Statement: If a and b are primitive nth and mth roots of unity then ab is a kth root of unity for some interger k.
I prove it as follows- a^n=1 and b^m=1 (By defn of primitive root of unity) Implies ab^(mn)=1 for mn=k and clearly k is the integer.
I want to know, whether my approach is right or not? Also, what is the smallest value of k? I don't have any clue how to find smallest k?
Please help?Thanks.
Your approach is fine.
In general, if $m$ and $n$ are relatively prime, then $mn$ is the smallest such $k$. (You can deduce this from the structure theorem for finite abelian groups, for example, or even just from the theorem that $|MN|=|M||N|/|M\cap N|$ for subgroups of abelian groups, where $M$ is the powers of $a$ and $N$ is the powers of $b$, and $M\cap N = \{1\}$ since any element has order a divisor of both $m$ and $n$.) But if $m$ and $n$ are not relatively prime, they don't uniquely determine the minimal $k$--it will depend on the specific $a$ and $b$. For example, let $\zeta$ be a primitive $3^{rd}$ root of unity. Then so is $\zeta^2$. But $\zeta\cdot\zeta$ is a primitive $3^{rd}$ root of unity whereas $\zeta\cdot\zeta^2$ is a primitive $1^{st}$ root of unity.