Abstract Algebra: Groups

Printable View

Feb 11th 2008, 01:34 AM
Jhevon

Abstract Algebra: Groups

(Hi) honorable mathematicians,

I was asked to verify that the set $H = \{ f : f: \mathbb{R} \mapsto \mathbb{R} \mbox{ and } f(x) \ne 0 \mbox{ for all } x \in \mathbb{R} \}$ is a group with respect to multiplication. i did that, no problem. but then the question asks, "How does this group differ from the group of invertible (real) mappings?"

i have no idea what this question wants from me... :confused:

Thanks
Feb 11th 2008, 02:17 AM
CaptainBlack

Quote:

Originally Posted by Jhevon

(Hi) honorable mathematicians,

I was asked to verify that the set $H = \{ f : f: \mathbb{R} \mapsto \mathbb{R} \mbox{ and } f(x) \ne 0 \mbox{ for all } x \in \mathbb{R} \}$ is a group with respect to multiplication. i did that, no problem. but then the question asks, "How does this group differ from the group of invertible (real) mappings?"

i have no idea what this question wants from me... :confused:

Thanks

Both the set and the group operation are different.

The first group includes things like f(x)=sin(x)+2, which is not invertible.
While the second includes f(x)=x, which is not in the set of the first group.

Also the group operation for the group of invertible real mappings is (usualy
though you have not said so) function composition, that is (f o g)(x)=f(g(x)).

RonL
Feb 11th 2008, 04:57 AM
kalagota

A question: does the "invertible" there the same as the existence of the inverse of f?
Feb 11th 2008, 05:56 AM
CaptainBlack

Quote:

Originally Posted by CaptainBlack

Both the set and the group operation are different.

The first group includes things like f(x)=sin(x)+2, which is not invertible.
While the second includes f(x)=x, which is not in the set of the first group.

Also the group operation for the group of invertible real mappings is (usualy
though you have not said so) function composition, that is (f o g)(x)=f(g(x)).

RonL

Oh yes nearly forgot, the first is Abelian (commutative) and the second is not.

(and so cannot be isomorphic)

RonL
Feb 11th 2008, 06:00 AM
ThePerfectHacker

Quote:

Originally Posted by CaptainBlank

Abelian

For some strange reason nobody ever capitalizes and instead writes abelian. All other algebraic and not even algebraic concepts are always capitalized: Noetherian, Artinian, Lagrangian mechanics, Newtonian fluid, .... I assume it is an error that has remained through the centuries.
Feb 11th 2008, 06:20 AM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

For some strange reason nobody ever capitalizes and instead writes abelian. All other algebraic and not even algebraic concepts are always capitalized: Noetherian, Artinian, Lagrangian mechanics, Newtonian fluid, .... I assume it is an error that has remained through the centuries.

I must say I did not think twice about it, it's named after a person and therefore should be capitalised.

RonL
Feb 11th 2008, 09:39 AM
Jhevon

Quote:

Originally Posted by kalagota

A question: does the "invertible" there the same as the existence of the inverse of f?

yes, i believe so. we defined the inverse of an element a to be an element b such that:

a*b = b*a = identity element

we call b a^{-1}

a similar notion is applied to mappings i suppose...