'Determine the properties of an operation $\circ$ in the set $A:=\{-1,0,1\}$, such that $(A,\circ)$ forms a group'.
Let $0$ be the identity element of the group, now $1\circ0=1$, $-1\circ0=-1$,$0\circ0=0$.
If $(A,\circ)$ forms a group, there must be ( assuming $0$ is the identity element ) one and only one inverse element for each $a\in A$, such that $a \circ a^{-1} = 0$. Now $-1\circ1=0$.
Since the identity and inverse elements must be unique, we can eliminate other results and state that $-1\circ-1=1$ and $1\circ1=-1$. At this point, we can create a table with respect to $\circ$, showing that this is an Abelian group - and therefore a group. What are the flaws in my work? Also, how can I show associativity for $(A,\circ)$, other than proving all possible combinations?
You are not required to prove there is a unique group of order 3, just that there is one. You know that there is a cyclic group of order 3.
So assuming 0 is the identity element you define $-1\circ-1=1,-1\circ1=0,-1\circ0=-1$, $1\circ1=-1$ and $1\circ0=1$. If you think of $0$ as $(-1)^3$ and $1$ as $(-1)^2$, then it is obvious that the group must be abelian and verification of associativity is trivial.