# Abstract Nonsense

## 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.