The set G=0, 1, 2. Define the composition + on G, for any a, b ∈ G by a + b = c where c is the least non-negative remainder obtained by dividing a + b by 3. Show that G forms an Abelian group under this composition.

For any $a, b \in G$, the result of the operation $a+b$ must also be in $G$.
The operation is defined as addition modulo 3. The possible sums $a+b$ for $a, b \in \{0, 1, 2\}$ range from $0+0=0$ to $2+2=4$.
When these sums are taken modulo 3:

• $0 \pmod 3 = 0 \in G$
• $1 \pmod 3 = 1 \in G$
• $2 \pmod 3 = 2 \in G$
• $3 \pmod 3 = 0 \in G$
• $4 \pmod 3 = 1 \in G$ Since all possible results are elements of $G$, the set $G$ is closed under the given operation.

For any $a, b, c \in G$, we must show that $(a+b)+c = a+(b+c)$.
Since the operation is addition modulo 3, and integer addition is associative, we have:

Therefore, $(a+b)+c = a+(b+c)$ holds for all $a, b, c \in G$.

An identity element $e \in G$ must exist such that for all $a \in G$, $a+e = a$ and $e+a = a$.
Let's test $e=0$:
For any $a \in G$, $a+0 = (a+0) \pmod 3 = a \pmod 3 = a$.
Similarly, $0+a = (0+a) \pmod 3 = a \pmod 3 = a$.
Since $0 \in G$, the identity element is 0.

For each element $a \in G$, there must exist an inverse element $a^{-1} \in G$ such that $a+a^{-1} = 0$ (the identity element).

• For $a=0$: $0+0 = 0 \pmod 3 = 0$. So, $0^{-1}=0$.
• For $a=1$: $1+2 = 3 \pmod 3 = 0$. So, $1^{-1}=2$.
• For $a=2$: $2+1 = 3 \pmod 3 = 0$. So, $2^{-1}=1$. Every element in $G$ has an inverse within $G$.

For any $a, b \in G$, we must show that $a+b = b+a$.
Since integer addition is commutative, $a+b = b+a$.
Therefore, $(a+b) \pmod 3 = (b+a) \pmod 3$.
Thus, $a+b = b+a$ holds for all $a, b \in G$.

Since all five properties (closure, associativity, identity, inverses, and commutativity) are satisfied, $G$ forms an Abelian group under the given composition.