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How to prove that a set is a group? related difficult/challenge quesyion.

by jessicaw
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jessicaw
#1
Oct19-10, 11:41 PM
P: 56
1.In proving that a set is a group from definition, we have to show there is an inverse element for each element, do we have to show uniqueness of it?
I believe this is unnecessary but people do this all the time.
My argument:
"For each [itex]a\in G[/itex], there exists a left inverse a' in G such that a'a=e." is enough.
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Simon_Tyler
#2
Oct20-10, 01:30 AM
P: 313
Provided you have shown the multiplication is associative then the inverse has to be unique.
jessicaw
#3
Oct20-10, 07:28 AM
P: 56
Quote Quote by Simon_Tyler View Post
Provided you have shown the multiplication is associative then the inverse has to be unique.
so in proving a set is a group it is redundant to show the uniqueness of inverse(just show existence is ok)?

Fredrik
#4
Oct20-10, 11:32 AM
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How to prove that a set is a group? related difficult/challenge quesyion.

It's only unnecessary if you have already proved that the axioms (xy)z=x(yz), ex=x and x-1x=e define a group. (The standard axioms are (xy)z=x(yz), ex=xe=x and x-1x=xx-1x=e). This is a bit tricky. Define a "left identity" to be an element e such that ex=x for all x, and a "left e-inverse" of x to be an element y such that yx=e. You can prove the theorem by proving all of the following, in this order.

1. There's at most one right identity.
2. If y is a left e-inverse to x, then x is a left e-inverse of y.
3. e is a right identity (which by #1 must be unique).
4. There's at most one left identity. (This means that e is an identity and is unique).
5. Every left-e inverse is a right e-inverse. (Hint: Use 2).
6. Every x has at most one left inverse. (This means that every element has a unique inverse).

If you've done this once, or if this is a theorem in your book, then all you have to do to verify that the structure you're considering is a group is to verify that the alternative axioms stated above are satisfied.
Landau
#5
Oct20-10, 03:14 PM
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It seems two different questions are asked:
(1) Is it enough to show that each element has an inverse, instead of in addition showing this inverse is unique?
(2) Is it enought to show that each element has a LEFT inverse, instead of showing that each element has both a right and a left inverse?

Or is it

(3) Is it enought to show that each element has a LEFT inverse, instead of showing that each element has both a right and a left inverse and that these are unique?
jessicaw
#6
Oct20-10, 11:01 PM
P: 56
Quote Quote by Landau View Post
It seems two different questions are asked:
(1) Is it enough to show that each element has an inverse, instead of in addition showing this inverse is unique?
(2) Is it enought to show that each element has a LEFT inverse, instead of showing that each element has both a right and a left inverse?

Or is it

(3) Is it enought to show that each element has a LEFT inverse, instead of showing that each element has both a right and a left inverse and that these are unique?
I use (2) as my argument for (1). I know that (2) is valid so (2) implys (1)?
Fredrik
#7
Oct21-10, 06:01 AM
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Quote Quote by jessicaw View Post
I use (2) as my argument for (1). I know that (2) is valid so (2) implys (1)?
Suppose that y and y' are left e-inverses of x, and z is a right e-inverse of x. Then y'x=yx, and if you multiply by z from the right, you get y'=y. Is that what you wanted to know? I'm not sure I understand what you're asking.

By the way, in both #1 and #6, you're referring to a mere statement as an "argument". It's not an argument if it doesn't include some evidence for the claim.


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