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The maximal subgroups are point stabilizers and plane stabilizers of order 24, and Singer cycle normalizers of order 21. A proper subgroup containing a 7-element has a unique Sylow 7-subgroup, since the other count gives an impossible index-three subgroup of the simple group. A subgroup of order dividing 24 has a fixed point or a three-point orbit on the seven nonzero vectors. A collinear orbit gives an invariant plane; an independent triple gives its fixed vector sum.

Choose a minimal nontrivial normal subgroup $L$ of $K$ itself. Every conjugate $L^g$ for $g\in G$ is also a minimal normal subgroup of $K$, since $K$ is normal in $G$. Distinct such subgroups intersect trivially: their intersection is normal in $K$ and minimality forces either equality or trivial intersection. Moreover $[L_i,L_j]\subseteq L_i\cap L_j=1$, so distinct ones commute.

Choose a maximal family of these conjugates whose product $M=L_1\times \cdots \times L_r$ is direct. For any further conjugate $L^g$, its intersection with $M$ is normal in $K$ and is either all of $L^g$ or trivial. In the latter case it commutes with all the existing factors and could be adjoined to the direct product of groups, contrary to maximality. Thus every conjugate is contained in $M$. The product of all conjugates is consequently $M$, which is nontrivial and normal in $G$. Minimality of $K$ as a $G$-normal subgroup gives $M=K$.

Any normal subgroup of a factor $L_i$ is normal in $K$: the other direct factors commute with it, and $L_i$ normalizes it. Minimality of $L_i$ as a $K$-normal subgroup makes it simple. All factors are isomorphic because they are conjugate to $L$. This proves the direct-product structure of a finite minimal normal subgroup:An abelian simple group is cyclic of prime order, so the abelian case is elementary abelian. Now let $H$ be a maximal proper subgroup of a nonabelian simple group $X$. It is nontrivial, since $X$ has nontrivial proper cyclic subgroups. Choose a minimal normal subgroup $K$ of $H$. Then $H\le N_X(K)$. This normalizer cannot be $X$, since that would make $1<K\le H<X$ normal in $X$. Maximality givesFor each of the two groups in this question, the allowed absence of proper nonabelian simple subgroups makes the factors of such a $K$ cyclic of prime order. Hence it suffices to examine normalizers of nontrivial elementary abelian subgroups.

In $A_5$, the possible primes are two, three and five. The Sylow subgroups for the last two primes are cyclic of orders three and five. Every involution is a double transposition; for $(12)(34)$ the centralizer in $A_5$ isthe Klein four-group fixing point five. Thus an elementary abelian $2$-subgroup is cyclic of order two or a Klein four-group. The normalizer of a cyclic order-two subgroup equals its centralizer and has order four, so is contained in a point stabilizer of order twelve. A Klein four-group's common fixed point is intrinsic, and the full point stabilizer $A_4$ normalizes it; its normalizer is therefore exactly that stabilizer.

An order-three subgroup has support a triple. Its normalizer preserves the triple and the remaining pair, and is $(S_3\times S_2)\cap A_5$, of order six. An order-five subgroup acts regularly on all five points. Label the points by ${\mathbb{F}}_5$, with the subgroup acting by translations. Its normalizer in $S_5$ is the twenty affine maps $x\mapsto ax+b$. The translations are even, whereas multiplication by a generator of ${\mathbb{F}}_5^{\times }$ is a four-cycle and is odd. Exactly half these maps are even, so the normalizer in $A_5$ has order ten.

The only candidates for maximal subgroups are thus the normalizers of orders twelve, ten and six. They are all maximal: none can lie in a larger candidate, except potentially an order-six subgroup in an order-twelve $A_4$. But an index-two subgroup of $A_4$ would contain every order-three element; there are eight such elements, plus the identity, already more than six. Hence that containment is impossible. Point stabilizers are conjugate; order-three and order-five subgroup normalizers each form one class by the Sylow theorems. Therefore the maximal subgroups of A5 form preciselyTheir numbers are five, six and ten respectively, since these proper maximal subgroups are self-normalizing in the simple group.

For $X=G{L}_3(2)$, choosing independent columns gives $|X|=(8-1)(8-2)(8-4)=168$. If $t$ is an involution, write $t=I+N$. Field characteristic two gives $N^2=0$, and $\mathrm{im}N\subseteq \ker N$ in dimension three forces its nonzero rank to be one. Thus $t=I+vf$, with nonzero vector $v$, nonzero linear functional $f$ and $f(v)=0$. This transvection fixes the plane $\ker f$ and has image line $⟨v⟩$. Every flag consisting of an incident line and plane occurs, giving $7\cdot 3=21$ involutions, all conjugate. Its centralizer, and hence the normalizer of its cyclic subgroup, is the flag stabilizer of order $168/21=8$. It lies in both the line and plane stabilizers of order $24$, so cannot be maximal.

For the elementary abelian subgroups in GL3 over F2, consider commuting transvections $I+vf$, $I+wh$. Their rank-one parts satisfyIf $v,w$ are independent, equality forces both sides zero; thus $f,h$ vanish on $⟨v,w⟩$ and are the same nonzero functional. If $v=w$, they have a common image line. Consequently the two types of order-four elementary abelian subgroup areNo extra transvection can commute with all elements of $K_L$: if its image were different, its functional would have to equal each of the two independent functionals occurring in $K_L$. Dually, the same holds for $K_W$. Thus these exhaust the elementary abelian $2$-subgroups of rank at least two. The normalizer of $K_L$ is the line stabilizer and the normalizer of $K_W$ is the plane stabilizer, each of order $168/7=24$. They give different conjugacy classes because their common fixed spaces have dimensions one and two. All line stabilizers are conjugate, as are all plane stabilizers.

For an order-three element, its minimal polynomial is $(X-1)(X^2+X+1)$: it has a fixed line and an invariant complementary irreducible plane. Its normalizer preserves both and is the full $G{L}_2(2)$ on that plane, of order six, since the order-three subgroup is normal in $G{L}_2(2)$. This normalizer lies in the corresponding line and plane stabilizers, so is not maximal.

An order-seven element is irreducible. Over the finite field ${\mathbb{F}}_2$ one has $X^7-1=(X-1)(X^3+X+1)(X^3+X^2+1)$, and both cubic factors are irreducible because neither has a root in ${\mathbb{F}}_2$. Its minimal polynomial divides this square-free product and cannot be just $X-1$, since the element is not the identity. It therefore contains a cubic factor; the three-dimensional space forces that irreducible cubic to be the whole minimal polynomial. Identify the space with ${\mathbb{F}}_8$ and the element with multiplication by a generator $\alpha$ of ${\mathbb{F}}_8^{\times }$. Its centralizer consists of the seven nonzero multiplication maps. A normalizer induces an ${\mathbb{F}}_2$-automorphism of this field, so it sends $\alpha$ to $\alpha$, ${\alpha }^2$ or ${\alpha }^4$. All three possibilities occur via the field automorphisms $x\mapsto x^{2^j}$. Therefore the normalizer of the Singer cycle has order $7\cdot 3=21$, and all such normalizers are conjugate by the Sylow theorems.

The maximal-normalizer reduction leaves exactly the two order-twenty-four stabilizer types and the order-twenty-one type. Each is maximal because it cannot be contained in any larger remaining candidate; equal-order subgroups cannot properly contain one another and $21$ does not divide $24$. Thus

Work in the finite-group setting. If $M$ is maximal in a nonabelian simple $G$, then $M\ne 1$: otherwise every nonidentity element would generate $G$, forcing a cyclic group of prime order. Choose a minimal normal subgroup $N\ne 1$ of $M$. Every characteristic subgroup of $N$ is normal in $M$, so minimality makes $N$ a characteristically simple group. Its normalizer contains $M$. By maximality it is $M$ or $G$, and the latter would make $N$ a nontrivial normal subgroup of the simple group $G$, impossible because $N\le M<G$. This proves the maximal subgroup normalizer criterion:

For $A_5$, stabilizer subgroups have order 12. Stabilizer subgroups of a two-element subset have order $|(S_2\times S_3)\cap A_5|=6$, and are isomorphic to $S_3$; for example $⟨(123),(12)(45)⟩$. The normalizer of a Sylow 5-subgroup has order 10: its normalizer in $S_5$ has order $5\cdot 4$, and exactly half its permutations are even, since a multiplier of order four on ${\mathbb{F}}_5$ is an odd four-cycle. This gives a dihedral group of order 10.

To prove completeness, an intransitive subgroup either fixes a point, or its orbits on five letters have sizes two and three. It therefore lies in one of the first two types. A transitive proper subgroup $H$ has order divisible by 5. Its number of Sylow 5-subgroups is one or six, since $A_5$ has six and the count is $1$ modulo 5. A count of six would make $30\mid |H|$, forcing an index-two subgroup and contradicting Simplicity of the alternating group A5. Therefore $H$ normalizes its unique Sylow 5-subgroup and lies in the order-ten normalizer.

Stabilizer subgroups are maximal because their index is prime. The order-ten group is transitive and cannot have a larger proper transitive overgroup by the preceding argument. The order-six group has no fixed point, cannot lie in a stabilizer subgroup, and cannot lie in the order-ten type because 3 does not divide 10. Thus all three types are maximal. $A_5$ is transitive on points and on unordered pairs, while Sylow conjugacy handles the normalizers. Consequently

Now let $G=G{L}_3(2)$. Choosing independent columns gives $|G|=7\cdot 6\cdot 4=168$. Its simplicity follows from the elementary projective-transvection proof in part 4(i), which does not use this subgroup classification.

A stabilizer subgroup and a plane stabilizer subgroup each have index seven and order 24. To construct the remaining type, identify ${\mathbb{F}}_2^3$ with ${\mathbb{F}}_8$ and take the Singer cycle $P$ of multiplications by nonzero field elements, of order seven. Any linear map commuting with a primitive multiplication commutes with all of ${\mathbb{F}}_8$, so its centralizer is precisely $P$. A normalizer sends a primitive element $a$ to another root of its minimum polynomial, namely $a,a^2,a^4$. The Frobenius map $x\mapsto x^2$ realizes those three choices. Hence

If $7\mid |H|$ for a proper $H<G$, its Sylow count is one or eight. Eight would make $56\mid |H|$, hence $|H|=56$. Its index-three coset action would inject the simple group $G$ into $S_3$: the kernel is normal and cannot be all of $G$. This is impossible. Thus $H$ normalizes a unique Sylow 7-subgroup and lies in the order-21 normalizer.

If $7\nmid |H|$, its order divides 24. In its action on seven nonzero vectors, an odd orbit has size one or three. A one-point orbit gives a fixed point. If a three-point orbit is collinear, it is the set of nonzero vectors of an invariant plane. If its three vectors are independent, their nonzero sum is fixed by $H$. Thus $H$ lies in a point or plane stabilizer subgroup.

The index-seven stabilizer subgroups are maximal, and any proper overgroup of the order-21 normalizer would again have a unique Sylow 7-subgroup and normalize that same subgroup, so the normalizer is maximal too. The point and plane stabilizer subgroups are not conjugate: a stabilizer subgroup fixes a point, whereas a plane stabilizer subgroup is transitive on its three internal points and its four external points, so fixes no point. Each type is one conjugacy class. We have proved maximal subgroups of GL3 over F2: