Isomorphism between Group of Units Ring of Integers Modulo 2^n and C2 x C2^(n-2)

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Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer and $n \ge 2$.

Let $R = \struct {\Z / 2^n \Z, +, \times}$ be the ring of integers modulo $2^n$.

Let $U = \struct {\paren {\Z / 2^n \Z}^\times, \times}$ denote the group of units of $R$.

Let $C_2$ be cyclic group of order $2$.

Let $C_{2^{n - 2} }$ be the cyclic group of order $2^{n - 2}$.


Then $U$ is isomorphic to the group direct product of $C_2$ and $C_{2^{n - 2} }$.


Proof


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Lemma

Let $\gen 5$ be the subgroup of $U$ generated by $5$.

Then $-1 \notin \gen 5$.

Proof of lemma

Aiming for a contradiction, suppose that $-1 \in \gen 5$, then

$-1 \equiv 5^k \pmod {2^n}$ for some $k$

then

$1 \equiv 5^{2 k} \pmod {2^n}$

so $2 k$ divides the order of $5$ in $U$ which is $2^{n - 2}$ by Order of $5$ in Units of Ring of Integers Modulo $2^n$.

hence

$k \divides 2^{n - 3}$

hence

$\paren {-1}^{\frac {2^{n - 3}} k} \equiv 5^{2^{n - 3}} \pmod {2^n}$

contradicting with $5^{2^{n-3}} \equiv 1 + 2^{n-1} \pmod {2^n}$ which is proved in Order of $5$ in Units of Ring of Integers Modulo $2^n$.

$\Box$

From the lemma, the elements

$\set {\paren {-1}^e 5^j: e=0,1; j=0,1, \ldots, 2^{n - 2}-1}$

are all distinct.

Since there are $2^{n - 1}$ of these elements and $\order U = 2^{n - 1}$ we have

$U \simeq \set{ \pm 1} \times \gen 5$

By Isomorphism between Ring of Integers Modulo 2 and Parity Ring, $\set{\pm 1} \simeq C_2$

By Order of $5$ in Units of Ring of Integers Modulo $2^n$, $\gen 5 \simeq C_{2^{n - 2} }$

$U \simeq C_2 \times C_{2^{n - 2} }$

$\blacksquare$


Source

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