8
$\begingroup$

I would like to prove that the group $(\mathbb{Z}/p^r\mathbb{Z})^{\ast}$ of the invertible elements of $\mathbb{Z}/p^r\mathbb{Z}$ with $p>2$ prime and $r>0$ is cyclic.

My text suggests to start proving that the kernel $W$ of the canonical homomorphism $(\mathbb{Z}/p^r\mathbb{Z})^{\ast}\to(\mathbb{Z}/p\mathbb{Z})^{\ast}$ is a cyclic group by verifying that $1+p$ has order $p^{r-1}$ in $W$.

I suppose, but I am not sure, that the said canonical homomorphism might be the projection $(\mathbb{Z}/p^r\mathbb{Z})^{\ast}\to(\mathbb{Z}/p\mathbb{Z})^{\ast}:a\mapsto\bar{a}$, and I would say that the kernel of the map is $\{\bar{1}, \overline{1+p},\overline{1+2p}...,\overline{1+(p^{r-1}-1)p}\}$, so it has order $p^{r-1}$ (and therefore if the group generated by $1+p$ has the same order, it must be the kernel itself).

But I haven't been able to prove that $p^{r-1}$ is the least natural number $m$ such that $(\overline{1+p})^{m}=\bar{1}$...

Furthermore, once proved that $W\subset(\mathbb{Z}/p^r\mathbb{Z})^{\ast}$ is cyclic, I don't know how to see that $(\mathbb{Z}/p^r\mathbb{Z})^{\ast}$ is cyclic too... Has anybody got any ideas?

I $\infty$-ly thank you in advance!!!

$\endgroup$
  • 1
    $\begingroup$ I broke up your text a bit (inserting line breaks) to make it a tad more readable. I hope you don't mind. $\endgroup$ – Namaste Dec 18 '13 at 14:08
  • 1
    $\begingroup$ See mathreference.com/num-mod,unm.html for a direct proof that $(\mathbb{Z}/p^r)^*$ is cyclic. If $[\zeta]$ is a generator of $(\mathbb{Z}/p)^*$, then $[\zeta+p]$ is a generator of $(\mathbb{Z}/p^r)^*$. $\endgroup$ – Martin Brandenburg Dec 18 '13 at 14:29
  • 1
    $\begingroup$ See www-groups.mcs.st-and.ac.uk/~neunhoef/Teaching/ff2013/… - Theorem 6.9 - Page 25. It may be help. $\endgroup$ – AHH Dec 18 '13 at 17:00
  • 2
    $\begingroup$ If you expand $(1+p)^m = 1 + \binom m1 p + \binom m2 p^2 + \cdots$ in a binomial expansion, you can show that the power of $p$ dividing $(1+p)^m-1$ is exactly the same as the power of $p$ dividing $\binom m1 p$. That will show you that $p^{r-1}$ is the least such natural number. After that, you should look to general group theory arguments - something like, a finite abelian group with a large cyclic subgroup must itself be cyclic. (By the way, there's an entirely different way to show that these groups are cyclic - see "primitive roots" in number theory.) $\endgroup$ – Greg Martin Dec 18 '13 at 18:21
  • $\begingroup$ at amWhy: thank you for editing! I am a beginner with formatting here... and Chag Urim sameach! at Martin: very interesting proof and site. at AHH: downloaded the paper, so interesting, but not only because of $\mathbb{F}_p$. at Greg: very elegant way to see that $p^k|(1+p)^m-1\iff p^k|mp$, useful to understand some points behind the mathreference proof. Thank you a lot everybody!!!! $\endgroup$ – Self-teaching worker Dec 18 '13 at 21:54
2
$\begingroup$

For $r=1$, you can see that $\big(\Bbb Z\big/p\Bbb Z\big)^\times$ is cyclic of order $p-1$ by realizing it as the multiplicative group of the finite field $\Bbb Z\big/p\Bbb Z$.

Now suppose $r > 1$, so that $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is an abelian group of order $p^{r-1}(p-1)$. Factor $p-1$ into primes as $$ p-1 = q_1^{s_1}\dotsb q_t^{s_t}. $$

We will prove that each Sylow subgroup of $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is cyclic and use the following lemma:

Lemma. Let $G$ be a finite group, and let $P_1,\dots,P_r$ be its nontrivial Sylow subgroups. If each $P_i$ is a normal subgroup of $G$, then $$ G \cong P_1\times \dotsb \times P_r. $$

Since $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is an abelian group, each of its Sylow subgroups is normal, so we will be able to use the Chinese Remainder Theorem (CRT) to conclude that $$ \big(\Bbb Z\big/p^r\Bbb Z\big)^\times \cong \Bbb Z\big/p^{r-1}\Bbb Z\times \Bbb Z\big/q_1^{s_1}\Bbb Z\times\dotsb\times\Bbb Z\big/q_t^{s_t}\Bbb Z \underbrace{\cong}_\text{CRT} \Bbb Z\big/p^{r-1}q_1^{s_1}\dotsb q_t^{s_t}\Bbb Z, $$ i.e. that $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is cyclic of order $p^{r-1}q_1^{s_1}\dotsb q_t^{s_t} = p^{r-1}(p-1)$, as desired.

Proof sketch that the Sylow subgroups are cyclic:

First show that the Sylow $p$-subgroup of $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is cyclic as follows. Use the Binomial Theorem and induction on $r$ to show that if $p$ is an odd prime then $(1+p)^{p^{r-1}}\equiv 1\bmod p^r$ but $(1+p)^{p^{r-2}}\not\equiv 1\bmod p^r$. Deduce that $1+p$ is an element of order $p^{r-1}$ in $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$. (Recall that if $g^n = 1$, then the order of $g$ divides $n$.) Now we know that the Sylow $p$-subgroup of $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is cyclic.

Consider the canonical homomorphism $\varphi\colon\big(\Bbb Z\big/p^r\Bbb Z\big)^\times\to \big(\Bbb Z\big/p\Bbb Z\big)^\times$ defined by $$ a + (p^r)\mapsto a+(p). $$ Note that $\varphi$ is a surjection, and that it is indeed the map you specified. By the first isomorphism theorem, $$ \frac{p^{r-1}(p-1)}{|\!\ker\varphi|} = \frac{\big|\big(\Bbb Z\big/p^r\Bbb Z\big)^\times\big|}{|\!\ker\varphi|} = \big|\big(\Bbb Z\big/p\Bbb Z\big)^\times\big|=p-1 \implies |\!\ker\varphi| = p^{r-1}, $$ hence the kernel of $\varphi$ is precisely the Sylow $p$-subgroup of $\big(\Bbb Z/p^r\Bbb Z\big)^\times$. For each prime $q\ne p$ dividing the order of $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$, the Sylow $q$-subgroup of $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ therefore maps injectively into $\big(\Bbb Z\big/p\Bbb Z\big)^\times$ under $\varphi$, and can thereby be identified with a (cyclic) subgroup of the cyclic group $\big(\Bbb Z\big/p\Bbb Z\big)^\times$. Hence we have shown that every Sylow subgroup of $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is cyclic. $\qquad\blacksquare$

Thus, by the lemma, we may write $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ as a direct product of its Sylow subgroups, each of which we know to be cyclic: $$ \big(\Bbb Z\big/p^r\Bbb Z\big)^\times \cong \Bbb Z\big/p^{r-1}\Bbb Z\times\Bbb Z\big/q_1^{s_1}\Bbb Z\times\dotsb\times\Bbb Z\big/q_t^{s_t}\Bbb Z. $$

By the Chinese Remainder Theorem, since all the primes $p,q_i$ are distinct, we have $$ \big(\Bbb Z\big/p^r\Bbb Z\big)^\times \cong \Bbb Z\big/p^{r-1}q_1^{s_1}\dotsb q_t^{s_t}\Bbb Z, $$ i.e., that $\big(\Bbb Z\big/p^r\Bbb Z\big)^\times$ is cyclic, as desired.

$\endgroup$
  • $\begingroup$ Where did you use the assumption $p>2$? $\endgroup$ – Pierre-Yves Gaillard Jul 7 '17 at 15:36
  • $\begingroup$ @Pierre-Yves Gaillard If $r\ge 3$, then $(\Bbb Z/2^r\Bbb Z)^\times$ is not cyclic. Indeed, it has two distinct subgroups of order 2. [cf. Dummit and Foote's Abstract Algebra, Exercise 23 of Section 2.3] $\endgroup$ – Alex Ortiz Jul 7 '17 at 15:45
  • $\begingroup$ @Pierre-YvesGaillard I realize my previous comment doesn't really answer your question. The assumption $p>2$ is needed to show that the order of $1+p$ is $p^{r-1}$ in $G$. $\endgroup$ – Alex Ortiz Jul 7 '17 at 15:56
  • $\begingroup$ Very elegant. I have been searching for a proof and I couldn't have found a better one. +1 $\endgroup$ – Cauchy Aug 14 '17 at 6:59
1
$\begingroup$

This is taken from Alan Baker's A Concise Introduction to the Theory of Numbers.

Let $p$ be an odd prime, and let $g$ be a primitive root ($\operatorname{mod}p$). We prove now that there exists an integer $x$ such that $g':=g+px$ is a primitive root ($\operatorname{mod}p^j$) for all prime powers $p^j$. We have $g^{p-1}=1+py$ for some integer $y$ and so, by the binomial theorem, $g'^{p-1}=1+pz$, where $$ z\equiv y+(p-1)g^{p-2}x\ (\operatorname{mod}p). $$ The coefficient of $x$ is not divisible by $p$ and so we can choose $x$ such that $(z,p)=1$. Then $g'$ has the required property. For suppose that $g'$ has order $d\ (\operatorname{mod}p^j)$. Then $d$ divides $\phi(p^j)=p^{j-1}(p-1)$. But $g'$ is a primitive root ($\operatorname{mod}\ p$) and thus $p-1$ divides $d$. Hence $d=p^k(p-1)$ for some $k < j$. Further, since is $p$ odd, we have $$ (1+pz)^{p^k}=1+p^{k+1}z_k, $$ where $(z_k,p)=1$. Now since $g'^d\equiv1\ (\operatorname{mod}p^j)$ it follows that $j=k+1$ and this gives $d=\phi(p^j)$, as required.

$\endgroup$

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.