Equivalence of definitions of finite characteristically simple group

The definitions that we have to prove as equivalent
We need to show that the following are equivalent for a nontrivial finite group:


 * 1) It is a characteristically simple group, i.e., it has no proper nontrivial characteristic subgroup.
 * 2) It can be expressed as an internal direct product of pairwise isomorphic finite simple groups.

Facts used

 * 1) uses::Normality is strongly intersection-closed
 * 2) uses::Normality is strongly join-closed
 * 3) uses::Direct factor implies transitively normal
 * 4) uses::Normality satisfies image condition
 * 5) uses::Normal subdirect product of perfect groups equals direct product

Proof of (1) implies (2)
Given: A finite nontrivial group $$G$$ that has no proper nontrivial characteristic subgroups.

To prove: We can find pairwise isomorphic finite simple subgroups $$H_1,H_2,\dots,H_n$$ in $$G$$ such that $$G$$ is an internal direct product of $$H_1,H_2,\dots,H_n$$. The proof will also show, as a by-product, that every minimal normal subgroup is a direct factor.

Proof:

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 $$G$$ that is an internal direct product of pairwise isomorphic simple groups $$H_1,H_2,\dots,H_n$$.

To prove: $$G$$ is characteristically simple.

Proof: We split in two cases. The simple groups$$H_i$$ are either all abelian (in which case $$G$$ is also abelian) or all non-abelian (in which case $$G$$ is also non-abelian).

Abelian case
In this case, all the $$H_i$$s are isomorphic to a group of prime order for some prime number $$p$$, and $$G$$ is an elementary abelian group of order $$p^n$$. 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).