Find a product of cyclic groups of prime power order isomorphic to the group of units in the ring of integers modulo 2016.

Question

So factoring gets me $2016 = 2^5 \times 3^2 \times 7$.

By the Chinese Remainder Theorem for Rings we have

$(\mathbb{Z}/2016\mathbb{Z})^\times = (\mathbb{Z}/2^5\mathbb{Z})^\times \times (\mathbb{Z}/3^2\mathbb{Z})^\times \times (\mathbb{Z}/7\mathbb{Z})^\times $

Here's whre I get confused. So I know each of these rings has order $\phi(p^\alpha)=p^{(\alpha-1)}(p-1)$ so that the three rings in the direct product have orders 16, 6 and 6, respectively.

But that's where I get stuck. How do I further factor these into a product of cyclic groups of prime order?

The solution gives me: $2016 = 2^5 \times 3^2 \times 7$, so the group of units is the product of the groups of units of the integers mod $2^5$, $3^2$, $7$, which are products of cyclic groups of orders $2$, $8$ and $2$, $3$ and $2$, 3. So the solution is that the group is a product of cyclic groups of orders $2$, $2$, $2$, $3$, $3$, $8$.

I'm really not following at all after the "which are products..." part here...

Answer 1

This problem is so last year!

An Abelian group of order $6$ will be isomorphic to $\Bbb Z/2\Bbb Z \times \Bbb Z/3\Bbb Z$.

You need the structure of $G=(\Bbb Z/2^n\Bbb Z)^\times$. For $n\ge 2$, $5$ always generates a cyclic subgroup of $G$ with order $2^{n-2}$. This always intersects the subgroup $\{1,-1\}$ in just the identity.

Answer 2

We have an isomorphism

\begin{align} \mathbb Z_{2016} &\to \mathbb Z_{32} \oplus \mathbb Z_9 \oplus \mathbb Z_7 \\ \bar x &\mapsto (\bar x, \bar x, \bar x) \end{align}

With inverse isomorphism (Using CRT)

$\begin{align} \mathbb Z_{32} \oplus \mathbb Z_9 \oplus \mathbb Z_7 &\to \mathbb Z_{2016}\\ (\bar u, \bar v, \bar w) &\mapsto \overline{-63u-224v+288w} \end{align} \tag 1$

The coefficients $-63, -224, 288$ behave very much line orthonormal vectors. Check out the multiplication table, modulo 2016, below.

$$\begin{array}{c|rrr|} \times \mod 2016 &-63 &-224 &288 \\ \hline -63 & -63 & 0 & 0 \\ -224 & 0 & -224 & 0 \\ 288 & 0 & 0 & 288 \\ \hline \end{array}$$

This can be used to show that the isomorphism $(1)$ is not just a group isomorphism, it's also a ring isomorphism. It follow that

$$\mathbf U_{2016} \cong \mathbf U_{32} \otimes \mathbf U_9 \otimes \mathbf U_7$$

The following mappings are isomorphisms from additive groups to multiplicative groups of units.

\begin{align} \mathbb Z_2 \oplus\mathbb Z_{8} &\to \mathbf U_{32} \\ (m,n) &\mapsto (-1)^m 3^n \\ (1,0) &\mapsto 31 \\ (0,1) &\mapsto 3 \end{align}

\begin{align} \mathbb Z_6 &\to \mathbf U_9 \\ m &\mapsto 2^m \\ 1 &\mapsto 2 \end{align}

\begin{align} \mathbb Z_6 &\to \mathbf U_7 \\ m &\mapsto 3^m \\ 1 &\mapsto 3 \end{align}

So, a generator of the multiplicative group $\mathbf U_{32} \otimes \mathbf U_9 \otimes \mathbf U_7$ and the corresponding multiplicative group $\mathbf U_{2016}$ would be \begin{align} (31,1,1) &\mapsto 127\\ (3,1,1) &\mapsto 1891 \\ (1,2,1) &\mapsto 1793 \\ (1,1,3) &\mapsto 577 \end{align}

So $\mathbf U_{2016} \cong \langle 127 \rangle \otimes \langle 1891 \rangle \otimes \langle 1793 \rangle \otimes \langle 577 \rangle$

Since the problem asks for prime-power orders, we are not done yet.

We know \begin{align} \operatorname{ord}(127) &= 2 \\ \operatorname{ord}(1891) &= 8 \\ \operatorname{ord}(1793) &= 6 \\ \operatorname{ord}(577) &= 6 \end{align}

and \begin{align} 1793^3 &\equiv 449 \pmod{2016} \\ 1793^2 &\equiv 1345 \pmod{2016} \\ 577^3 &\equiv 1441\pmod{2016} \\ 577^2 &\equiv 289 \pmod{2016} \\ \end{align}

So \begin{align} \operatorname{ord}(127) &= 2 \\ \operatorname{ord}(1891) &= 8 \\ \operatorname{ord}(449) &= 2 \\ \operatorname{ord}(1345) &= 3 \\ \operatorname{ord}(1441) &= 2 \\ \operatorname{ord}(289) &= 3 \end{align}

and $\mathbf U_{2016} \cong \langle 127 \rangle \otimes \langle 1891 \rangle \otimes \langle 449 \rangle \otimes \langle 1345 \rangle \otimes \langle 1441 \rangle \otimes \langle 289 \rangle$