Hi!

Just want to clarify that the following is not a group because G2 fails

(Z,o), where x o y = 3x+y

G2 identity

e o x = 3x + e = x

that is 2x + e = 0

Take e = 0, then 0 is not an element of Z and so this is not a group.

Thanks.

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- Mar 9th 2011, 05:19 AMArronGroup Axioms
Hi!

Just want to clarify that the following is not a group because G2 fails

(Z,o), where x o y = 3x+y

G2 identity

e o x = 3x + e = x

that is 2x + e = 0

Take e = 0, then 0 is not an element of Z and so this is not a group.

Thanks. - Mar 9th 2011, 05:35 AMSwlabr
You are correct that the problem is lack of inverse, however I think your argument is slightly flawed.

0 is not necessarily your identity. I mean, take Z under the operation a.b = a+b-1. Then 1 is your identity.

Here, you have that e=-2x for all $\displaystyle x \in \mathbb{Z}$, and so e is not fixed (and so no identity exists). Alternatively, you could note that $\displaystyle e\circ y = e+y=y$ so e=0, and so then you can use the argument you used above. But my point is that you need to prove that e=0, you can't just assume it is. - Mar 9th 2011, 05:38 AMTheChaz
There are a few things worth mentioning... (I'll use "*" for the operation, FYI)

When you say:

"e * x = 3x + e = x",

you haven't correctly used the definition.

e * x := 3e + x

We want this to equal x, so e must be 0, which IS in Z! No contradiction (yet).

But if e = 0, then

x * e = 3x + e = 3x + 0 = 3x =/= x!

THAT is the problem! - Mar 9th 2011, 05:45 AMSwlabr
- Mar 9th 2011, 06:07 AMpoirot
- Mar 9th 2011, 06:07 AMTheChaz
(I know what e is...)

e*x = x explains the second "=", not the first.

The definition of a*b is not symmetric!

His statement:

e * x = 3x + e = x

My revision:

e * x = 3e + x = x - Mar 9th 2011, 06:10 AMSwlabr
- Mar 9th 2011, 06:15 AMArron
Thanks for your help everyone.