3
$\begingroup$

I need to show that the group $(\mathbb{Z}/p^a\mathbb{Z})^{\times}$ is cyclic for odd prime $p$ and for $a\in \mathbb{N}^+$. I have already shown that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ has a generator $g$ and so is cyclic. I have also shown that either $g$ or $g+p$ is a generator for $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$.

How can I proceed to show that $(\mathbb{Z}/p^a\mathbb{Z})^{\times}$ is also cyclic for higher powers?

For the life of me I can't get it.

Thanks for any help you can provide!

  • 0
    Is the edit okay with you?2012-11-09

1 Answers 1

4

I think the clearest way to proceed is to use the following lifting lemma.

Lemma: Let $p$ be an odd prime and let $x$ be coprime to $p$. If for some exponent $k$ we have $x\equiv 1 \pmod{p^k},\ \ \ \ \ \ x^p \not\equiv 1 \pmod{p^{k+1}}$ then we also have $x^p\equiv 1 \pmod{p^{k+1}},\ \ \ \ \ \ x^p \not\equiv 1 \pmod{p^{k+2}}$ Proof: Let us consider the binomial theorem. Since $x\equiv 1 \pmod{p^k}$, there is some $n\in\mathbb{Z}$ such that $x = 1 + np^k$. The second congruence then requires $n\nmid p$. Taking the expression to the $p$th power then gives us $x^p=(1+np)^p = \sum_{i=0}^p\binom{p}{i}n^ip^{ki}$ In the above expression, all terms with $i\ge 2$ are necessarily divisible by $p^{k+2}$, therefore only the first two terms survive the congruence $x^p \equiv 1 + np^{k+1} \not\equiv 1 \pmod{p^{k+2}}$ but clearly, the second term is divisible by $p^{k+1}$ so we have $x^p \equiv 1 \pmod{p^{k+1}}$ The result is as desired. $\square$

From your work with $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$, you know that there is some $g$ such that $g^{p-1} \equiv 1 \pmod p,\ \ \ \ \ \ g^{p-1} \not\equiv 1\pmod{p^2}$ Now use the above lemma to inductively lift the congruence into $p^a$.

  • 0
    Sorry for not markin$g$ correct right away. I was out of town for a while!!2012-11-13