^{2}}&=\frac{\frac{q^{m(m-1)}(q^{m}+1) }{{\textrm{gcd} (4, q^{m}+1))}} \prod _{i=1}^{m-1} (q^{2i}-1)}{\left( \frac{q^{{m(m-1)}} }{ {\textrm{gcd} (4, q^{m}+1))}} (q-1)^m \right) ^2 }\\&= \frac{ {\textrm{gcd} (4, q^{m}+1))} (q^{m}+1) \prod _{i=1}^{m-1} (q^{2i}-1)}{ q^{m(m-1)} (q-1)^{2m}}\\&< \frac{{\textrm{gcd} (4, q^{m}+1))} (q^{m}+1) }{q^{m(m-1)}} \frac{\prod _{i=1}^{m-1} q^{2i}}{(q-1)^{2m} } \\&= {\textrm{gcd} (4, q^{m}+1))} (q^{m}+1) \frac{q^{m(m-1)}}{q^{m(m-1)} (q-1)^{2m}} \\&< \frac{4 (2q^{m})}{(q-1)^{2m}} \\&< 8 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ( \text {by Remark}\, 4) \end{aligned}$$

  This implies that,

  $$\begin{aligned} \Big (\frac{|H |}{\vert H_1 \vert } \Big )^2< 8\,|H |\implies \frac{|H |}{\vert H_1 \vert } < 3\,\sqrt{|H |}. \end{aligned}$$

• 3.2 $H= ~^2D_m(q^2)$; $m \ge 4$, $q=2$ (Method 2) We know that the group $^2D_m(2^2)$(or $\mathrm{P\Omega }_{2\,m}^{-}(2)$) is of order $2^{m(m-1)}(2^{m}+1) \prod _{i=1}^{m-1} (2^{2i}-1)/\textrm{gcd}(4, 2^{m}+1).$ The group $\mathrm{P\Omega }_{2\,m}^{-}(2)$ has a maximal subgroup of index $(2^{m}+1) (2^{m-1}-1)$ (see [25, p. 175]). Let $H_1$ be one such subgroup then its order is,

  $$\begin{aligned} |H_1 |&= \frac{ \vert ~^2D_m(2^2) \vert }{(2^{m}+1) (2^{m-1}-1)}\\&=\frac{2^{m(m-1)}(2^{m}+1) }{\textrm{gcd}(4, 2^{m}+1)} \frac{\prod _{i=1}^{m-1} (2^{2i}-1) }{(2^{m}+1) (2^{m-1}-1) }\\&= 2^{m(m-1)} (2^{m-1}+1)\prod _{i=1}^{m-2} (2^{2i}-1). \end{aligned}$$

  Let $H_2$ be the Sylow 2-subgroup of $H_1,$ then $|H_2 |= 2^{m(m-1)}$ and $\vert H_{2} \vert ^2 < \vert ~^2D_m(2^2) \vert .$ Thus,

  $$\begin{aligned} \frac{ |H_1 |}{|H_2 |}= (2^{m-1}+1)\prod _{i=1}^{m-2} (2^{2i}-1). \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{\Big (\frac{ |H_1 |}{|H_2 |}\Big )^2}{ |~^2D_m(2^2) |}&= \frac{((2^{m-1}+1)\prod _{i=1}^{m-2} (2^{2i}-1))^2}{\frac{2^{m(m-1)}(2^{m}+1) }{\textrm{gcd} (4, 2^{m}+1)} \prod _{i=1}^{m-1} (2^{2i}-1) }\\&= \frac{(2^{m-1}+1)^2 \prod _{i=1}^{m-2} (2^{2i}-1)}{2^{{m(m-1)}} (2^{2m-2}-1) (2^{m}+1) }\\&= \frac{(2^{m-1}+1) \prod _{i=1}^{m-2} (2^{2i}-1)}{2^{{m(m-1)}} (2^{m-1}-1) (2^{m}+1) }\\&< \frac{(2^{m-1}+1) \prod _{i=1}^{m-2} 2^{2i}}{2^{{m(m-1)}} (2^{m-1}-1) (2^{m}+1) }\\&= \frac{(2^{m-1}+1) 2^{(m-2)(m-1)} }{2^{{m(m-1)}} (2^{m-1}-1) (2^{m}+1) }\\&= \frac{(2^{m-1}+1) }{(2^{m-1}-1) (2^{m}+1)} \cdot \frac{ 2^{(m-2)(m-1)} }{2^{{m(m-1)}}}\\&< 1.\\ \Big (\frac{ |H_1 |}{|H_2 |}\Big )^2&< |~^2D_m(2^2)|. \end{aligned}$$

  Now,

  $$\begin{aligned} \Big (\frac{ \vert ~^2D_m(2^2) \vert }{|H_1 |}\Big )^2&= ((2^{m}+1) (2^{m-1}-1))^2\\&< 2^{m(m-1)} (2^{m}+1) (2^{m-1}-1) (2^{m-1}+1) \prod _{i=1}^{m-2} (2^{2i}-1) \\&<2^{m(m-1)}(2^{m}+1) \prod _{i=1}^{m-1} (2^{2i}-1) \\&= |~^2D_m(2^2) |. \end{aligned}$$

1.3 Exceptional Group of Lie Type

• (1) $H= G_2(q)$; $q\ge 3$ (Method 1) The group $G_2(q)$ is simple for all $q\ge 3$. It has order (see [24, p. 252]),

  $$\begin{aligned} \vert G_2(q) \vert = q^{6} (q^{6}-1)(q^{2}-1). \end{aligned}$$

  Thus, $G_2(q)$ has a Sylow p-subgroup $H_2$ of order $q^{6}$ and $|H_2 |^2 \le |G_2(q) |.$ Also, it has the Borel subgroup $H_1$ of order $q^{6}(q-1)^2$ (see [23, p. 124]). Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ q^{6}(q-1)^2}{q^{12} }\\&= \frac{(q-1)^2}{q^{6} }\\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2 \le |H_1 |\le |H |. \end{aligned}$$

  Now,

  $$\begin{aligned} \frac{|H |}{\vert H_1 \vert ^2}&=\frac{ q^{6} (q^{6}-1)(q^{2}-1)}{( q^{6}(q-1)^2)^2 }\\&< \frac{q^{6+2}}{ q^{6}(q-1)^4}\\&< \frac{q^{2}}{(q-1)^4}\\&< 1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ( \text {by Remark}\, 4) \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{\vert H\vert }{|H_1 |} \Big )^2 < |H |. \end{aligned}$$

• (2) $H=F_4(q)$ (Method 2) The finite simple group $F_4(q)$ has order (see [24, p. 252]),

  $$\begin{aligned} \vert F_4(q) \vert = q^{24} (q^{12}-1)(q^{8}-1)(q^{6}-1)(q^{2}-1). \end{aligned}$$

  It is known that (see [23, p. 156]) the group $F_4(q)$ has a maximal subgroup $q^{1+14}: Sp_{6}(q).C_{q-1}$ of order $q^{24}(q^{6}-1)(q^4-1) (q^{2}-1) (q-1)$ say $H_1$. This subgroup has a Sylow p-subgroup say $H_2$ of order $q^{24}$ and $|H_2 |^2 \le \vert F_4(q) \vert .$ Now,

  $$\begin{aligned} \frac{\vert H_1 \vert }{|H_2 |^2}&= \frac{ q^{24} (q^{6}-1)(q^4 -1) (q^{2}-1) (q-1) }{q^{48} }\\&< \frac{q^{1+2+4+6} }{q^{24} }\\&<1.\\ \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{\vert H_1 \vert }{|H_2 |} \Big )^2< \vert H_1 \vert < |H |. \end{aligned}$$

  Now,

  $$\begin{aligned} \frac{|H |}{\vert H_1 \vert ^2}&=\frac{q^{24} (q^{12}-1)(q^{8}-1)(q^{6}-1)(q^{2}-1)}{( q^{24} (q^{6}-1)(q^4 -1) (q^{2}-1) (q-1))^2 }\\&= \frac{ (q^{4} -1)^2 (q^8+q^4+1) (q^4+1) }{ q^{24} (q^{6}-1)(q^4 -1)^2 (q^{2}-1) (q-1)^2}\\&= (1+ \frac{1}{q^4} + \frac{1}{q^8}) (1+ \frac{1}{q^4}) \frac{1 }{ q^{12} (q^{6}-1) (q^{2}-1) (q-1)^2} \\&< 1. \end{aligned}$$

  Thus, we get

  $$\begin{aligned} \Big (\frac{|H |}{\vert H_1 \vert } \Big )^2 < |H |. \end{aligned}$$

• (3) $H=E_6(q)$; $q>2$ (Method 1) The group $E_{6}(q)$ is a finite simple group. The order of $H=E_6(q)$ is (see [24, p. 252]),

  $$\begin{aligned} \vert E_6(q) \vert= & {} \frac{q^{36}}{{\textrm{gcd}(3,q-1)}} (q^{12}-1)(q^{9}-1)(q^{8}-1)\\{} & {} \quad (q^{6}-1)(q^{5}-1)(q^{2}-1). \end{aligned}$$

  Clearly, it has a Sylow p-subgroup $H_2$ of order $q^{36}$ and $|H_2 |^2 \le \vert E_6(q) \vert .$ Let $H_1$ be the Borel subgroup of H then the order of $H_1$ is $q^{36} (q-1)^6$ (see [22, 23]). Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2} = \frac{ q^{36} (q-1)^6}{q^{72}}< \frac{q^6 }{q^{36} }< 1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |}\Big )^2< |H_1 |< |H |. \end{aligned}$$

  Now,

  $$\begin{aligned} \frac{|H |}{|H_1 |^2}&=\frac{q^{36} (q^{12}-1)(q^{9}-1)(q^{8}-1)(q^{6}-1)(q^{5}-1)(q^{2}-1)}{{\textrm{gcd}(3,q-1)} ( q^{36} (q-1)^8 )^2 }\\&< \frac{q^{12+9+8+6+5+2}}{{\textrm{gcd}(3,q-1)} q^{36} (q-1)^{16} } \\&= \frac{q^{42}}{ {\textrm{gcd}(3,q-1)} q^{36} (q-1)^{16} } \\&= \frac{ q^{6}}{ {\textrm{gcd}(3,q-1)} (q-1)^{16} } \\&=\frac{1}{{\textrm{gcd}(3,q-1)} (q-1)^{4}} \frac{ q^{6}}{(q-1)^{12} } \\&< \frac{ q^{6}}{(q-1)^{12} } \\&<1. \quad \quad \quad \quad \quad \quad \quad \quad \quad (\text {by Remark}\, 4)\\ \end{aligned}$$

  This implies that,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 \le |H |. \end{aligned}$$

  Notice that the group $E_6(2)$ is of constant order. However, we can use Method 2 to reduce the constants $b_1$ and $b_2$ to 1. By taking $H_1$ to be maximal subgroup of order (see [31]) $2^{36}\cdot 3^{3}\cdot 5\cdot 7\cdot 31$ and $H_2$ to be its Sylow 2-subgroup of order $2^{36}$.

• (4) $H= ~^2E_6(q)$ (Method 1) The group $^2E_6(q)$ is simple for all q and it has order (see [24, p. 252]),

  $$\begin{aligned} \vert ~^2E_6(q) \vert = \frac{q^{36}(q^2-1)(q^5+1)(q^6-1)(q^8-1)(q^9+1)(q^{12}-1)}{\textrm{gcd}(3,q+1)}. \end{aligned}$$

  The group $~^2E_6(q)$ has a Sylow p-subgroup $H_2$ of order $q^{36}$ and $|H_2 |^2 \le \vert H \vert .$ It is well known that the order of the Borel subgroup is a product of the order of the Sylow p-subgroup and the maximal torus. The order of the maximal torus in H is $(q+1)^6$(see [23, p. 173]). Thus, the order of the Borel subgroup $H_1$ of H is $q^{36} (q+1)^6$. Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ q^{36} (q+1)^6}{q^{72} }\\&< \frac{(q+1)^6}{q^{36} }\\&< \frac{(q+q)^6 }{q^{36} } < 1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{\vert H_1 \vert }{|H_2 |} \Big )^2< |H_1 |< |H |. \end{aligned}$$

  Also,

  $$\begin{aligned} \frac{|H |}{\vert H_1 \vert ^2}&=\frac{ q^{36} (q^{2}-1)(q^{5}+1)(q^{6}-1)(q^{8}-1)(q^{9}+1)(q^{12}-1)}{( q^{36}(q+1)^6 )^2 }\\&< \frac{(q^{5}+1)(q^9+1)}{q^8(q+1)^{12} } \\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 < |H |. \end{aligned}$$

• (5) $H = ~^3D_4(q)$ (Method 1) The group $^3D_4(q)$ is simple for all q. It is known that the order of the group $^3D_4(q)$ is (see [24, p. 252]),

  $$\begin{aligned} \vert ~^3D_4(q) \vert = q^{12} (q^{8}+q^4+1)(q^{6}-1)(q^{2}-1). \end{aligned}$$

  Clearly, it has a Sylow p-subgroup $H_2$ of order $q^{12}$ and the Borel subgroup $H_1$ of order $q^{12}(q^3-1) (q-1)$ (see [23, p. 141]). Also, notice that $|H_2 |^2 \le |H |.$ Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ q^{12} (q^3-1)(q-1)}{q^{24} }\\&= \frac{(q^3-1)(q-1) }{q^{12} }\\&< \frac{q^4 }{q^{12} } < 1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{\vert H_1 \vert }{|H_2 |} \Big )^2< \vert H_1\vert < |H |. \end{aligned}$$
  $$\begin{aligned} \frac{|H |}{\vert H_1 \vert ^2}&=\frac{ q^{12}(q^{8}+q^4+1)(q^{6}-1)(q^{2}-1)}{( q^{12} (q^3-1)(q-1) )^2 }\\&= \frac{(q^{8}+q^4+1)(q^3+1)(q+1)}{q^{12}(q^3-1)(q-1) }\\&= \frac{(1+\frac{1}{q^4}+\frac{1}{q^8})(1+\frac{1}{q^3})(1+\frac{1}{q})}{(q^3-1)(q-1)} \\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 < |H |. \end{aligned}$$

• (6) $H= E_7(q)$, $q>2$ (Method 1) The simple group $E_7(q)$ has order (see [24, p. 252]),

  $$\begin{aligned} \vert E_7(q) \vert= & {} \frac{q^{63}}{{\textrm{gcd}(2,q-1)}}(q^{18}-1)(q^{14}-1)(q^{12}-1)(q^{10}-1)\\{} & {} \quad (q^{8}-1)(q^{6}-1)(q^{2}-1). \end{aligned}$$

  Clearly, it has a Sylow p-subgroup $H_2$ of order $q^{63}$ and $|H_2 |^2 \le |H |.$ Let $H_1$ be the borel subgroup of H then its order is (see [22, 23]),

  $$\begin{aligned} |H_1 |= q^{63} (q-1)^7. \end{aligned}$$

  Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ q^{63} (q-1)^7}{q^{126}}\\&< \frac{q^7 }{q^{63} }\\&< 1 \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2< |H_1 |< |H |. \end{aligned}$$
  $$\begin{aligned} \frac{|H |}{|B |^2}&=\frac{q^{63}}{{\textrm{gcd}(2,q-1)}}\frac{{\displaystyle {\prod _{i\in \{2,6,8,10,12,14,18\}}}} ({q^{i}}-1)}{ ( q^{63} (q-1)^7 )^2 }\\&< \frac{q^{18+14+12+10+8+6+2}}{{\textrm{gcd}(2,q-1)} q^{63} (q-1)^{14} } \\&= \frac{q^{70}}{ {\textrm{gcd}(2,q-1)} q^{63} (q-1)^{14} } \\&= \frac{1}{{\textrm{gcd}(2,q-1)}} \frac{q^{7}}{ (q-1)^{14}}\\&< 1. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (q \ne 2). \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 < |H |. \end{aligned}$$

  Notice that the group $E_7(2)$ is of constant order. However, we can use Method 2 to reduce the constants $b_1$, $b_2$ to 1. By taking $H_1$ to be maximal subgroup of order (see [31]) $2^{63}\cdot 3^{4}\cdot 7^{2}\cdot 5$ and $H_2$ to be its Sylow 2-subgroup of order $2^{63}$.

• (7) $H= E_8(q)$, $q>2$ (Method 1) We know that the group $E_8(q)$ is simple for all q and has order (see [24, p. 252]),

  $$\begin{aligned} \vert H \vert = q^{120} (q^{30}-1)(q^{24}-1)(q^{20}-1)\\ (q^{18}-1)(q^{14}-1)(q^{12}-1)(q^{8}-1)(q^{2}-1). \end{aligned}$$

  The group $E_8(q)$ has a Sylow p-subgroup $H_2,$ then $|H_2 |= q^{120}$ and $|H_2 |^2 \le \vert H \vert .$ Let $H_1$ be the Borel subgroup then order $|H_1 |= q^{120} (q-1)^8$ (see [23, p. 176]). Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&= \frac{ q^{120} (q-1)^8}{q^{240} }\\&< \frac{q^8 }{q^{120} }\\&< 1. \end{aligned}$$

  Threfore,

  $$\begin{aligned} \Big (\frac{\vert H_1 \vert }{|H_2 |} \Big )^2< \vert H_1\vert < |H |. \end{aligned}$$
  $$\begin{aligned} \frac{|H |}{\vert H_1 \vert ^2}&=\frac{q^{120} {\displaystyle {\prod _{i\in \{2,8,12,14,18,20,24,30\}}} ({q^{i}}-1)}}{( q^{120} (q-1)^8 )^2 }\\&= \frac{(q^{8}-1)}{(q-1)^{16} } {\displaystyle \prod _{i\in \{2,12,14,18,20,24,30 \}} (1- \frac{1}{q^{i}}) }\\&\le \frac{q^{8}}{(q-1)^{16} } \\&\le 1. \quad \quad \quad \quad \quad \quad \quad \quad \quad ( \text {by Remark}\, 4) \end{aligned}$$

  This implies that,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 \le |H |. \end{aligned}$$

  Notice that the group $E_8(2)$ is of constant order. However, we can use Method 2 to reduce the constants $b_1$, $b_2$ to 1. By taking $H_1$ to be maximal subgroup of order (see [31]) $2^{119}\cdot 3^{4}\cdot 5\cdot 7^{2}\cdot 31$ and $H_2$ to be its Sylow 2-subgroup of order $2^{119}$.

• (8) $H =~^2B_2(q)$, $q=2^{2t+1}$ and $t\ge 1$ (Method 1) The finite simple group $^2B_2(q)$ has order (see [24, p. 252]),

  $$\begin{aligned} \vert H \vert = q^2(q^2+1)(q-1). \end{aligned}$$

  Thus, it has a Sylow 2-subgroup $H_2$ of order $q^2$ and $|H_2 |^2 \le |H |.$ Let $H_1$ be the Borel subgroup of $^2B_2(q)$ then the order $H_1$ is $q^{2}(q-1)$ (see [23, p. 117]). Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&=\Big (\frac{q^{2}(q-1)}{q^{4}}\Big )\\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2< |H_1 |< |H |. \end{aligned}$$

  Now,

  $$\begin{aligned} \frac{|H |}{\vert H_1 \vert ^2}&=\frac{q^{2}(q^2+1)(q-1)}{(q^{2}(q-1))^2}\\&= \frac{(q^2+1)}{q^{2}(q-1)}\\&= \frac{1+\frac{1}{q^2}}{q-1}\\&< 1. \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 < \vert H \vert . \end{aligned}$$

• (9) $H= ~^2G_2(q)$; where, $q=3^{2t+1}$ and $t\ge 1$ (Method 1) The group $^2G_2(q)$ has order (see [24, p. 252]),

  $$\begin{aligned} |~^2G_2(q) |= q^3(q^3+1)(q-1). \end{aligned}$$

  Thus, $^2G_2(q)$ has a Sylow 3-subgroup of order $q^3$ and $|H_2 |^2 < |H |.$ Also, $^2G_2(q)$ has the Borel subgroup $H_1$ of order $q^3(q-1)$ (see [23, p. 137]). Consider,

  $$\begin{aligned} \frac{|H_1 |}{|H_2 |^2}&=\frac{q^{3}(q-1)}{q^{6}}\\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2< |H_1 |< |H |. \end{aligned}$$

  Now,

  $$\begin{aligned} \frac{|H |}{ |H_1 |^2}&=\frac{q^3(q^3+1)(q-1)}{(q^{3}(q-1))^2}\\&=\frac{(q^3+1)}{q^{3}(q-1)}\\&= \frac{1+\frac{1}{q^3}}{q-1}\\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H |}{|H_1 |} \Big )^2 < |H |. \end{aligned}$$

• (10) $H= ~^2F_4(q)$, $q=2^{2t+1}$ and $t\ge 1$ (Method 1) The order of the group $^2F_4(q)$ is $q^{12}(q^6 +1)(q^4-1)(q^3+1)(q-1)$ (see [24, p. 252]). Thus, it has a Sylow 2-subgroup of order $q^{12}$ and $|H_2 |^2 < \vert H \vert .$ Also, $^2F_4(q)$ has the Borel subgroup $H_1$ of order $q^{12}(q-1)^2$ (see [23, p. 165]). Consider,

  $$\begin{aligned} \frac{|H_1 |}{ |H_2 |^2}&=\frac{q^{12}(q-1)^2}{q^{24}}\\&< 1. \end{aligned}$$

  Therefore,

  $$\begin{aligned} \Big (\frac{|H_1 |}{|H_2 |} \Big )^2< |H_1 |< |H |. \end{aligned}$$
  $$\begin{aligned} \frac{|H |}{ |H_1 |^2}&=\frac{q^{12}(q^6+1)(q^4-1)(q^3+1)(q-1)}{(q^{12}(q-1)^2)^2}\\&= \frac{(q^6+1)(q^4-1)(q^3+1)(q-1)}{q^{12}(q-1)^4}\\&=\frac{(q^6+1)(q^4-1)(q^3+1)}{q^{12}(q-1)^3}\\&< \frac{q^4 (1+\frac{1}{q^6})(1+\frac{1}{q^3})}{q^3(q-1)^3}\\&= \frac{q (1+\frac{1}{q^6})(1+\frac{1}{q^3})}{(q-1)^3}\\&<1. \end{aligned}$$

  This implies that,

  $$\begin{aligned} \Big (\frac{\vert H) \vert }{|H_1 |} \Big )^2 < |H |. \end{aligned}$$

  Table 2 Table representing the constant factor and Method used for choosing suitable subgroups

  Table 3 Order of the simple groups (case (iii) of Theorem 8) and order of its subgroups $H_2, H_1$

  Table 4 Order of the simple groups (case (iv) of Theorem 8) and order of its subgroups $H_2, H_1$

• (11) $H= ~^2F_4(2)';$ (Method 2) The simple group $H= ~^2F_4(2)';$ has order 17971200. It is known that H has a maximal subgroup of order 11232. We take $H_1$ to be this maximal subgroup and $H_2$ to be the Sylow 2-subgroup of $H_1$ which has order 32. Thus, we get $b_{1}=b_{2}=1$.

1.4 Tables

In this section, we cover the details of Sporadic simple groups (Table 2) and the order of all the simple groups that we define in cases (ii)-(iv) of Theorem 8 in Table 3 and 4. These tables also contain the order of the subgroups $H_2$ and $H_1$. Table 2 represents the information about the subgroups $H_2$ and $H_1$ of the Sporadic simple groups.