June 08, 2026
ABSTRACT: A finite group is said to be quartic-free if its order is not divisible by \(p^4\) of any prime \(p\). A finite group is called an \(A\)-group if all of its Sylow subgroups are abelian. Objective of this paper is to provide explicit structure of a quartic-free \(A\)-group. Further in the process of providing the explicit structure we also determine the derived length of a solvable quartic-free \(A\)-group.
Keywords: quartic-free groups, \(A\)-groups, general linear groups, solvable groups, non-solvable groups, nilpotent groups.
Mathematics Subject Classification-MSC2020: 20E28, 20E34, 20E45, 20F99.
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In 1893, Otto H\(\ddot{\text{o}}\)lder described groups of order \(p^3\) and \(p^4\). Soon after, he arrived at a formula for the number of groups of order \(n\) using the structure of groups of order \(n\) when \(n\) is square-free, that is, the square of no prime divides \(n\). (see [1], [2]). Result of H\(\ddot{\text{o}}\)lder, Burnside and Zassenhaus [3] shows that every finite group with cyclic Sylow subgroups of order \(n\) is metacyclic with odd-order derived subgroup \(G' \cong \mathbb{Z}_m\) and cyclic quotient \(G/G'\) of order \(l = n/m\).
The structure of cube-free groups was studied by Heiko Dietrich and Bettina Eick. They also presented an algorithm to construct cube-free groups up to isomorphism of a given order using \({\rm GAP}\) (see [4]). Later, S. Qiao and C. H. Li gave a more explicit structure of cube-free groups (see [5]).
Objective of this paper is to expand the work of S. Qiao and C. H. Li to the quartic-free \(A\)-groups and obtain the explicit structure of quartic-free \(A\)-groups. Further we would like to point out that though the paper is specifically oriented on quartic-free \(A\)-groups however the methodology developed in this paper may allow to obtain the structure of \(A\)-groups of higher degree, that is \(A\)-groups whose orders are not divisible by \(p^n\) for every prime \(p\) where \(n\geq 5\).
Throughout the paper, \(p\) is a prime, \(q\) is a power of \(p\) and \(\mathbb{F}_q\) is the finite field of order \(q\). Let \(D(n,q)\) denote the subgroup of diagonal matrices of \(GL(n,q)\). Let \(M(n,q) = D(n,q) \rtimes S_n\) be the subgroup of monomial matrices in \({\rm GL}(n,q)\). Let \(N(n,q)\) be the normaliser of \(S(n,q)\) where \(S(n,q) \cong \mathbb{Z}_{q^n-1}\) is a Singer cycle in \({\rm GL}(n,q)\). The Borel subgroup \(B(n,q)\) of \({\rm GL}(n,q)\) is defined as
\[B(n,q) = \left\{ \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{pmatrix} \mid a_{ii} \in {\mathbb{F}_q}^*, a_{ij} \in \mathbb{F}_q \right \}\]
Note that \(B(2,q) \cong \mathbb{Z}_q \rtimes (\mathbb{Z}_{q-1} \times \mathbb{Z}_{q-1})\).
Let \(G\) be a group. We denote \({\rm sol}(G)\) to be the maximal solvable normal subgroup of \(G\).
Now we shall state the main results of our paper.
Theorem 1 (Main Theorem 1). Let \(G\) be a solvable quartic-free \(A\) group. Then \(G = A \rtimes ((B \rtimes (C \rtimes D)))\) where \(A, B, C, D\) are abelian quartic-free subgroups \(G\).
Theorem 2 (Main Theorem 2). Let \(G\) be a non-solvable quartic-free \(A\)-group. Then \(G = L \rtimes S\) where \(L\) is a solvable and \(S\) is a simple. Further if \(S\) acts non trivially on \(L\) then \(5 \mid |S|\) and \(S \cong A_5\).
In this section we prove Theorem 1. In the process of proving the result we classify solvable quartic-free \(A\)-subgroups of \({\rm GL}(2,q)\) and solvable quartic-free \(p'\) \(A\)-subgroups of \({\rm GL}(3,q)\). Additionally we also provide the structure of solvable cube-free group of a given order.
The proof of Theorem 1 is divided into several results. We begin our proof by stating a standard result.
Lemma 3. Let \(G\) be a solvable group and let \(d(G)\) denote the derived length of \(G\). Then
if \(H \leq G\) then \(d(H) \leq d(G)\).
if \(N \trianglelefteq G\) then \(d(G) \leq d(G/N) + d(N)\)
\(d(H_1 \times \cdots \times H_k) = max_{1\leq i \leq k}\{d(H_i)\}\) where \(H_i\)’s are solvable.
Now we recall an important result by D. Taunt (see [6]).
Lemma 4. Let \(G\) be a solvable \(A\)-group. Then \(G = G' \rtimes N_1\) where \(N_1\) is the system normalizer of \(G\).
Since a subgroup of a solvable \(A\)-group is also a solvable \(A\)-group we can apply Lemma [Strctre95of95sol95A95grps] to \(G'\) and obtain that \(G' = G'' \rtimes N_2\) where \(N_2\) is the system normalizer of \(G'\). So from Lemma [Strctre95of95sol95A95grps] we have \(G = (G'' \rtimes N_2) \rtimes N_1\). Continuing in this fashion we obtain the following corollary.
Corollary 5. Let \(G\) be a solvable \(A\)-group. Then \(G = ((G^{m-1} \rtimes N_{m-1}) \rtimes \cdots )\rtimes N_1\) where \(N_i\) is the system normalizer of \(G^{i-1}\) and \(m\) is the derived length of \(G\).
Since a system normalizer of a solvable group is nilpotent therefore a system normalizer of a solvable \(A\)-group is abelian. Thus by the Corollary 5 a solvable \(A\)-group is largely a chain of semi-direct products of abelian groups.
Now let \(G\) be a solvable \(A\)-group and let \(F\) be the Fitting subgroup of \(G\). Since \(F\) is nilpotent, \(F = P_1 \times \cdots \times P_s\) where \(P_i\) is the Sylow \(p_i\)-subgroup of \(F\). Therefore \[\label{eq95i} G/F = N_G(F)/C_G(F) \leq {\rm Aut}(F) \cong \prod_{i=1}^{s}{\rm Aut}(P_i).\tag{1}\]
Let \(K_i\) be the \(i^{th}\) projection of \(G/F\) into \({\rm Aut}(P_i)\). Then \(G/F\) is embedded into \(K_1 \times \cdots \times K_s\). Therefore by the properties of derived length we have \[\label{eq95ii} d(G) \leq max_{1 \leq i \leq s}\{d(K_i)\} + 1. \tag{2}\]
Therefore in order to obtain the derived length of a solvable quartic-free \(A\)-group we now examine the structure of subgroups of \({\rm Aut}(P)\) where \(P\) is an abelian quartic-free \(p\)-group.
Proposition 6. Let r be a prime. Let \(H\) be a solvable irreducible \(A\)-subgroup of \({\rm GL}(r,q)\). Then one of the following holds.
\(H\) is imprimitive and \(H\) conjugates to a subgroups of \(M(r,q)\).
\(H\) is primitive and \(H\) conjugates to a subgroup of \(N(r,q)\).
Proof. If \(H\) is an imprimitive subgroup of \({\rm GL}(r,q)\) then by [7], \(H\) conjugates to a subgroup of \(M(r,q)\).
So assume that \(H\) is a primitive subgroup of \({\rm GL}(r,q)\). If \(H\) is abelian then by [8], \(H\) conjugates to a subgroup of \(S(r,q) \leq N(r,q)\). So let \(H\) be a non abelian subgroup of \({\rm GL}(r,q)\) and let \(F\) the Fitting subgroup of \(H\). Let \(V\) be an irreducible \(\mathbb{F}_qH\)-module. Then by Clifford’s Theorem \(V\) is a semi-simple \({\mathbb{F}_qF}\)-module of dimension \(r\). Since \(r\) is a prime either \(V\) is a direct sum of isomorphic \(1\)-dimensional \({\mathbb{F}_qF}\)-submodules or \(V\) is an irreducible \({\mathbb{F}_qF}\)-module. If \(V\) is a direct sum of \(1\)-dimensional isomorphic \({\mathbb{F}_qF}\)-submodules, then \(F\) is a subgroup of \({\rm GL}(r,q)\) of scalar matrices but this contradicts that \(C_H(F) = F\). Therefore \(V\) must be an irreducible \({\mathbb{F}_qF}\)-module. Thus by [8], \(F\) is conjugate to a subgroup of \(S(r,q)\). Since \(F \trianglelefteq H\) by [8], \(H\) is conjugate to a subgroup of \(N(r,q)\). ◻
Next we study the structure of quartic-free \(A\)-subgroups of \({\rm GL}(r,q)\) for \(r \in \{2,3\}\).
Lemma 7. Let \(H\) be a solvable quartic-free \(A\)-subgroup of \({\rm GL}(2,q)\). Then one of the following holds.
\(H\) conjugates to a subgroup of \(B(2,q)\). Further \(H \cong P \rtimes (\mathbb{Z}_l \times \mathbb{Z}_s)\) where \(P\) is the Sylow \(p\)-subgroup of \(H\) and \(l \mid q-1\) and \(s \mid q-1\).
\(H\) conjugates to a subgroup of \(D(2,q)\) and \(H \cong \mathbb{Z}_l \times \mathbb{Z}_s\) where \(l \mid q-1\) and \(s \mid q-1\).
\(H\) conjugates to a subgroup of \(M(2,q)\) and \(H = K \rtimes P\) where \(K\) is Hall \(2'\)-subgroup of \(H\) contained in \(D(2,q)\) and \(P\) is a Sylow \(2\)-subgroup of \(H\).
\(H\) conjugates to a subgroup of \(N(2,q)\). \(H = K \rtimes P\) where \(K\) is Hall \(2'\)-subgroup of \(H\) contained in \(S(2,q)\) and \(P\) is a Sylow \(2\)-subgroup of \(H\).
Proof. If \(H\) is reducible but not completely reducible then \(H\) conjugates to a subgroup of \(B(2,q)\cong \mathbb{Z}_q \rtimes (\mathbb{Z}_{q-1} \times \mathbb{Z}_{q-1})\). In particular \(H \cong P \rtimes (\mathbb{Z}_l \times \mathbb{Z}_s)\) where \(P\) is the Sylow \(p\)-subgroup of \(H\) and \(l \mid q-1\) and \(s \mid q-1\).
If \(H\) is reducible and \(p\nmid |H|\), then by Mashke’s Theorem \(H\) is completely reducible. Thus the underlying \(\mathbb{F}_qH\)-module is a direct sum of two one dimensional \(\mathbb{F}_qH\)-submodule. Therefore \(H\) conjugate to a subgroup of \(D(2,q) \cong \mathbb{Z}_{q-1} \times \mathbb{Z}_{q-1}\).
If \(H\) is primitive then by Proposition 6, \(H\) conjugates to a subgroup of \(M(2,q)\). Let \(H' = H \cap D(2,q)\) and \(K\) be a Sylow \(2'\)-subgroup of \(H'\). Then clearly \(H = K \rtimes P\).
If \(H\) is primitive then by Proposition 6, \(H\) conjugates to a subgroup of \(N(2,q)\). Let \(H' = H \cap S(2,q)\) and \(K\) be a Sylow \(2'\)-subgroup of \(H'\). Clearly \(H = K \rtimes P\). ◻
Remark 1. Note that an odd order \(p'\) quartic-free \(A\)-subgroup of \({\rm GL}(2,q)\) is abelian.
Lemma 8. Let \(H\) be a \(p'\) quartic-free \(A\)-subgroup of \({\rm GL}(3,q)\). Then one of the following holds.
\(H\) is reducible and \(H\) is conjugate to a subgroup of \(D(3,q)\). In particular \(H \cong \mathbb{Z}_l \times \mathbb{Z}_m \times \mathbb{Z}_s\) where \(l \mid q-1\), \(m \mid q-1\) and \(s \mid q-1\).
\(H\) is reducible and \(H\) is conjugate to a subgroup of \({\mathbb{F}_q}^* \times M(2,q)\). In particular \(H = H_{2'} \rtimes H_2\) with \(H_{2'}\) abelian.
\(H\) is reducible and \(H\) is conjugate to a subgroup of \({\mathbb{F}_q}^* \times N(2,q)\). In particular \(H = H_{2'} \rtimes H_2\) with \(H_{2'}\) abelian.
\(H\) is irreducible and \(H\) is conjugate to a subgroup of \(M(3,q)\) with \(3 \mid |H|\). In particular if \(|H|\) is odd then \(H = H_{3'} \rtimes H_3\). If \(|H|\) is even then \(H = H_{\{2,3\}'} \rtimes H_{\{2,3\}}\) where \(H_{\{2,3\}'} \leq D(3,q)\).
\(H\) is reducible and \(H\) is conjugate to a subgroup of \(N(3,q)\). In particular either \(H\) is cyclic or \(H = H_{3'} \rtimes H_3\) with \(H_{3'}\) cyclic.
Proof. Let \(V\) be the underlying \(\mathbb{F}_qH\)-module. If \(V\) is reducible and direct sum of three \(1\)-dimensional \(\mathbb{F}_H\)-submodules of \(V\). Then \(H\) conjugates into a subgroup of \(D(3,q)\).
If \(V\) is reducible and direct sum of \(V_1\) and \(V_2\) where \(V_1\) is a \(1\)-dimensional \(\mathbb{F}_qH\)-submodule and \(V_2\) is a \(2\)-dimensional irreducible \(\mathbb{F}_qH\)-submodule. Then by Lemma [cor95to95subgrps95of95GL], either \(H\) conjugates to a subgroup of \({\mathbb{F}_q}^* \times M(2,q)\) or to a subgroup of \({\mathbb{F}_q}^* \times N(2,q)\). Now first assume that \(H \leq {\mathbb{F}_q}^* \times M(2,q)\). Let \(L = {\mathbb{F}_q}^* \times D(2,q)\). Then \(H \cap L\) is normal in \(H\). Thus \(H_{2'} \leq H \cap L\) is normal in \(H\). Hence \(H = H_{2'} \rtimes H_2\). Similarly we can show that \(H = H_{2'} \rtimes H_2\) when \(H\) is conjugate to a subgroup of \({\mathbb{F}_q}^* \times N(2,q)\).
Now assume that \(V\) is irreducible. If \(V\) is imprimitive then by Proposition 6, \(H\) is conjugate to a subgroup of \(M(3,q)\). Further since \(H\) is irreducible it permutes three \(1\)-dimensional subspaces of \(V\) transitively, therefore \(3 \mid |H|\). Now assume that \(H \leq M(3,q)\). Let \(\pi \in M(3,q)\) be the permutation matrix corresponding to a \(3\)-cycle. Then \(H \leq D(3,q) \rtimes \langle \pi \rangle\). Therefore \(H = H_{3'} \rtimes H_3\). Now let \(K = H\cap D(2,q)\). Then clearly \(H = K_{\{2,3\}'} \rtimes H_{\{2,3\}} = H_{\{2,3\}'} \rtimes H_{\{2,3\}}\).
So assume that \(V\) is irreducible and primitive. Then by Proposition 6, \(H\) is conjugate to a subgroup of \(N(3,q)\). If \(H\) is abelian then \(H\) is cyclic. So suppose that \(H\) is non abelian. Let \(K = S(3,q)\cap H\). Then \([H:K] = 3\). Thus \(H = K_{3'} \rtimes H_3 = H_{3'} \rtimes H_3\). ◻
Lemma 9. Let \(P \cong {\mathbb{Z}}_{p^2} \times {\mathbb{Z}_{p}}\). Then \({\rm Aut}(P) \cong R \rtimes ({\mathbb{Z}_{p-1}\times \mathbb{Z}_{p-1}})\) where \(R\) is the Sylow \(p\)-subgroup of \({\rm Aut}(P)\) order \(p^3\). In particular a \(p'\)-subgroup of \({\rm Aut}(P)\) is isomorphic to \(\mathbb{Z}_l \times \mathbb{Z}_s\) where \(l \mid p-1\) and \(s \mid p-1\).
Proof. Let \(G = {\rm Aut}(P)\). A standard argument shows that \(|G| = p^3(p-1)^2\). Thus by Sylow’s Theorem it can be seen that \(R\) is normal in \(G\). So \(G\) is a semi-direct product of \(R\) by \(G/R\). Now since \(P_p = \{x \in P \mid x^p = 1\} \cong {\mathbb{Z}_p \times \mathbb{Z}_p}\) is characteristic in \(P\), there is a homomorphism from \(G/R\) to \({\rm GL}(2,p)\). Futher since the subgroup \(P^p = \{x^p \mid x \in P\} \cong \mathbb{Z}_p\) is also characteristic in \(P\) by Maschke’s Theorem the image of \(G/R\) conjugates into a subgroup of \(D(2,p)\). Therefore \(G/R \cong \mathbb{Z}_{p-1}\times \mathbb{Z}_{p-1}\). ◻
Note that it is clear that an odd order quartic-free \(A\)-subgroup of \({\rm Aut}(P)\) where \(P\) is an abelian group of order \(p^{\alpha}\) with \(\alpha \in \{1,2,3\}\) is meta abelian. Thus from Lemma [Strctre95of95sol95A95grps] and equation (2 ), we have the following structure of an odd order quartic-free \(A\)-group.
Corollary 10. Let \(G\) be a quartic-free \(A\)-group of odd order. Then \(G = (A \rtimes B) \rtimes C\) where \(A,B\) and \(C\) are abelian subgroups of \(G\) of suitable orders.
Before moving forward we provide the structure of a solvable cube-free group. Li and Qiao have already shown that a solvable cube-free group is largely a semi-direct product of cube-free abelian groups of suitable orders (see [5]). But our result provides description of these abelian groups. In particular we show that an odd order cube-free group \(G\) is metaabelian with \(G' \cong {\mathbb{Z}_a}\times \mathbb{Z}_b\) and \(G/G' \cong {\mathbb{Z}_c} \times \mathbb{Z}_d\) where \(a,b,c,d\) are suitable cube-free integers. Further in addition we prove that all the compliments of \(G'\) in \(G\) are conjugate.
Proposition 11. Let \(G\) be a cube-free group of even order and let \(H\) be a Hall \(2'\)-subgroup of \(G\). Then
\(H = H' \rtimes N\) where \(H'\) is abelian and \(N\) is a system normaliser of \(H\) containing a Sylow \(3\)-subgroup of \(H\). Further all compliments of \(H'\) in \(H\) are conjugate.
\(G = H \rtimes P\) or \(G = (P \times H') \rtimes N\) where \(P\) is a Sylow \(2\)-subgroup of \(G\).
Proof. Let \(F\) be the Fitting subgroup of \(H\). Let \(\{p_1,\ldots,p_k\}\) be the set of all prime divisors of \(|F|\) and let \(F = P_1 \times \cdots \times P_k\) where \(P_i\) is a Sylow \(p_i\) subgroup of \(F\). Then \(H/F\) is embedded in \(\prod_{i=1}^{k} {\rm Aut}(P_i)\). If is clear form Corollary [cor95to95subgrps95of95GL] that \(H/F\) is abelian and hence \(H' \leq F\). Thus by Lemma [Strctre95of95sol95A95grps], \(H = H' \rtimes N\) with \(H'\) and \(N\) abelian. Let \(P\) be the Sylow \(3\)-subgroup of \(H\). Let \(r \neq 3\) be a prime dividing \(|H|\) and let \(R\) be a Sylow \(r\)-subgroup of \(H\). Then \(r \nmid |{\rm Aut}(P)|\). Thus by Burnside’s compliment Theorem \(P\) normalizes \(R\) and hence \(P\) conjugates to a subgroup of \(N\).
Now if \(N'\) is another compliment of \(H'\) in \(H\) then it is not difficult to see that \(N'\) is also a system normalizer of \(H\). Therefore \(N'\) is conjugate to \(N\) in \(H\).
Part \({\rm (ii)}\) of this result follows from [5]. ◻
Remark 2. The structure of non-solvable cube-free groups will be discussed in the next section.
Now we investigate the structure of quartic-free \(A\)-groups whose order divides only two primes. First we consider the case when \((p,r) \neq (2,3)\).
Lemma 12. Let \(p\) and \(r\) be distinct primes with \(p < r\) and \((p,r) \neq (2,3)\). Let \(G\) be a quartic-free \(A\)-group of order \(p^\alpha r^\beta\). Let \(P\) be a Sylow \(p\)-subgroup of \(G\) and \(R\) be a Sylow \(r\)-subgroup of \(G\). Then either \(G = P \rtimes R\) or \(G = R \rtimes P\).
Proof. Let \(N = N_G(R)\). Then by Sylow’s theorem \([G: N] \equiv 1\bmod{r}\). If \([G:N] = 1\) then \(R \trianglelefteq G\) and \(G = R \rtimes P\). If \([G: N] = p\) then \(N \trianglelefteq G\) and therefore \(G = R \rtimes P\). If \(\alpha > 1\) and \([G: N] = p^2\) then \(r \mid p^2 - 1\). Since \(r > p\) and \((p,r) \neq (2,3)\) this case is not possible. If \(\alpha = 3\) and \([G : N] = p^3\) then \(N = R\) and by Burnside’s compliment theorem [3] we have \(G = P \rtimes R\). ◻
Now we deal with the case \((p,r) = (2,3)\). This case require case by case analysis of automorphism group of Sylow \(2\)-subgroup of \(G\).
Lemma 13. Let \(G\) be a quartic-free \(A\)-group of order \(2^\alpha 3^\beta\). Let \(P\) and \(R\) be Sylow \(2\) and Sylow \(3\)-subgroups of \(G\) respectively. Then one of the following holds.
\(G = R \rtimes P\).
\(G = P \rtimes Q\) where \(P \in \{\mathbb{Z}_2 \times \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2\}\).
\(P \cong \mathbb{Z}_2 \times \mathbb{Z}_2\) and \(G \cong \mathbb{Z}_9 \times (A_4 \times \mathbb{Z}_3)\).
\(P \cong \mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2\) and \(G \cong (\mathbb{Z}_9 \rtimes \mathbb{Z}_2) \times (A_4 \times \mathbb{Z}_3)\).
\(P \cong \mathbb{Z}_2 \times \mathbb{Z}_2\) and \(G \cong (\mathbb{Z}_3 \times \mathbb{Z}_3) \times (A_4 \times \mathbb{Z}_3)\).
\(P \cong \mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2\) and \(G \cong (\mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2) \times (A_4 \times \mathbb{Z}_3)\).
\(G \cong (\mathbb{Z}_2 \times \mathbb{Z}_2) \rtimes (R \rtimes \mathbb{Z}_2)\) where \(P \cong \mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2\).
Proof. If \(P \not \in \{\mathbb{Z}_2 \times \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2\}\) then \({\rm Aut}(P)\) is a \(2\)-group. Therefore \(N_G(P) = C_G(P)\) and by Burnside’s compliment Theorem \(G = R \rtimes P\). So assume that \(P\) is elementary abelian. Let \(N = N_G(R)\). If \([G:N] \neq 4\) then as in the proof Lemma [2-primes] either \(G = R \rtimes P\) or \(G = P \rtimes R\). Suppose that \([G: N] = 4\).
If \(\beta = 1\) then \([G : N_G(P)] \in \{1,3\}\). If \([G: N_G(P)] = 1\) then \(P \trianglelefteq G\) and \(G = P \rtimes R\). If \([G : N_G(P)] = 3\) then \(N_G(P) = C_G(P)\) and \(G = R \rtimes P\).
If \(\beta = 2\) and \([G:N_G(P)] \neq 3\) then the above arguments can be repeated to show that \(G = P \rtimes R\) or \(G = R \rtimes P\). Suppose \([G: N_G(P)] = 3\) then there is a homomorphism \(\phi:G \rightarrow S_3\) with \(\ker(\phi) \leq N_G(P)\). If \(8\mid |\ker(\phi)|\) then \(P \leq \ker(\phi)\) and \(G = P \rtimes R\). Thus \(4 \mid \ker(\phi)\) and by Frattini’s argument \(G = P\cap \ker(\phi) \rtimes N_G(R) \cong (\mathbb{Z}_2 \times \mathbb{Z}_2) \rtimes (R \rtimes \mathbb{Z}_2)\).
If \(\beta = 3\) and \([G: N_G(P)] \neq 9\) then the structure of \(G\) can be obtained as in previous cases. So we assume that \([G: N_G(P)] = 9\) and \([G:N] = 4\). As \([G: N] = 4\) there is a homomorphism \(\psi: G \rightarrow S_4\) with \(\ker(\psi) \leq N\). If \(R \leq \ker(\psi)\) then \(G = R \rtimes P\). So assume that \(R\) is not contained in \(\ker(\psi)\). Then \(K = P\ker(\phi)\) is a subgroup of \(G\) of index \(3\). Thus there is a homomorphism \(\eta: G \rightarrow S_3\) such that \(\ker(\eta) \leq K\). If \(\ker(\eta) = K\) and \(P \trianglelefteq K\). Then \(G = P\rtimes R\). If \(P\) is not normal in \(K\). Then by Frattini’s argument we have \(G = KN_G(P) = \hat{R} \rtimes N_G(P)\) where \(\hat{R}\) is a Sylow \(3\)-subgroup of \(K\) of order \(9\). Thus either \(G \cong \mathbb{Z}_9 \rtimes (P \rtimes \mathbb{Z}_3)\) or \(G \cong (\mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes (P \rtimes \mathbb{Z}_3)\).
So assume that \(\ker(\eta) \lneq K\). Then \(M = R\ker(\eta)\) is a subgroup of \(G\) of index \(2\). Thus \(M \trianglelefteq G\). If \(R \trianglelefteq M\) then \(G = R \rtimes P\). Otherwise by Frattini’s argument \(G = MN_G(R) \cong (\mathbb{Z}_2 \times \mathbb{Z}_2) \rtimes (R \rtimes \mathbb{Z}_2)\) this completes the proof. ◻
Lemma 14. Let \(G\) be a quartic-free \(A\)-group of even order. Let \(P\) be a Sylow \(2\)-subgroup of \(G\) and let \(P \not \in \{{\mathbb{Z}}_2\times \mathbb{Z}_2, {\mathbb{Z}}_2\times \mathbb{Z}_2 \times \mathbb{Z}_2\}\). Then \(G = H \rtimes P\) where \(H\) is a Hall \(2'\)-subgroup of \(G\).
Proof. Since \(P\) is not elementary abelian \({\rm Aut}(P)\) is a \(2\)-group. Thus \(P\) normalizes the Sylow system of a Hall \(2'\)-subgroup of \(G\). Hence \(G = H \rtimes P\). ◻
In this section we prove Theorem [Main95theorem952]. We begin by classifying the simple quartic-free \(A\)-groups. Here we assume that the classification of finite simple groups holds.
Lemma 15. Let \(G\) be a simple quartic-free \(A\) group. Then \(G \cong {\rm PSL}(2,q)\) for some \(q = p^\alpha\) with \(\alpha \in \{ 1,2,3\}\) and \(q+1\) and \(q-1\) quartic-free or \(G \cong J_1\) first Janko group of order \(2^3.3.5.7.11.19\).
Proof. By [9] a simple non-abelian group with abelian Sylow-2 subgroup is isomorphic to \(J_1\) or \(L_2(q)\) for \(q>3\) and \(q \equiv 0,3\) or \(5\bmod{8}\). But by [10], \(L_2(q) = {\rm PSL}(2,q)\). ◻
Now we prove Theorem [Main95theorem952]. That is we now discuss the structure of a non-solvable quartic-free \(A\)-group.
Recall that we denote \({\rm sol}(G)\) to be the largest normal solvable subgroup of a group \(G\).
Proof of Theorem 2. Suppose that the statement of the Theorem is not true. Let \(G\) be a minimal counter example and let \(H = {\rm sol}(G)\). If \(H ={1}\) then \(G\) is semi-simple (simple in our case) and the result holds trivially. So assume that \(H \not = {1}\) and let \(N\) be the minimal normal subgroup of \(G\) contained in \(H\). Then \(N\) is elementary abelian. By the minimal choice of \(G\) we have \(G/N = S/N \rtimes T/N\) where \(S/N\) is solvable and \(T/N\) is non-abelian simple. If \(G \not = T\) then by minimal choice of \(G\) we have \(T = N \rtimes U\) where \(U\) is non-abelian simple. Consequently \(G = S \rtimes U\) contrary to the assumption. Thus \(G = T\) and \(G/N\) is simple. Therefore \(G/N = (G/N)' = G'N/N\). If \(G' \lneq G\) then by minimality \(G' = R \rtimes K\) where \(R\) is solvable and \(K\) is simple. Consequently \(G = (NR) \rtimes K\) contradicting that \(G\) is a counter example. So we assume that \(G = G'\). Since \(G/N\) is semi-simple we have \(C_G(N) = N\) or \(C_G(N) = G\). If \(C_G(N) = G\) then \(Z(G) = N\) contradicting [3]. Hence \(C_G(N) = N\) but then \({\rm gcd}(|N|,[G:N]) = 1\) and by Schur-Zassenhaus [3], \(G = N \rtimes M\) where \(M\) is simple this contradiction completes the proof.
Corresponding author, Dr. B. R. Ambedkar University Delhi, Delhi 110006, India; E-mail: prashun07kumar@gmail.com.↩︎