Frattini subset

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

=

⟨

S

⟩

.

Definition 1

.

An

element

g

∈

A

is said to be a

non-generator

if given a subset

S

of

A

such that

A

=

⟨

S

∪

{

g

}

⟩

then in fact

A

=

⟨

S

⟩

. The set of all non-generators is called the

Frattini subset

of

A

.

Example 2

.

The only non-generator of

Z

, the group of

integers

under

addition

,
is 0.

Proof.

Take

n

≠

0

. Without loss of generality,

n

is

positive

. Then take

1

<

m

some integer relatively prime to

n

. Thus by the

Euclidean algorithm

we know there are integers

a

,

b

such that

1

=

a

⁢

m

+

b

⁢

n

.
This shows

1

∈

⟨

m

,

n

⟩

so in fact

{

m

,

n

}

generates

ℤ

.

However,

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

ℤ

is

0

.
∎

Example 3

.

In the ring

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,

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

⟨

M

,

a

⟩

is a larger

subgroup

than

M

. Thus

G

=

⟨

M

,

a

⟩

. But

a

is a non-generator so

G

=

⟨

M

⟩

=

M

. This contradicts the assumption that

M

is a maximal
subgroup and therefore

a

∈

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

=

⟨

S

,

a

⟩

, then set

M

=

⟨

S

⟩

. 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

⟨

S

,

a

⟩

=

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

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