595: For Group and Its Subgroup, Subgroup Is Normal Subgroup if Its Conjugate Subgroup by Each Element of Group Is Contained in It|T.B.P.

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description/proof of that for group and its subgroup, subgroup is normal subgroup if its conjugate subgroup by each element of group is contained in it

Topics

About: group

Starting Context

The reader knows a definition of normal subgroup.
The reader knows a definition of conjugate subgroup of subgroup by element.

Target Context

The reader will have a description and a proof of the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Structured Description

Here is the rules of Structured Description.

Entities:

G^{'}

\in {the groups}

G

\in {the subgroups of G^{'}}

//

Statements:

\forall p \in G^{'} ∖ G (p G p^{- 1} \subseteq G)

⟹

G \in {the normal subgroups of G^{'}}

2: Natural Language Description

For any group,

G^{'}

, and any subgroup,

G

, of

G^{'}

, if for each

p \in G^{'} ∖ G

p G p^{- 1} \subseteq G

, then,

G

is a normal subgroup of

G^{'}

3: Proof

Let us suppose that for each

p \in G^{'} ∖ G

p G p^{- 1} \subseteq G

.

Obviously, for each

p \in G

p G p^{- 1} \subseteq G

. So, for each

p \in G^{'}

p G p^{- 1} \subseteq G

.

For each

p \in G^{'}

G \subseteq p G p^{- 1}

? For any

p^{'} \in G

, let us take

p^{″} := p^{- 1} p^{'} p \in G

, because

p^{- 1}

can be taken in lieu of

p

p^{'} = p p^{″} p^{- 1} \in p G p^{- 1}

. So,

G \subseteq p G p^{- 1}