593: Finite Product of Subgroups Is Associative|T.B.P.

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description/proof of that finite product of subgroups is associative

Topics

About: group

The table of contents of this article

Starting Context
Target Context
Orientation
Main Body
1: Structured Description
2: Natural Language Description
3: Note
4: Proof

Starting Context

The reader knows a definition of group.

Target Context

The reader will have a description and a proof of the proposition that for any group, the product of any finite number of subgroups is associative.

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_{1}, . . ., G_{n}}

\subseteq {the subgroups of G}

S

= G_{1} . . . G_{n}

with any association

S^{'}

= G_{1} . . . G_{n}

with any association
//

Statements:

S = S^{'}

2: Natural Language Description

For any group,

G

, and any subgroups,

G_{1}, . . ., G_{n}

, of

G

S = G_{1} . . . G_{n}

with any association is

S^{'} = G_{1} . . . G_{n}

with any association.

3: Note

We are not claiming that

S

S^{'}

is a group;

S = S^{'}

is just set-wise.

4: Proof

For each

p \in S

p = g_{1} . . . g_{n}

with the corresponding association. But as multiplications in

G

is associative, that multiplication can be done in the association that corresponds to that of

S^{'}

, and the result is on

S^{'}

. So,

S \subseteq S^{'}

. Likewise,

S^{'} \subseteq S