0
$\begingroup$

Let $p$ be an odd prime number.
I want to prove that $(\mathbf Z/p^r \mathbf Z)^*$ is an cyclic group.
I have known that $\overline {p-1} \in (\mathbf Z/p^r \mathbf Z)^*$ is of order $p^{r-1}$.
Since the order of the group is $(p-1)p^{r-1}$, if we find an element with order $p-1$, we conclude that $(\mathbf Z/p^r \mathbf Z)^*$ must have an element with order $(p-1)p^{r-1}$.

$\because$ Every finite commutive group has an element with order of its exponent.

So how to I find the existence of an element with order $p-1$?

$\endgroup$
1
$\begingroup$

I think it is possible by induction, for $r=1$ it is the classical result I assume to be true.

Take $0\leq x\leq p-1$ such that $x$ is of order $p-1$ in $\mathbb{Z}/p \mathbb{Z}^*$. We will be looking for an element of order $p-1$ in $\mathbb{Z}/p^2 \mathbb{Z}^*$ under the form :

$$x_2=x+\alpha p $$

Where $\alpha$ is to be determined, we have :

$$x_2^{p-1}=x^{p-1}+(p-1)\alpha x^{p-2} p\text{ mod } p^2 $$ $$x_2^{p-1}=x^{p-1}-\alpha x^{p-2} p\text{ mod } p^2 $$

Now because $x$ is of order $p-1$ we have $x^{p-1}=1+t_1p$ hence :

$$x_2^{p-1}=1+(t_1-\alpha x^{p-2}) p\text{ mod } p^2 $$

Hence $\alpha$ must verify the equation $t_1-\alpha x^{p-2}=0$ mod $p$. This is clearly solvable because $x^{p-2}$ can be inversed mod $p$, hence one can construct $\alpha$ such that :

$$x_2:=x+\alpha p\text{ is of order dividing } p-1 $$

Now $x_2=x$ mod $p$ so its order must be divided by the order of $x$ mod $p$ which is $p-1$ hence $x_2$ is exactly of order $p-1$.

Assume now that $x_{r-1}$ is of order $p-1$ in $\mathbb{Z}/p^{r-1} \mathbb{Z}^*$ then define :

$$x_r:=x_{r-1}+\alpha p^{r-1} $$

Likewise, this define an element in $\mathbb{Z}/p^{r} \mathbb{Z}^*$ and :

$$x_r^{p-1}=x_{r-1}^{p-1}+(p-1)\alpha x_{r-1}^{p-2} p^{r-1}\text{ mod } p^r $$

$$x_r^{p-1}=x_{r-1}^{p-1}-\alpha x_{r-1}^{p-2} p^{r-1}\text{ mod } p^r $$

But $x_{r-1}^{p-1}=1+t_{r-1}p^{r-1}$ hence we have that :

$$x_r^{p-1}=1+(t_{r-1}-\alpha x_{r-1}^{p-2}) p^{r-1}\text{ mod } p^r $$

Now the equation in $\alpha$ : $t_{r-1}-\alpha x_{r-1}^{p-2}=0$ mod $p$ is solvable hence one can find $\alpha$ such that $x_r$ is of order dividing $p-1$, now by construction $x_r$ mod $p^{r-1}$ is $x_{r-1}$ hence its order must be divided by $p-1$, finally $x_r$ is an element of order $p-1$ in $\mathbb{Z}/p^r \mathbb{Z}^*$.

$\endgroup$
  • $\begingroup$ I appreciate your answer. It is very good. $\endgroup$ – Jeong Jun 12 '15 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.