Equivalence of definitions of finite characteristically simple group
This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
View other semi-basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
This article gives a proof/explanation of the equivalence of multiple definitions for the term finite characteristically simple group
View a complete list of pages giving proofs of equivalence of definitions
Contents
The definitions that we have to prove as equivalent
We need to show that the following are equivalent for a nontrivial finite group:
- It is a characteristically simple group, i.e., it has no proper nontrivial characteristic subgroup.
- It can be expressed as an internal direct product of pairwise isomorphic finite simple groups.
Definitions used
Term | Definition used |
---|---|
characteristic subgroup | A subgroup of a group is termed characteristic if every automorphism of the whole group restricts to an automorphism of the subgroup. |
simple group | A nontrivial group is said to be simple if it has no proper nontrivial normal subgroup. |
internal direct product (case of finitely many subgroups) | We'll use the following definition of internal direct product, which is a little different from the usual one, but equivalent: is an internal direct product of normal subgroups if and, for all with , is trivial. |
Facts used
- Normality is strongly intersection-closed
- Normality is strongly join-closed
- Direct factor implies transitively normal
- Normality satisfies image condition
- Normal subdirect product of perfect groups equals direct product
Proof
Proof of (1) implies (2)
Given: A finite nontrivial group that has no proper nontrivial characteristic subgroups.
To prove: We can find pairwise isomorphic finite simple subgroups in such that is an internal direct product of . The proof will also show, as a by-product, that every minimal normal subgroup is a direct factor.
Proof:
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | Let be a minimal normal subgroup of (note that always has a minimal normal subgroup). | is finite and nontrivial. | Since is nontrivial, the set of nontrivial normal subgroups of is non-empty (in particular, it includes ). Since is finite, this set must have a minimal element (not necessarily unique). Let be one such minimal normal subgroup. | ||
2 | is the characteristic closure of in , i.e., is generated by the set of all subgroups of the form . | is characteristically simple. | The group , i.e., the characteristic closure of in , is characteristic in by definition. It's nontrivial, because it contains the nontrivial normal subgroup . Combining with the fact that is characteristically simple gives that this must be all of . | ||
3 | Construct , subgroups of , all normal subgroups automorphic to that together generate , as follows. Set . Having constructed , check if they generate . If not, then set to be any subgroup of the form (for some automorphism of ) that is not contained in the subgroup . The process terminates in finitely many steps, because the subgroup generated keeps growing strictly at each stage, and is a finite group. Let be the total number of subgroups used. | is finite | Steps (1), (2) | Step (2) guarantees that if the subgroups so far do not generate , then at least some subgroup automorphic to has been missed from the collection, thus making this process valid. | |
4 | For any automorphism of and any normal subgroup of , either or is trivial. | Step (1) | Since is minimal normal, so is . Thus, by Fact (1), is normal. Since is minimal normal, either is trivial or (which means ). | ||
5 | Each join intersects trivially for . | Fact (2) | Steps (3), (4) | All the s are images of under automorphisms, hence are normal. Thus, by Fact (2), their join is also normal (and equals their product). Further, by Step (4), this normal subgroup must either intersect or contain . However, it cannot contain because of the way we construct . | |
6 | is an internal direct product of , which are all normal subgroups isomorphic to and hence to each other (in fact, automorphic to each other in ). | Steps (3), (5) | Steps (3) and (5) show that all the conditions for an internal direct product are satisfied. | ||
7 | is a direct factor of . | Steps (3), (6) | Recall that we had set in Step (3), and then discovered that it is part of an internal direct product, hence it is a direct factor. | ||
8 | is simple. | Fact (3) | Step (1) | Since is a direct factor of , Fact (3) tells us that any normal subgroup of is normal in . Sine is minimal normal, it cannot contain any strictly smaller nontrivial normal subgroups of , and hence, by the preceding sentence, it cannot contain any proper nontrivial normal subgroups of itself. Thus, is simple. |
Steps (6) and (8) together are what we wanted to prove.
Proof of (2) implies (1)
Note that this implication actually works even for infinite groups, but we will restrict our proof to finite internal direct products.
Given: A nontrivial group that is an internal direct product of pairwise isomorphic simple groups .
To prove: is characteristically simple.
Proof: We split in two cases. The simple groups are either all abelian (in which case is also abelian) or all non-abelian (in which case is also non-abelian).
Abelian case
In this case, all the s are isomorphic to a group of prime order for some prime number , and is an elementary abelian group of order . This group is clearly characteristically simple, because it is a group whose automorphism group is transitive on non-identity elements (on account of being the additive group of a vector space).
Non-abelian case
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | Each of the s is a perfect group. | Fact (4) | The s are simple and non-abelian. | given+fact-direct. | |
2 | For any normal subgroup of , is a subdirect product of some subset of the set of subgroups . | Fact (4) | Consider the set of s such that the projection from to has a non-identity image. Each image is normal in by fact (1). Since each is simple, each such image must be equal to . Thus, is a subdirect product of those s. | ||
3 | For any normal subgroup of , is the full subgroup generated by the s that figure in the subdirect product obtained in Step (2), i.e., it is the internal direct product of some subset of . | Fact (5) | Steps (1), (2) | Step-fact combination direct. | |
4 | The only characteristic subgroups of are and the trivial subgroup. | Step (3) | Fix a system of isomorphisms that identify all the direct factors . The symmetric group of degree embeds inside via its natural action by permutation of the direct factors since they are pairwise isomorphic. The only subsets of that are invariant under this action are the whole set and the empty subset. Thus, the only normal subgroups of that are invariant under all automorphisms induced by these permutations are the trivial subgroup and the whole group. Thus, is characteristically simple. |