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 $\displaystyle x \in \mathbb{Z}$, 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.
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!