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In her paper Periodic Decimal Fractions, Bennet uses the phrase "the exponent to which $10$ belongs modulo $p$" to mean "the least positive integer $k$ such that $10^k\equiv 1\pmod p$."

After looking around, it seems that this language is somewhat commonplace, and it can be found on documents by Wolfram and other papers. As a native English speaker, I have never encountered this language and it took me a while to figure out the author's intent.

  • Does anybody know where this language came from?
  • How can we explain the use of "belongs to" here?
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    This is very old language. Nowadays it is much more common to say that $10$ has (multiplicative) order $k$ in $\,\Bbb Z_{10},\,$ (i.e. modulo $10)$. This is consistent with language used in groups and monoids. – 2017-02-27
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    You mean 10 has (multiplicative) order k in $\mathbb{Z}_p$ (i.e. modulo $p$). Just to avoid any confusion – 2017-03-06
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    @mdave16 Yes. So my best guess is that $10$ "belongs to" a cyclic subgroup of size $k$. – 2017-03-06

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It comes from Gauss's Disquisitiones Arithmeticae, and his development of primitive roots.

In the section on primitive roots he proves, amongst others, two results:

  1. If $a^k\equiv 1 \pmod p$, then $k|(p-1)$

  2. The number of elements that (Gauss's word) 'belong' to $k$ is $\phi(k)$, with $\phi(k)$ being the totient function.

For example, using $p=19$ and primitive root $3$ we get the following table.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18\\ \hline 3&9&8&5&15&7&2&6&18&16&10&11&14&4&12&17&13&1\\ \hline \end{array}

The divisors of $p-1=18$ are $1,2,3,6,9,18$, with totient values $1,1,2,2,6,6$.

The order of an element is the lowest value of $k$ such that $a^k\equiv 1\pmod p$, which gives the following table:

\begin{array}{|c|c|c|} \hline k&\phi(k)&elements\\ \hline 1&1&18\\\hline 2&1&9\\\hline 3&2&7,11\\\hline 6&2&8,12\\\hline 9&6&4,5,6,9,16,17\\\hline 18&6&2,3,10,13,14,15\\ \hline \end{array}

At this point, Gauss says that, for example, '$16$ belongs to the exponent $9$, or '$8$ belongs to the exponent $6$'.

To clear up, $7$ belongs to exponent $3$ because $7^3\equiv 1\pmod{19}$. If we look up $7$ in the first table, we can see that $3^6\equiv 7\pmod{19}$, and then that $(3^6)^3=3^{18}\equiv 1\pmod{19}$.