Group Theory

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neonash7777
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Group Theory

Post by neonash7777 » Wed Jan 31, 2007 3:11 am

Okay I have a question about group theory.
To be a group these must be true:
Closure
Associativity
Existence of Identity
Existence of Inverse
~~~~~
Then if it's also commutative, it's an abelian group.

Which this has all been fine for me, as the majority of all groups I've worked with have been...but there in lies my problem.

What if it isn't commutative. To what extent must the identity exist?
We have always said that
x • Identity = x

But must there also be (Identity • x = x) for it to be a group?
Do we need the identity to work both ways for it to be an identity? Or do we need only one of the two ways to work? Or do we need the first way I stated to work and the second way is insignificant?
Also if there is an identity both ways, do they have to be the same for it to be a true identity?

These same questions hold true an inverse. What if we find two inverses depending on if the inverse comes first or second. Or worse, what if we found two identities, and then 2 inverses for each of those giving us 4 inverses?![/i]
Phi, it's a whole "h" of a lot cooler than pi!

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neonash7777
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Post by neonash7777 » Thu Feb 01, 2007 1:58 am

Ok I found a little more info but still not complete.

It came to my attention that it would be impossible to have 2 Identites as what would
IDa • ID b =??
It couldn't equal both ID's unless they were the same.
Phi, it's a whole "h" of a lot cooler than pi!

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Alvaro
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Post by Alvaro » Thu Feb 01, 2007 2:40 am

I haven't done this kind of thing in more than 10 years, but I think you define inverse as something that is the inverse on both sides. As for the neutral element, you can prove that if the group has a neutral element on one side, it is also an element on the other side.

Confundus
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Post by Confundus » Tue Feb 13, 2007 9:34 pm

As for the identities, the identity in a group G is *defined* as the element 1 such that 1*x=x*1=x, for every x[isin]G
i guess you could work with just left-(or right-) handed identities.
consider this:
1*x=x
x*(1*x)=x*x after left-multiplying
(x*1)*x=x*x by associativity
x*1=x after right-multiplying the inverse of x

so to have only left-handed inverses, you'd have to remove some nice properties, like associativity or right-handed inverses.

however, that would be losing pretty much every nice theorem and property of a group, like lagranges theorem for example.

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daniel.is.fischer
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Re:

Post by daniel.is.fischer » Mon Sep 03, 2007 4:41 pm

Confundus wrote:As for the identities, the identity in a group G is *defined* as the element 1 such that 1*x=x*1=x, for every x[isin]G
i guess you could work with just left-(or right-) handed identities.
consider this:
1*x=x
x*(1*x)=x*x after left-multiplying
(x*1)*x=x*x by associativity
x*1=x after right-multiplying the inverse of x

so to have only left-handed inverses, you'd have to remove some nice properties, like associativity or right-handed inverses.

however, that would be losing pretty much every nice theorem and property of a group, like lagranges theorem for example.
Theorem: If you have an associative composition on a set, and this composition has a left unit, e (i.e. forall x: e*x = x), and each element has a left inverse (forall x. exists y: y*x = e), then this composition makes the set a group.

Of course, you might replace left by right.

Exercise: find a proof, but not in an algebra textbook.
Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.

hyperdex
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Re: Group Theory

Post by hyperdex » Tue Sep 04, 2007 4:40 pm

It's actually easy to prove that if a set has both a left identity and a right identity, then they are equal. Let L and R be the left and right identities, and consider LR. LR=R since L is a left identity, and LR=L since R is a right identity. Hence L=R.

Once you have the existence of a unique identity, showing that left and right inverses are equal is also easy. G be a group with identity I, and let X be an element with left and right inverses L and R, that is, I=LX=XR. Then

L = LI = L(XR) = (LX)R = IR = R

Cheers,

Dave

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