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Take the set of integers modulo $n$, written $\Bbb{Z}/(n)$ and any prime $p$ and consider the multiplicative subgroup specified by $H^{\times} = \{ x^p = 1: x \in \Bbb{Z}/(n)^{\times}\}$, then $|H^{\times}| = p^k$ for some $k$. Is this already known?

Generalizing, from what I can tell looking at the results of some simple python code. I think the conjecture could be:

Let $p \gt 0$ be any natural with prime factors $p_1, \dots, p_r$, then $|H^{\times}| = p_1^{e_1}\cdots p_r^{e_r} = $ some composition of powers of the prmes making up $p$.

Daniel Donnelly
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2 Answers2

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Yes, in fact, the structure of the group $\mathbb{Z}/(n)^\times$ is known completely (see Wikipedia or the answers to this question; in particular, Hurkyl's answer sketches an elegant proof using the $p$-adic logarithm).

In particular, from these descriptions one can easily deduce the following values for $k$. Let $a_p(n)$ denote the number of distinct prime factors $q$ of $n$ such that $p\mid q-1$. If $p$ is odd, $k=a_p(n)$ if $p^2\not\mid n$, and $k=a_p(n)+1$ if $p^2\mid n$. If $p=2$, $k=a_2(n)$ if $4\not\mid n$, $k=a_2(n)+1$ if $4\mid n$ but $8\not\mid n$, and $k=a_2(n)+2$ if $8\mid n$.

Community
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Eric Wofsey
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Yes, it is. There is a theorem saying that given G is a finite group, G is a p-group iff |G|=p^k for some k. The proof requires Lagrange Theorem and Cauchy's Theorem.

Deepleeqe
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  • Oh, I thought the question was asking what the value of $k$ is. But I see that it could be read this way too... – Eric Wofsey Nov 26 '15 at 04:40