Abstract Nonsense

Crushing one theorem at a time

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.

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

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