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Show that the set $ G = \{2^n / n \in Z\} $ is an abelian group under multiplication.

1 Answer

  • A non-empty set G, together with an operation $*$ i.e., $(G, *)$ is said to be a group if it satisfies the following axioms
  • (1) Closure axiom : $a,b\in G\Rightarrow a*b\in G$
  • (2) Associative axiom : $\forall a, b, c \in G, (a * b) * c = a * (b * c)$
  • (3) Identity axiom : There exists an element $e \in G$ such that $a * e = e * a = a, \forall a ^{-1} G.$
  • (4) Inverse axiom : $\forall a \in G$ there exists an element $a^{-1}\in G$ such that $a^{-1}*a=a*a^{-1}=e.$
  • e is called the identity element of $G$ and $a^{-1}$ is called the inverse of a in $G.$
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 $2^n,2^0\in G(n,0\in z)$
$\therefore 2^0=1$ 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.
answered Sep 16, 2013 by sreemathi.v

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