Step 1:

$G=\{2^n/n\in z\}$ to show that $(G,.)$ is an abelian group.

Step 2:

Closure :

Let $2^n,2^m\in G(n,m\in z)$

$2^n.2^m=2^{n+m}\in G$(Since $n+m \in z)$

Closure is satisfied.

Step 3:

Associativity :

The elements of $G$ are all integers.Multiplication of integers is associative property is satisfied.

Step 4:

Existence of identity :

Consider $2^n,2^0\in G(n,0\in z)$

$2^n.2^0=2^0.2^n=2^n$

$\therefore 2^0=1$ is the identity element.

Step 5:

Existence of inverse :

Consider $2^n\in G$

$\therefore n\in z\Rightarrow -n\in z$

Consider $2^n.2^{-n}=2^{-n}.2^n=2^0=1$

$\therefore 2^{-n}$ is the inverse of $2^n$

Every element in $G$ has an inverse.

Step 6:

Commutative property :

The elements of $G$ are integers and multiplication of integers is commutative.$\therefore$ commutative property is satisfied.

The four group axioms being satisfied and the commutative property being satisfied,$(6,.)$ is an abelian group.