Suppose is a set with a binary operation. Then we say a subset of generates if finite iterations of products (commonly called words over ) from the set eventually produce every element of . We write this property as .
However, is not generated by , as . Therefore cannot be removed form the generating set and so indeed the only non-generator of is . ∎
In the ring , the element is a non-generator.
Check the possible generating sets directly. ∎
The set of positive integers under addition, , has no non-generators.
Apply the same proof as done to in Example 2. ∎
So we see not all sets with binary operations have non-generators. In the case that a binary operation has an identity then the identity always serves as a non-generator due to the convention that the empty word be defined as the identity. However, without further assumptions on the product, such as associativity, it is not always possible to treat the Frattini subset as a subobject in the category of the orignal object. For example, we have just shown that the Frattini subset of a semi-group need not be a semi-group.
In the category of groups, the Frattini subset is a fully invariant subgroup.
To prove this, we prove the following strong re-characterize the Frattini subset.
For a group , given a non-generator , and any maximal subgroup of . If is not in then is a larger subgroup than . Thus . But is a non-generator so . This contradicts the assumption that is a maximal subgroup and therefore . So the Frattini subset lies in every maximal subgroup.
In contrast, if is in all maximal subgroups of , then given any subset of for which , then set . If , then is a non-generator. If not, then lies in some maximal subgroup of . Since lies in all maximal subgroup, lies in , and thus contains . As is maximal, this is impossible. Hence and is a non-generator. ∎