NIELIT 2022 April Scientist B | Section B | Question: 65

Question

Let $\text{G}$ be a multiplicative group and $a \in \text{G.}$ If the order of $a$ is $6,$ then the order of $a^{5}$ is equal to :

• A. $1$
• B. $5$
• C. $6$
• D. $30$

Answer 1

Theorem:- Let ‘a’ be an element in a group such that order of a, ord(a)=n. If ‘p’ is a positive integer then,

$ord(a^{p})=\frac{n}{\text{GCD}(n,p)}$

Here,

$ord(a)=6$

$p=5$

so, $\text{GCD}(n,p)=\text{GCD}(5,6)=1$

$\implies ord(a^{5})=\frac{6}{\text{GCD}(5,6)}=6$

So, correct answer is option (C).

Ref:-https://gateoverflow.in/341566/Gate-overflow-discrete-mathematics-3-1

Answer 2

We know –

Order of an element of a group = Order of subgroup generated by that element.

Subgroup generated by an element "a" of a group = Subgroup generated by "$a^{-1}$".

Now, Order of a = 6 --- given.

Thus, $a^6$ = e (identity element).

Left multiplying by $a^{-1}$ on both sides.

Thus, a$^5$ = $a^{-1}$.

Thus, subgroup generated by a = subgroup generated by $a^5$.

Thus, Order of a = Order of $a^5$ = 6.

Answer :- C. 6