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 $A=\langle S\rangle$.

Definition 1.

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

Example 2.

The only non-generator of $\mathbb{Z}$, the group of integers under addition, is 0.

Proof.

Take $n\neq 0$. Without loss of generality, $n$ is positive. Then take $1 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 $1\in\langle m,n\rangle$ 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 $\mathbb{Z}_{4}$, the element $2$ is a non-generator.

Proof.

Check the possible generating sets directly. ∎

Example 4.

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

Proof.

Apply the same proof as done to in Example 2. ∎

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.

Proof.

For a group $G$, given a non-generator $a$, and $M$ any maximal subgroup of $G$. If $a$ is not in $M$ then $\langle M,a\rangle$ is a larger subgroup   than $M$. Thus $G=\langle M,a\rangle$. But $a$ is a non-generator so $G=\langle M\rangle=M$. 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 $G=\langle S,a\rangle$, then set $M=\langle S\rangle$. 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 $\langle S,a\rangle=G$. As $H$ is maximal, this is impossible. Hence $G=M$ and $a$ is a non-generator. ∎

Title Frattini subset FrattiniSubset 2013-03-22 16:30:56 2013-03-22 16:30:56 Algeboy (12884) Algeboy (12884) 7 Algeboy (12884) Definition msc 20D15 FrattiniSubgroup