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A part of Aluffi's "Algebra: Chapter 0" exercise II.4.12 suggests computing the order of $[9]_{31}$ in $(\mathbb{Z}/31\mathbb{Z})^*$. Sure, I could just multiply $9$ a few times until I get $1$ as a remainder (and thus derive that the order in question is 15), but is there a better way?

A few thoughts of mine:

  • Firstly, $[9] = [3]^2$, so it'd be sufficient to prove that $[3]$ is a generator (and indeed it is). But I was unable to do this efficiently.
  • Another attack direction is that, since $31$ is prime, one might note that $(\mathbb{Z}/31\mathbb{Z})^*$ is cyclic and, having $30$ elements, is isomorphic to $\mathbb{Z}/30\mathbb{Z}$. Maybe we could derive something meaningful by inspecting some isomorphism $\varphi$ between the two? I tried deriving what should it do to the elements of $(\mathbb{Z}/31\mathbb{Z})^*$, and I was able to figure out how it behaves on the powers of $[2]$, but it didn't bring me closer to understanding what it does to $[3]$ or $[9]$.

So shall I just accept my fate and consider this to be an exercise in multiplication and division with remainder?

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  • $\begingroup$ It might help to consider $$\operatorname{Aut}(\Bbb Z_{31})\cong (\Bbb Z/31\Bbb Z)^*.$$ Just as another perspective . . . $\endgroup$ – Shaun Apr 21 at 22:10
  • $\begingroup$ Could you please expand on that? I tried myself and it wasn't particularly fruitful for me, but anything involving morphisms looks promising and elegant! $\endgroup$ – 0xd34df00d Apr 22 at 0:04
  • $\begingroup$ The idea wasn't fully fledged, @0xd34df00d, I'm afraid; it would just have been the first place I'd look. Perhaps if you examine a proof of the isomorphism, something'd pop out. Sorry :) $\endgroup$ – Shaun Apr 22 at 5:39
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By lil' Fermat and Lagrange's theorem, all non-zero elements in $\mathbf Z/31\mathbf Z$ have order a divisor of $30$. So the order of $9$ is among $\;\{2, 3,5,6,10,15,30\}$.

It is not very long to check that, $\bmod 31$, \begin{gather}9^2\equiv -12, \quad 9^3\equiv -12\cdot 9=-108\equiv 16,\quad 9^5\equiv -12\cdot 16=-192\equiv -6,\\ 9^6\equiv-6\cdot 9=-54\equiv8, \quad 9^{10}\equiv 36\equiv 5,\quad 9^{15}\equiv 5\cdot -6=-30\equiv 1, \end{gather} so $9$ has order $15$.

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  • $\begingroup$ Since $9=3^2$ and since $3$ must have one of the orders you listed, $9$ must have order $3$ or $5$ or $15$. So you only need to do half of the checking in this answer. $\endgroup$ – Andreas Blass Apr 21 at 23:08
  • $\begingroup$ Sure, but anyway, most of these computations are useful steps to compute the other powers. $\endgroup$ – Bernard Apr 21 at 23:18
  • $\begingroup$ Agreed, but there are other ways to do the computations. Here's how I did them (I'm not claiming it's better, just different). Modulo 31, we have $3^3=27\equiv-4$, so $9^3\equiv(-4)^2=16$. And $3^5=243\equiv-5$ so $9^5\equiv(-5)^2=25\equiv-6$. In other words, I just kept taking advantage of the facts that $9=3^2$ and that I know some powers of $3$. (And I got lucky in that $3^5=243$ is very close to an obvious multiple $248$ of $31$.) $\endgroup$ – Andreas Blass Apr 21 at 23:29
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This is a variant on Bernard's answer, mainly to show one way to reduce the amount of computation (which isn't a lot to begin with) needed to identify the order.

As noted in Bernard's answer, the only possible orders for the nonzero elements of $\mathbb{Z}/31\mathbb{Z}$ are the divisors of $30$, i.e., $1$, $2$, $3$, $5$, $6$, $10$, $15$, and $30$. Since $31\equiv3$ mod $4$, $-1$ is not a square mod $31$. Therefore, since $9$ is a square, its order cannot be even. (If $9^{2n}\equiv1$ mod $31$, then $9^n\equiv\pm1$ mod $31$.) So in order to conclude that its order is $15$, it suffices to rule out $3$ and $5$ as orders. (The only element of order $1$ is $1$.)

Note that $9\cdot7=63\equiv1$ mod $31$. Now $9^2=81\equiv-12$ mod $31$, so $9^4\equiv144\equiv20$. Since $20$ is neither $9$ nor $7$ mod $31$, we can conclude that neither $9^3$ nor $9^5$ is $1$ mod $31$.

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  • $\begingroup$ @OP "...it suffices to rule out $3,5$" is a special case of the Order Test. $\endgroup$ – Bill Dubuque Apr 22 at 2:23

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