Let be a group. A subgroup of is called characteristic if for every automorphism of of Compare that to the definition of a normal subgroup: a subgroup of is normal if for every inner automorphism of So every characteristic subgroup is normal. A few examples of characteristic subgroups of are: the trivial subgroup the group itself, the center of and the commutator subgroup of
Remark. Let be a subgroup of a group If is an automorphism of then is also an automorphism of So is characteristic if and only if for all automorphisms of
Definition 1. A group is called characteristically simple if the only characteristic subgroups of are and itself.
In this post, we characterize all finite characteristically simple groups. But first, let’s give a few examples of those groups. The trivial ones are simple groups; it is clear that every simple group is characteristically simple because every characteristic subgroup is normal. A highly non-trivial example is given in Example 2 in this post. A slightly non-trivial examples of characteristically simple groups are finite elementary abelian -groups. Recall that a finite elementary abelian –group, where is a prime number, is any finite direct product of copies of the cyclic group Looking at the group additively, it is clear that finite elementary abelian -groups are just finite dimensional vector spaces over the finite field (or you can call this one if you have trouble distinguishing between as a cyclic group and as a field).
Example 1. Every finite elementary abelian -group is characteristically simple.
Proof. So, considering additively, is a finite dimensional vector space over the field Suppose, to the contrary, that has a characteristic subgroup Let be a basis for and extend it to a basis of Now, the -linear map defined by
is an automorphism of and so, since is characteristic, contradiction.
Definition 2. A minimal normal subgroup of a group is a normal subgroup of such that the only subgroups of that are normal in are and itself.
Example 2. Every characteristic subgroup of a normal subgroup is normal. In particular, every minimal normal subgroup of a group is characteristically simple.
Proof. Let be a characteristic subgroup of a normal subgroup of a group and let Then, since is a normal subgroup of the map defined by is an automorphism of Since is a characteristic subgroup of we have and so for all i.e. is normal in
We are now ready to characterize all finite characteristically simple groups.
Theorem. A finite group is characteristically simple if and only if is the direct product of some isomorphic simple groups.
Proof. Suppose first that
where is a simple group. We need to show that is characteristically simple. Since is simple, either is abelian or the center of is trivial. If is abelian, then for some prime and hence is characteristically simple by Example 1. So we may assume that Suppose that is a characteristic subgroup of and let Then and so there exists such that Let Then, since is normal,
Notice that since we have So we have shown the set
which is a normal subgroup of is non-trivial and hence i.e., for all So for all and all because the map defined by
is an automorphism of and is characteristic. Thus and the proof of one side of the Theorem is complete. Notice that for this part of the Theorem, we did not need to be finite.
Conversely, suppose that is characteristically simple. We need to show that is the direct product of some isomorphic simple groups. Choose a normal subgroup of such that and is as small as possible. Let be subgroups of such that and the positive integer is as large as possible.
Claim 1. Let be an automorphism of Then for all
Proof. First notice that since for all each hence is a normal subgroup of and for all Suppose now, to the contrary, that for some So is a proper subgroup of and thus Hence, since is a normal subgroup of we have by minimality of So are two normal subgroups of whose intersection is trivial. That implies which contradicts maximality of This contradiction completes the proof of Claim 1.
Claim 2.
Proof. By Claim 1, for all and so for all automorphisms of Therefore is a characteristic subgroup of and hence because and is characteristically simple. The proof of Claim 2 is now complete.
We can now finish the proof of the Theorem. By Claim 2, for some normal subgroups such that for all So the only thing left is to show that each is simple. Suppose that is a normal subgroup of some Then is also a normal subgroup of because is a direct summand of But and so, by minimality of we must have or Hence either or which proves that is simple.
Corollary. Let be a minimal normal subgroup of a finite group Then either is an elementary abelian -group or is the direct product of some isomorphic non-abelian simple groups.
Proof. By Example 2, is characteristically simple and hence, by the Theorem, is the direct product of some isomorphic simple groups We are done if each is non-abelian. Suppose now that is abelian for some Then for some prime because is simple and abelian. Hence for all and thus is an elementary abelian -group. in this case.
Note. The reference for the only if part of the Theorem and the Corollary is chapter 2, section 1, of the book Finite Groups by Daniel Gorenstein. Strangely, Gorenstein doesn’t even mention the if part of the Theorem in his book; the proof of that was added by me.
Exercise 1. Show that every characteristically simple group of order is simple. Hint. If is not simple, then, by the Theorem, for some simple group and some integer Show that the equation has no solution for integers by, for example, proving that has alway a prime divisor of exponent
Exercise 2. Show that a finite solvable group is characteristically simple if and only if it is an elementary abelian -group.Hint. Since is solvable, and so if is characteristically simple, then i.e. is abelian. Now use the Theorem.
Exercise 3. Show that the symmetric group is not characteristically simple.Hint. First Proof: The alternating group is a normal subgroup of and so is not simple hence not characteristically simple, by Exercise 1. Second Proof: is the unique subgroup of order in and hence for any automorphism of i.e. is a characteristic subgroup of
