Simple group

A nontrivial simple group has no normal subgroups other than the identity subgroup and itself.

For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

For a nontrivial simple soluble group, the commutator subgroup is either the identity or the whole group. It cannot be the whole group because the derived series would never terminate. The group is therefore Abelian. Every nonidentity element then generates a normal cyclic subgroup, which must be the whole group. An infinite cyclic group has proper nontrivial subgroups; a finite cyclic group is simple exactly when its order is prime. Thus the soluble simple groups are precisely the cyclic groups of prime order.

A nontrivial finite group whose only normal subgroups are the trivial group and the whole group. A cyclic group of prime number order gives an abelian example; $A_5$ gives a nonabelian alternating group example. Finite versions of classical groups can produce further examples after taking suitable subgroups and quotients; small parameters require separate checks.

For a finite simple group of this order, $e=0$ gives a normal Sylow $5$-subgroup, and $e=1$ gives a nontrivial sign homomorphism from the regular action, because an involution swaps fifteen pairs. Thus $e\ge 2$. The Sylow $2$-count divides $15$; it cannot be one or three. If it is five, a Sylow subgroup's normalizer has index five. If it is fifteen, the strengthened Sylow congruence from intersections produces two Sylow subgroups meeting in index two. Their common subgroup is normal in both, so its normalizer contains both and has index three or five. Simplicity excludes index three. The resulting faithful degree-five action embeds $G$ in $A_5$ and forces order $60$, proving the conclusion.

For a faithful primitive group action, suppose a point stabilizer has an abelian normal subgroup whose conjugates generate the whole group. Every nontrivial normal subgroup then contains the derived subgroup. Indeed the normal subgroup is transitive, and all those conjugates have the same image in the quotient, making the quotient abelian. If the whole group is nontrivial and perfect, it is simple.

A normal subgroup of $A_6$ has order dividing 360 and consists of the identity plus whole conjugacy classes of the alternating group on six letters. If it omits the odd-sized class of size 45, its order is an odd divisor and no nontrivial class sum fits. If it includes that class, the possible class sums not exceeding 180 are $46,86,118,126,136,158,176$, none dividing 360. Every proper subgroup has order at most 180; therefore $A_6$ is a simple group.

The alternating group $A_n$ is simple for every $n\ge 5$. In particular it has no proper subgroup of index two, since every index-two subgroup is normal.

A finite nonabelian simple group is a finite group that is a nonabelian group and a simple group. Every such group is a perfect group because its commutator subgroup is a nontrivial normal subgroup.

Let $p$ be the smallest prime divisor of the order of a finite nonabelian simple group. A cyclic Sylow subgroup would give a normal p-complement, so if $p^3$ does not divide the order, the Sylow subgroup must be $C_p^2$. Conjugation embeds its normalizer modulo centralizer into $G{L}_2(p)$, and this quotient has order prime to $p$. For odd $p$, every other prime divisor of $|G{L}_2(p)|=p(p-1{)}^2(p+1)$ is smaller than $p$, again forcing trivial conjugation and a contradiction to the Burnside transfer theorem. For $p=2$, nontrivial conjugation must have order three, giving $12\mid |G|$. The groups $PS{L}_2(5^{2m+1})$ give infinitely many examples with exactly two factors of two in their orders.

The natural actions of these five sporadic simple groups have degrees 11, 12, 22, 23 and 24, and respective transitivity degrees 4, 5, 3, 4 and 5. They arise from exceptional finite designs and codes.

These simple groups have order $q^3(q^3+1)(q-1)$ and a two-transitive action on $q^3+1$ points. The restriction excludes the smallest nonsimple parameter; this family differs from the Ree groups of type $F_4$.

These simple groups have order $q^2(q^2+1)(q-1)$ and a two-transitive action on a Suzuki ovoid of $q^2+1$ points. The point-stabilizer root group is regular on the remaining points.

The conjugacy classes of $A_5$ have sizesA normal subgroup is a union of these classes containing the identity. No proper nontrivial such union has size dividing $60$, so Lagrange's theorem proves that $A_5$ is simple.

Every finite nonabelian simple group has order at least 60; the alternating group A5 attains this bound.