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I'm stuck on that question, because I'm trying to find a simple way to solve it.

I'm pretty sure there exists such elements because the group is cyclic and $|\mathbb{Z}_{31}^{*}| = 30$, hence there exists a sub-group of any order dividing $30$. So I know there is a sub-group of order $10$ and it's cyclic because any sub-group of a cyclic group is cyclic, and it's generator is an elements of order $10$.

Now, how can I proceed from here? How can one find the generator of this sub-group?

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    If $g$ is a generator for $\mathbb{Z}_{31}^\star$, then $g^3$ has order $10$. – 2017-02-28
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    You should write "So I **know** there is a subgroup of order $10 \ldots$". – 2017-02-28

2 Answers 2

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Let's see: $2^5 = 32 \equiv 1 \mod 31$ so that's no good. Ah, but then $(-2)^5 \equiv -1 \mod 31$, so $(-2)^{10} \equiv 1$, and the order of $-2$ is not $2$ or $5$...

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Choose a random element $b\neq 0$ and cube it $\,a = b^{\large 3}.\,$ Then $\,a^{\large 10} = b^{\large 30} = 1.\,$ By the Order Test below, it follows that $a$ has order $10$ iff $\,a^{\large 2}\neq 1\neq a^{\large 5}.\,$ Keep testing till you find one of order $10$.

Order Test $\,\ \,a\,$ has order $\,n\iff a^{ n} = 1\,$ but $\,a^{n/p} \not= 1\,$ for every prime $\,p\mid n.\,$

Proof $\ (\Leftarrow)\ $ If $\,a\,$ has $\,\rm\color{#c00}{order\ k}\,$ then $\,k\mid n.\,$ If $\,k < n\,$ then $\,k\,$ is proper divisor of $\,n\,$ therefore $\,k\,$ must omit at least one prime $\,p\,$ from the unique prime factorization of $\,n,\,$ hence $\,k\mid n/p,\,$ say $\, kj = n/p,\,$ so $\,a^{n/p} = (\color{#c00}{a^k})^j= \color{#c00}1^j= 1,\,$ contra hypothesis. $\ (\Rightarrow)\ $ By definition of order.

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    Thanks, that exactly what I was looking for. – 2017-03-01
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