Non-associative: set with a binary operation, but has inverses and identity I've been thinking about an example of some set with a binary operation which would satisfy all axioms of groups except for associativity. I'm new to Group Theory, so I would appreciate your knowledgeable insight.
My example:
$S={\mathbb{C} - \{0\}}$; $H = (S, \text^)$, where ^ is exponentiation. Can we consider this as a correct example? The identity under ^ appears to be 1. The inverse appears to be $\frac{2i\pi}{\ln(x)}$.
 A: The object you are describing is called a loop. If you want a high powered example that fits in one mouthful, try the multiplicative group loop of non-zero octonions. As a more elementary alternative, I thought I'd give you a finite loop to ponder on (although you may need to work a bit to verify all the properties). I present Loop 8.1.4.0:
$$\begin{array}{c|c|c|c|c|c|c|c|}
\circ& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline
1 & 1 & 7 & 5 & 0 & 6 & 2 & 4 & 3\\\hline
2 & 2 & 6 & 7 & 5 & 0 & 3 & 1 & 4\\\hline
3 & 3 & 0 & 6 & 7 & 5 & 4 & 2 & 1\\\hline
4 & 4 & 5 & 0 & 6 & 7 & 1 & 3 & 2\\\hline
5 & 5 & 2 & 3 & 4 & 1 & 7 & 0 & 6\\\hline
6 & 6 & 4 & 1 & 2 & 3 & 0 & 7 & 5\\\hline
7 & 7 & 3 & 4 & 1 & 2 & 6 & 5 & 0\\\hline
\end{array}$$
Here $0$ is the identity, and the inverses are $0,3,4,1,2,6,5,7$ for $0\dots7$ respectively. Consider $1\circ 1\circ 2$ to show non-associativity and $1\circ 2$ for non-commutativity.

The above is actually an example of a Bol loop, which satisfies the more complicated weak associativity property $a(b(ac))=(a(ba))c$. For general loops, there are smaller examples; the smallest non-associative loop has order $5$ - here is one of them:
$$\begin{array}{c|c|c|c|c|}
\circ & 0 & 1 & 2 & 3 & 4\\\hline
0 & 0 & 1 & 2 & 3 & 4\\\hline
1 & 1 & 4 & 0 & 2 & 3\\\hline
2 & 2 & 3 & 4 & 1 & 0\\\hline
3 & 3 & 0 & 1 & 4 & 2\\\hline
4 & 4 & 2 & 3 & 0 & 1\\\hline
\end{array}$$
Note that this does not have two-sided inverses, since we have identities like $1\circ 2=0$ and $3\circ 1=0$, so that the left inverse of $1$ is $2$ and the right inverse is $3$. For non-associativity just consider $1\circ 1\circ 1$ (this loop is not even power-associative).
A: Consider the set $\{0,1,2\}$ with the binary operation $a*b = |a-b|$. It has an identity element (namely $0$) and every element has an inverse (namely itself) and it is not associative: $(1*1)*2 = 0*2 = 2$ but $1*(1*2) = 1*1 = 0$.
