# Group Axioms

• Mar 9th 2011, 05:19 AM
Arron
Group 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 AM
Swlabr
Quote:

Originally Posted by Arron
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.

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 $x \in \mathbb{Z}$, and so e is not fixed (and so no identity exists). Alternatively, you could note that $e\circ y = e+y=y$ 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 AM
TheChaz
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 AM
Swlabr
Quote:

Originally Posted by TheChaz
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

No-what he has said here is fine, as e*x=x because e is the identity (by assumption).
• Mar 9th 2011, 06:07 AM
poirot
Quote:

Originally Posted by Arron
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.

0 is an integer btw. Your argument could be [LaTeX ERROR: Convert failed] so $$e=-2x$$ but this is not fixed (dependant on x) which the identity must be therefore no identity element so not a group.
• Mar 9th 2011, 06:07 AM
TheChaz
(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 AM
Swlabr
Quote:

Originally Posted by TheChaz
(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

Ah-touche-I didn't spot that error!
• Mar 9th 2011, 06:15 AM
Arron