# Frattini subset

Suppose $A$ is a set with a binary operation^{}. Then we say a subset $S$ of $A$
*generates* $A$ if finite iterations of products^{} (commonly called words over $S$) from the set $S$ eventually produce every element of $A$. We write
this property as $A=\u27e8S\u27e9$.

###### Definition 1.

An element $g\mathrm{\in}A$ is said to be a *non-generator* if given a subset $S$
of $A$ such that $A\mathrm{=}\mathrm{\u27e8}S\mathrm{\cup}\mathrm{\{}g\mathrm{\}}\mathrm{\u27e9}$ then in fact
$A\mathrm{=}\mathrm{\u27e8}S\mathrm{\u27e9}$. The set of all non-generators is called the
*Frattini subset* of $A$.

###### Proof.

Take $n\ne 0$. Without loss of generality, $n$ is positive. Then take
$$ some integer relatively prime to $n$. Thus by the
Euclidean algorithm^{} we know there are integers $a,b$ such that $1=am+bn$.
This shows $1\in \u27e8m,n\u27e9$ so in fact $\{m,n\}$ generates
$\mathbb{Z}$.

However, $\mathbb{Z}$ is not generated by $m$, as $m>1$. Therefore $n$ cannot be removed form the generating set $\{m,n\}$ and so indeed the only non-generator of $\mathbb{Z}$ is $0$. ∎

###### Example 3.

In the ring ${\mathrm{Z}}_{\mathrm{4}}$, the element $\mathrm{2}$ is a non-generator.

###### Proof.

Check the possible generating sets directly. ∎

###### Example 4.

The set of positive integers under addition, $\mathrm{N}$, has no non-generators.

###### Proof.

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.

###### Proposition 5.

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.

###### Theorem 6.

In a group $G$, the intersection^{} of all maximal subgroups is the
Frattini subset.

###### Proof.

For a group $G$, given a non-generator $a$, and $M$ any maximal subgroup
of $G$. If $a$ is not in $M$ then $\u27e8M,a\u27e9$ is a larger subgroup^{}
than $M$. Thus $G=\u27e8M,a\u27e9$. But $a$ is a non-generator so
$G=\u27e8M\u27e9=M$. This contradicts the assumption that $M$ is a maximal
subgroup and therefore $a\in M$. So the Frattini subset lies in every
maximal subgroup.

In contrast, if $a$ is in all maximal subgroups of $G$, then given any subset $S$ of $G$ for which $G=\u27e8S,a\u27e9$, then set $M=\u27e8S\u27e9$. If $M=G$, then $a$ is a non-generator. If not, then $M$ lies in some maximal subgroup $H$ of $G$. Since $a$ lies in all maximal subgroup, $a$ lies in $H$, and thus $H$ contains $\u27e8S,a\u27e9=G$. As $H$ is maximal, this is impossible. Hence $G=M$ and $a$ is a non-generator. ∎

Title | Frattini subset |
---|---|

Canonical name | FrattiniSubset |

Date of creation | 2013-03-22 16:30:56 |

Last modified on | 2013-03-22 16:30:56 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 7 |

Author | Algeboy (12884) |

Entry type | Definition |

Classification | msc 20D15 |

Related topic | FrattiniSubgroup |