Cyclic Group of Order 8 is not isomorphic to Group of Units of Integers Modulo n/Proof 2

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Theorem

Let $n \in \Z_{\ge 0}$ be an integer.

Let $\struct {\Z / n \Z, +, \cdot}$ be the ring of integers modulo $n$.

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

Let $C_8$ denote the cyclic group of order $8$

Then:

$U$ and $C_8$ are not isomorphic.

Proof

Aiming for a contradiction, suppose $U$ and $C_8$ are isomorphic.

$\order U = \order {C_8} = 8$

From Order of Group of Units of Integers Modulo n we have that

$8 = \order U = \map \phi n$

where $\phi$ denotes the Euler $\phi$-function.

$\map \phi 1 = \map \phi 2 = 1$, so $n > 2$.

By Cyclicity Condition for Units of Ring of Integers Modulo n, either:

$n = p^\alpha$

or:

$n = 2 p^\alpha$

where $p \ge 3$ is prime and $\alpha \ge 1$.

In either case, we get

$8 = \map \phi n = p^{\alpha - 1} \paren{p - 1}$

so $p - 1 \divides 8$, but $p \ge 3$, so $p \in \set {3,5}$.

If $p = 3$, we get:

$8 = 3^{\alpha - 1}\times 2$

contradiction.

If $p = 5$, we get:

$8 = 5^{\alpha - 1}\times 4$

contradiction.

$\blacksquare$