Q113 / 180 · COMEDK 2013 · Mathematics

Which of the following is true?

  • Correct answer
    A
    The set of all fourth roots of unity is a multiplicative group
  • B
    The set of all cube roots of unity is an additive group
  • C
    $(a b)^{-1}=a^{-1} b^{-1}$ for all $a, b$ in any group $G$
  • D
    If $(a b)^{2}=a^{2} b^{2}$ for all $a, b$ in any group $G$, then the group $G$ is non-abelian

Solution

(a) Set of all fourth roots of unity is $$ \begin{aligned} &a=\{1,-1, i,-i\} \\ &\text { Now, } 1 \times(-1)=-1 \in G \\ &\quad 1 \times i=i \in G \\ &1 \times(-i)=-i \in G \\ &(-1) \times i=-i \in G \\ &(-1) \times(-i)=i \in G \\ &i \times(-i)=-i^{2}=1 \in G \end{aligned} $$ So, it is true. (b) Set of all cube root of unit is $$ H=\left\{1, w, w^{2}\right\} $$ Now, $\quad 1+w=-w^{2} \notin H$ So, it is false. (c) It is not true, as this is possible only, when $a$ is an abelian group, i.e. $a b=b a$ (d) We have, $$ \begin{aligned} (a b)^{2} &=a^{2} b^{2} \\ \Rightarrow \quad(a b)(a b) &=(a a)(b b) \\ \Rightarrow \quad(b)(a) &=(a)(b) \Rightarrow b a=a b \end{aligned} $$ $\Rightarrow a$ is an abelian group. So, it is false.

Hence, the correct answer is option: 'A'.

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