Multiplication modulo powers of primes

The main goal of this post is to summarise a few fundamental statements for working with groups of type \mathbb{Z}_{p^e}^*. It is going to be very technical and maybe difficult to learn for the first time you see it, but give it a chance and you will see that it actually makes sense.


Drapal part 2.10


First, let us recall some consequences of the Chinese remainder theorem: Let n=n_1\cdot\ldots\cdot n_k, such that n_i’s are pairwise coprime. Then

  • \mathbb{Z}_n\cong\mathbb{Z}_{n_1}\times\ldots\times\mathbb{Z}_{n_k}.
  • \mathbb{Z}_n^*\cong\mathbb{Z}_{n_1}^*\times\ldots\times\mathbb{Z}_{n_k}^*.

Except of the Chinese remainder theorem, we have two more new (rather technical) propositions dealing with properties of groups \mathbb{Z}_{p^e}^*, namely with the group structure and elements of order 2:

\textbf{Lemma:} Let p be a prime number and let e\in\mathbb{N}, e\geq 2. Then

  • Set P=\{1+a\cdot p|\ 0\leq a<p^{e-1}\}. Then P is a subgroup of \mathbb{Z}_{p^e}^* of order p^{e-1}.
  • An element p^e-1 is of order 2 in \mathbb{Z}_{p^e}^*.
  • If p=2, then 2^{e-1}-1 and 2^{e-1}+1 are also of order 2 in \mathbb{Z}_{2^e}^*.

\textbf{Theorem:} Again, let p be a prime number and let e\in\mathbb{N}, e\geq 2. Then

  • If p=2, e\geq 3, then \mathbb{Z}_{2^e}^*\cong\mathbb{Z}_2\times \mathbb{Z}_{2^{e-2}}^*.
  • If p=2, e=2, then \mathbb{Z}_{p^e}^*=\mathbb{Z}_4^*=\{1,3\}\cong \mathbb{Z}_2.
  • Finally, if p is odd, then \mathbb{Z}_{p^e}^* is a cyclic group of order \varphi(p^e)=(p-1)\cdot p^{e-1}, i.e., \mathbb{Z}_{p^e}^* \cong\mathbb{Z}_{(p-1)\cdot p^{e-1}}\cong\mathbb{Z}_{p-1}\times\mathbb{Z}_{p^{e-1}}.

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