# Abstract Nonsense

## The Unit and Automorphism Group of Z/nZ (pt. II)

Point of Post: This is a continuation of this post.

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We now have to deal with the case $p=2$ separately.

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Theorem: Let $n\in\mathbb{N}$ be arbitrary. Then, $U(\mathbb{Z}/2\mathbb{Z})\cong\{e\}$, $U(\mathbb{Z}/4\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$ and for $n\geqslant 3$ one has that $U(\mathbb{Z}/2^n\mathbb{Z})\cong \left(\mathbb{Z}/2^{n-2}\mathbb{Z}\right)\times(\mathbb{Z}/2\mathbb{Z})$

Proof: The cases for $n=1,2$ are trivial by just counting the order of the respective groups (since there is only one group of order $1$ and likewise for $2$). So, let $n\geqslant 3$, we claim that $5^{2^{n-3}}\cong 1+2^{n-1}\text{ mod }2^n$. Indeed, it’s trivially true for $n=3$ and if we assume it’s true for $n$ then we see that

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$1+2^n\equiv 1+2^n+2^{2n-2}=(1+2^{n-1})^2=5^{2^{n-2}}\text{ mod }2^{n+1}$

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where we used the easily verifiable fact that $a\equiv b\text{ mod }p^\ell$ implies that $a^p\equiv b^p\text{ mod }p^{\ell+1}$ for primes $p$ (just apply the Binomial Theorem). Thus, the induction is complete. But,  this clearly implies that $5^{2^{n-2}}\equiv 1\text{ mod }2^n$ and $5^{2^{n-3}}\not\equiv1\text{ mod }2^n$ and so (using similar logic to the case for odd primes) we have that $|5|=2^{n-2}$. So, now consider the reduction map $f:U(\mathbb{Z}/2^n\mathbb{Z})\to U(\mathbb{Z}/4\mathbb{Z})$. Note that since this is obviously surjective we have that $|\ker f|=2^{n-2}$ and since $5\in ker f$ we have that $\ker f=\langle 5\rangle$. Now, note that $\{-1,1\}$ are a subgroup of $U(\mathbb{Z}/2^n\mathbb{Z})$ which are disjoint from $\langle 5\rangle$ (they are not in $\ker f$) and so the product $\{-1,1\}\langle 5\rangle$ is a subgroup of $U(\mathbb{Z}/2^n\mathbb{Z})$ of order $2^{n-1}$ and so equal to $U(\mathbb{Z}/2^n\mathbb{Z})$. It follows then from basic group theory that $U(\mathbb{Z}/2^n\mathbb{Z})\cong \langle 5\rangle\times\{-1,1\}\cong (\mathbb{Z}/2^{n-2}\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ as desired. $\blacksquare$

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With this and the Chinese Remainder Theorem we may thus finally conclude that

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Theorem: Let $p$ be a prime number and $n\in\mathbb{N}$. Then,

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$U(\mathbb{Z}/p^n\mathbb{Z})\cong\begin{cases}\mathbb{Z}/p^{n-1}(p-1) & \mbox{if}\quad p\text{ is odd}\\ \{e\} &\mbox{if}\quad p=2\text{ and }n=1\\ \mathbb{Z}_2 & \mbox{if}\quad p=2\text{ and }n=2\\ (\mathbb{Z}/2^{n-2}\mathbb{Z})\times (\mathbb{Z}/2\mathbb{Z}) &\mbox{if}\quad p=2\text{ and }n\geqslant 3\end{cases}$

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and if $m\in\mathbb{Z}$ has prime decomposition $N=p_1^{a_1}\cdots p_m^{a_m}$ then

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$U\left(\mathbb{Z}/N\mathbb{Z}\right)\cong U(\mathbb{Z}/p_1^{a_1})\times\cdots\times U(\mathbb{Z}/p_m^{a_m}\mathbb{Z})$

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The Automorphism Group of $\mathbb{Z}/n\mathbb{Z}$

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So, how does this help us figure out what $\text{Aut}(\mathbb{Z}/n\mathbb{Z})$ is? Well, the following theorem gives us a hint:

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Theorem: Let $n\in\mathbb{N}$. Then, $\text{Aut}\left(\mathbb{Z}/n\mathbb{Z}\right)\cong U\left(\mathbb{Z}/n\mathbb{Z}\right)$.

Proof: Define $f:U(\mathbb{Z}/n\mathbb{Z})\to \text{Aut}\left(\mathbb{Z}/n\mathbb{Z}\right)$ by $f_a(z)=az$ for $z\in\mathbb{Z}/n\mathbb{Z}$. We begin by showing that this is a well-defined mapping (in the sense that $f_a\in\text{Aut}(\mathbb{Z}/n\mathbb{Z})$ for each $a\in\mathbb{Z}/n\mathbb{Z})$. To see that $f_a$ is an automorphism we first note that $f_a(b+c)=ab+ac=f_a(b)+f_a(c)$ so that $f$ is a homomorphism, it’s an injection since $f_a(b)=f_a(c)$ implies $ab=ac$ and since $a$ is invertible this implies $b=c$, and it’s surjective since for any $b\in\mathbb{Z}/n\mathbb{Z}$ we have that $f_a(a^{-1}b)=b$ (note, we already knew it was surjective since $f_a$ is an injection from a finite set to itself, from where the rest follows from basic set theory).

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Now, to show that $f$ itself is an isomorphism. To see that $f$ is a homomorphism we just compute that $f_{ab}(c)=(ab)c=a(bc)=f_a(bc)=f_a(f_b(c))$ and since $c$ was arbitrary we thus have that $f_{ab}=f_a\circ f_b$ and so a homomorphism. Now, injectivity of $f$ is trivial since $f_a=f_b$ implies $a=f_a(1)=f_b(1)=b$.  To prove surjectivity let $\pi\in\text{Aut}\left(\mathbb{Z}/n\mathbb{Z}\right)$ we claim that $\pi=f_{\pi(1)}$. Indeed, $\pi(a)=\pi(a\cdot 1)=a\cdot\pi(1)=a\pi(1)$ (where we’ve used $\cdot$ to distinguish the $a$-fold sum of an element) from where the conclusion follows. $\blacksquare$

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Remark: The more sophisticated, shorter proof of the above (which we spared for those who don’t know as much ring theory) is that, by definition, $\text{Aut}(A)=U(\text{End}(A))$ for an abelian group $A$, where $A$ is the endomorphism ring but it is easy to show that $\text{End}(A)\cong A$ if $A$ is cyclic.

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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.