Abstract Nonsense

Crushing one theorem at a time

The Unit and Automorphism Group of Z/nZ (Pt. I)


Point of Post: In this post we find the general form of U\left(\mathbb{Z}/n\mathbb{Z}\right) and \text{Aut}\left(\mathbb{Z}/n\mathbb{Z}\right) in terms of products of cyclic groups.

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Motivation

There are two groups naturally associated to \mathbb{Z}/n\mathbb{Z} one which is important for ring theory and one for group theory. Namely, when dealing with \mathbb{Z}/n\mathbb{Z} as a ring the unit group plays a pivotal role, encompassing a large part of the important information about the multiplicative structure of \mathbb{Z}/n\mathbb{Z}. Similarly, when thinking of \mathbb{Z}/n\mathbb{Z} as an additive group we know that a structure paramount to understanding it is its automorphism group \text{Aut}(\mathbb{Z}/n\mathbb{Z}). Thus, it would be nice to find a nice description for each of these groups. The first step in this process is to prove that finding either U(\mathbb{Z}/n\mathbb{Z}) or \text{Aut}(\mathbb{Z}/n\mathbb{Z}) is sufficient to finding both since, as we shall see, they are isomorphic. Thus, it suffices to find a nice description of U(\mathbb{Z}/n\mathbb{Z}). The pivotal piece of this puzzle was recently discussed in a post on ring theory. Namely, from information from recent posts, in particular the Chinese Remainder Theorem we have that U(\mathbb{Z}/n\mathbb{Z})\cong U(\mathbb{Z}/p_1^{a_1}\mathbb{Z})\times\cdots\times U(\mathbb{Z}/p_m^{a_m}\mathbb{Z}) thus reducing the problem to finding U(\mathbb{Z}/p^a\mathbb{Z}) for primes p. This post picks up where the CRT leaves off by doing just this, finding an explicit description for U(\mathbb{Z}/p^a\mathbb{Z}).

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The Unit Group of \mathbb{Z}/n\mathbb{Z}

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We begin out study with a seemingly simple lemma:

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Lemma: Let p be an odd prime. Then, (1+p)^{p^{n-1}}\equiv 1\text{ mod }p^n and (1+p)^{p^{n-2}}\not\equiv1\text{ mod }p^n for all n\geqslant 2.

Proof: By the Binomial Theorem we have that

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\displaystyle (1+p)^{p^{n-1}}=1+\sum_{j=1}^{p^{n-1}}{p^{n-1} \choose j}p^j\equiv 1+\sum_{j=1}^{n-1}{p^{n-1} \choose j}p^j\text{ mod }p^n

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What we claim now is that p^n\mid {p^{n-1}\choose j}p^j for each j=1,\cdots,n-1. To see this first note that we may rewrite this as

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\displaystyle {p^{n-1} \choose j}p^j=\frac{(p^{n-1})!p^j}{j!(p^{n-1}-j)!}=\frac{p^j p^{n-1}(p^{n-1}-1)\cdots (p^{n-1}-(j-1))}{j(j-1)\cdots 1}=p^n\frac{(p^n-p)\cdots(p^n-p(j-1))}{j(j-1)\cdots 1}

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Now, if we can prove that the fraction in this last expression is an integer we’ll be done. But, since p is a prime the only way that factoring out the power of p could make this not an integer is if the resulting fraction has a numerator which is divisible by a higher power of p than the denominator. To see why this isn’t true we first note that if p^\ell\mid j-k for k=1,\cdots,j-1 then p^{\ell+1}\mid p^n-p(j-k) provided \ell\leqslant n-1. But, evidently \ell\leqslant n-1 for otherwise \ell\geqslant n and thus p^n\leqslant p^\ell\leqslant j-k\leqslant k-1\leqslant n-1 and p^n\mid n-1 is never true for any prime p and n\geqslant 1. Thus, we see that if p^\ell\mid (j-1)\cdots 1 we have that p^{\ell+j-1} divides the numerator, thus if we can show that p^t\nmid j for any t\geqslant j we’ll be done. But, this is clear since p^j>j for every prime p. Thus, p^n\mid{p^{n-1} \choose j}p^j and so

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\displaystyle (1+p)^{p^{n-1}}=1+\sum_{j=1}^{n-1}{p^{n-1} \choose j}p^j\equiv 1\text{ mod }p

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To show that (1+p)^{p^{n-2}}\not\equiv1\text{ mod }p^n we again expand by the Binomial Theorem to get

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\displaystyle (1+p)^{p^{n-2}}\equiv 1+p^{n-1}+\sum_{j=2}^{n-2}{p^{n-2} \choose j}p^j\text{ mod }p^n

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and, similar to before, we claim that p^n\mid{p^{n-2} \choose j}p^j. Indeed, using the same idea it suffices to show that the maximal power of p dividing (p^{n-1}-p)\cdots (p^{n-1}-p(j-1)) is greater than that of j(j-1)\cdots 1. Now, using the same idea we see that p^\ell\mid j-k implies that p^{\ell+1}\mid p^{n-1}-p(j-k) assuming that \ell\leqslant n-2. That said, if this were not true then p^{n-1}\leqslant p^\ell\leqslant k-j\leqslant k-1\leqslant n-2 which is impossible for n\geqslant 2 and primes p. Thus, by an analogous argument to before we see that we may conclude that p^n\mid {p^{n-2}\choose j}p^j as long as we can prove that p^t\nmid j for t\geqslant j-1. That said, it’s trivial to check that p^{j-1}>j for all j\geqslant 2 and odd primes p (this is where we need p=2, because this is false otherwise since 2=2^{2-1}). Thus, we have that p^n\mid {p^{n-2}\choose j}p^j and so

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\displaystyle (1+p)^{p^{n-2}}\equiv 1+p^{n-1}+\sum_{j=2}^{n-2}{p^{n-2} \choose j}p^j\equiv 1+p^{n-1}\not\equiv1\text{ mod }p^n

\blacksquare

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Corollary: For odd primes p and n\geqslant 1 one has that 1+p has order p^{n-1} in \mathbb{Z}/p^n\mathbb{Z}

Proof: We merely note that by the previous lemma we have that |1+p|\mid p^{n-1} but that |1+p|\nmid p^{n-2} (since otherwise (1+p)^{p^{n-2}} would be 1 modulo p^n) and so evidently |1+p|=p^{n-1}. \blacksquare

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We are now able to complete the first ‘half’ of identifying U(\mathbb{Z}/p^n\mathbb{Z})

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Theorem: Let p be an odd prime, then U\left(\mathbb{Z}/p^n\mathbb{Z}\right)\cong\mathbb{Z}/p^{n-1}(p-1)\mathbb{Z}.

Proof: The result has already been proven for n=1 and so it suffices to prove this for n\geqslant 2. To see this note that there is a natural morphism f:U\left(\mathbb{Z}/p^n\mathbb{Z}\right)\to U(\mathbb{Z}/p\mathbb{Z}) which is just the reduction map. This is an epimorphism since f(a)=a for each a\in\mathbb{Z}/p\mathbb{Z} (where obviously the two a‘s are ‘different’). We thus have by the first isomorphism theorem that |\ker f|=p^{n-1} and since 1+p\in\ker f and |1+p|=p^{n-1} we must have that \ker f=\langle 1+p\rangle. Now, choose a\in U(\mathbb{Z}/p^n\mathbb{Z}) such that f(a) has order p-1 in U(\mathbb{Z}/p\mathbb{Z}) (we can do this since U(\mathbb{Z}/p\mathbb{Z}) is cyclic of order p-1 and f is surjective). We then have from elementary group theory that p-1\mid |a|. So, note that since a^{p-1}\in \ker f we have that a^{p-1}=(1+p)^j for some j. Now, since (p-1,p^{n-1})=1 we have that p-1 is invertible in \mathbb{Z}/p^n\mathbb{Z}, so let q=(p-1)^{-1} and let x=-jq. Note then that

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(a(1+p)^x)^{p-1}=a^{p-1}(1+p)^{-(p-1)jq}=(1+p)^{-(p-1)jq+j}=1

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since -(p-1)jq+j\equiv -j+j\equiv 0\text{ mod }p^{n-1}. So, \left|a(1+p)^x\right|\mid p-1 but since f(a(1+p)^x)=f(a) we have by the same argument as before that p-1\mid |a(1+p)^x| and so |a(1+p)^x|=p-1. Note then that since (p-1,p^{n-1})=1 and a(1+p)^x and 1+p commute we have, from basic group theory, that |a(1+p)^{x+1}|=\text{lcm}(p-1,p^{n-1})=p^{n-1}(p-1). And, since |U(\mathbb{Z}/p^n\mathbb{Z})|=\varphi(p^n)=p^{n-1}(p-1) (where \varphi is Euler’s totient function). The conclusion follows. \blacksquare

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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September 10, 2011 - Posted by | Algebra, Group Theory | , , , , , , , , ,

2 Comments »

  1. […] The Unit and Automorphism Group of Z/nZ (pt. II) Point of Post: This is a continuation of this post. […]

    Pingback by The Unit and Automorphism Group of Z/nZ (pt. II) « Abstract Nonsense | September 10, 2011 | Reply

  2. […] the primes so that this is true) we then have by assumption that is cyclic that (recalling theclassification of the automorphism groups of cyclic groups). But, by assumption this evidently implies that and so […]

    Pingback by A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. II) « Abstract Nonsense | September 13, 2011 | Reply


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