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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.

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    $\begingroup$ Hint In particular, $a$ and $b$ are both $35$th roots of unity. $\endgroup$
    – Travis Willse
    Commented Oct 25, 2015 at 20:39
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    $\begingroup$ Saying $o(a) = 5$ is two statements : it means $a^5 = 1$ and that $a^k \ne 1$ for $0 < k < 5$. $\endgroup$
    – Joel Cohen
    Commented Oct 25, 2015 at 20:40

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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 $o(b)=4$.

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    $\begingroup$ If I had never seen such an argument before I might wonder why $(ab)^k=1 \Rightarrow a^k=1$. $\endgroup$
    – Nex
    Commented Oct 25, 2015 at 20:42
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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 $35$.

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