TIFR2010-Maths-B-9

Question

Let $$G=\left \{ z \in \mathbb{C} \mid z^n = 1 \text{ for some positive integer } n \right \}$$. Then under multiplication of complex numbers,

• A. $G$ is a group of finite order
• B. $G$ is a group of infinite order, but every element of $G$ has finite order
• C. $G$ is a cyclic group
• D. None of the above

Answer 1

Order of a group is "number of elements in the set".

Order of an element $a$ of group is smallest positive integer $n$ such that $a^n=e$, where $e$ is identity element of group.

Now identity element of group given in question is $1$, because operator is multiplication, and $z*1=z$ for any complex number $z$ in the group.

So order of each element $z$ of group is smallest positive integer $n$ such that $z^n=1$, but in the question it is given that this property is satisfied by each $z$ for some positive integer $n$, hence order of each element is finite.

Order of given group is infinite i.e. there are infinite number of elements in group, because for each positive $n$, we can write $z=1^{1/n}$, which gives us $n$ $n^{th}$ roots of unity. Since number of choices for $n$ is infinite, therefore number of all roots will also be infinite, hence infinite number of elements in the group.

So option (b) is correct.

Comments

closure property satisfied: As all real numbers are also considered complex.

Associative property: Yes complex number multiplication satisfy associativity.

Identity: Yes, 1.

Inverse: $z^{n-1}$ where n is the order of the element.

Plus the above conclusion to select option ‘b’.