The problem given in the image defines a set $G$ as follows:

$G = \{ a \in \mathbb{Z} \mid a \text{ is a multiple of 2, or } a \text{ is a multiple of 3} \}.$

The question asks: Is $G$ a group under the usual addition? If yes, you are to prove your answer.

Solution Outline

To determine if $G$ is a group under addition, we must check whether $G$ satisfies the four group axioms:

1. Closure: If $a, b \in G$, then $a + b \in G$.
2. Associativity: Addition in $\mathbb{Z}$ is associative, so this property will automatically hold.
3. Identity: There exists an element $0 \in G$ such that $a + 0 = a$ for all $a \in G$.
4. Inverses: For every $a \in G$, there exists an element $-a \in G$ such that $a + (-a) = 0$.

Let me know if you'd like a detailed, step-by-step proof for each property.

Do you have any questions, or would you like further clarification on any part?

Further Questions

1. What would change if $G$ were defined under multiplication instead of addition?
2. Can we find an example of an element in $G$ and check if it has an inverse in $G$?
3. How does closure under addition depend on the multiples of 2 and 3?
4. What would happen if we restricted $G$ to only multiples of 2?
5. Can you list some examples of elements in $G$?

Tip

When proving closure for subsets defined by conditions like "multiple of 2 or 3," consider testing sums of examples to see if the property holds.