Proving K as a Subgroup of G: A Comprehensive Guide

In group theory, one of the fundamental concepts is that of subgroups. A subset ( K ) of a group ( G ) is considered a subgroup if it satisfies three main criteria: closure under the group operation, the presence of the identity element, and the existence of an inverse for each element within ( K ). This article delves into the proof that ( K ) is a subgroup of ( G ) by demonstrating these criteria.

The Criteria for a Subgroup

For ( K ) to be a subgroup of ( G ), it must satisfy the following conditions:

closed under the group operation: for any ( f, g in K ), the product ( f circ g ) is also in ( K ). contain the identity element of ( G ): the identity element of ( G ) must also be in ( K ). contain inverses of all its elements: the inverse of any element in ( K ) must also be in ( K ).

Proving ( K ) is Closed Under Composition

The first criterion deals with closure under the group operation. We need to show that if ( f ) and ( g ) are elements of ( K ), then their composition ( f circ g ) is also in ( K ).

Given ( f, g in K ), we have that ( f(0) 0 ) and ( g(0) 0 ). The composition of ( f ) and ( g ), denoted as ( f circ g ), can be calculated as follows:

f circ g (0)  f(g(0))  f(0)  0

Hence, ( f circ g ) also fixes 0, implying that ( f circ g in K ). This establishes that ( K ) is closed under composition.

Proving the Identity of ( G ) Belongs to ( K )

The second criterion requires that the identity element of ( G ) must be in ( K ). The identity element, denoted as ( e ), is defined as a function that leaves every element of the group unchanged. In the context of our problem, the identity function ( e ) for ( G ) is the function that maps every element to itself and, in particular, ( e(0) 0 ).

Since ( e(0) 0 ), it follows that the identity function ( e ) of ( G ) is in ( K ). This ensures that ( K ) contains the identity element of ( G ).

Proving the Inverse of Each Element in ( K ) is in ( K )

The final criterion is to verify that the inverse of every element in ( K ) is also in ( K ). Consider an element ( f in K ), which satisfies ( f(0) 0 ). The inverse of ( f ), denoted as ( f^{-1} ), is the function such that ( f circ f^{-1} e ) and ( f^{-1} circ f e ), where ( e ) is the identity element of ( G ).

By the definition of ( f ), we have:

f^{-1}(0)  0

This shows that the inverse function ( f^{-1} ) also fixes 0, implying that ( f^{-1} in K ). Therefore, ( K ) contains the inverse of every element in ( K ).

Conclusion

We have demonstrated that ( K ) is closed under the group operation, contains the identity element of ( G ), and contains the inverse of every element in ( K ). These three conditions are the defining requirements for ( K ) to be considered a subgroup of ( G ). Thus, ( K ) is indeed a subgroup of ( G ).

Further Reading and References

For a deeper understanding of group theory and subgroups, you may refer to the following resources:

Wikipedia: Subgroup High-School Abstract Algebra by J.A. Gallian (Chapter on Subgroups) Abstract Algebra by Dummit and Foote (Chapter on Subgroups)

These sources provide a more detailed exploration of subgroup properties and their applications.