Question

Let G={3^n;n element of Z} prove that (G,•)
is an abelian group

Answer 1

Given $G=\{3^n : n \ element \ of \ \Z\}$ .

Closure: Let any $a\in G, b\in G$ , so $a = 3^n, b = 3^m$ where $n \in \Z, m\in \Z$ .

Now $a \times b = 3^n \times 3^m = 3^{n m}$ where $n m \in \Z$ $\implies ab \in G$ .

Associative: Let $a, b, c \in G$ , so $a = 3^n, b = 3^m, c = 3^k$ ,

hence $(a \times b) \times c = (3^{nm}) \times 3^k = 3^{nmk} = 3^n \times (3^m \times 3^k) = a \times (b\times c)$ .

Identity: For all $a \in G$ , $a = 3^m$ where $m \in \Z$ .

There exists $e = 1 = 3^0$ where $0 \in \Z$ , such that $a \times e = e\times a = a$ .

Inverse: For all $a \in G$ , $a = 3^m$ where $m \in \Z$ .

There exists $b = 3^{-m} : a\times b = b\times a = 1$ .

Commutative: $a \times b = 3^{nm} = b \times a \ \ \forall \ a, b \in G$ .