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Let $R$ be a commutative ring and $R^{\times}$ its multiplicative group.

Let $H \leqslant R^{\times}$ and $H' \subset \Bbb{Z}$ be the set of exponents $k$ such that the solutions to $X^k = 1$ form the subgroup $H$ of $R^{\times}$. $H'$ is a subgroup of $\Bbb{Z}$: $a^{k} = a^{\ell} = 1$ for all $a \in H \implies a^{k-\ell} = 1 $ for all $a \in H$.

I am interested a sufficient condition for there to exist a nontrivial $H$ induced by some $n \in \Bbb{Z}$. Would this have to do with divisors of the group or ring orders involved?

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    $H'$ is not a subgroup of $\Bbb Z$. If you add $k$ and $-k$ together you get $0$, and the solutions to $X^0=1$ may be bigger than $H$. Also, if you add $k$ to itself you get $2k$, and the equation $X^{2k}=1$ may have more solutions than $X^k=1$. – 2017-01-07
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    @arctictern Ah, thanks. Means I need to rethink this. – 2017-01-07
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    You said $H'$ is the set of all $k$ such that $X^k=1$ if *and only if* $X\in H$. Thus given $k,\ell\in H'$, in order to secure $k-\ell\in H'$ you would need to conclude $a^{k-\ell}=1$ if *and only if* $a\in H$. – 2017-01-07
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    https://en.wikipedia.org/wiki/Discrete_logarithm – 2017-01-07

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