I have this math problem:
i) Suppose that $a$ and $b$ are roots of unity. Suppose that $o(a)=5$ and $o(b)=7$. Prove that $o(ab)=35$.
ii) Give an example such that $a$ and $b$ are roots of unity, but $o(ab) \ne o(a) \cdot o(b)$.
I'm not 100% how to start this problem. I know that $o(a)=5$ means that $a^5=1$. I also know that $o(b)=7$ means that $b^7=1$. I also know that $a^{5m} = 1$ and $b^{7j}=1$. But I'm not sure how to answer these questions. Thanks.
On
As the orders of $a$ and $b$ are coprime, Lagrange's theorem implies $\langle\, a\,\rangle\cap\langle\, b\,\rangle=\{1\}$.
Now if $(ab)^k=a^kb^k=1$, $ak=b^{-k}=1$.
There results that $k$ is a multiple of both $5$ and $7$, i.e. of $35$. As $(ab)^{35}=(a^5)^7(b^7)^5=1$, this proves the order of $ab$ is equal to $35$.
For $i)$ it is easy to see that $(ab)^{35}=(a^5)^7(b^7)^5=1$, and it remains to show that no lower power suffices. Suppose $(ab)^k=1$ where $1\le k\le 34$. Then $a^k=1$ so $k$ is a multiple of $5$. Similarly, $k$ is a multiple of $7$. Thus $k$ is at least $lcm(5,7)=35$.
For $ii)$ consider $o(a)=2$ and $o(b)=4$.