Theorem: Cyclic Group of Order 8 is not isomorphic to Group of Units of Integers Modulo n

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:
=== Lemma ===
{{:Cyclic Group of Order 8 is not isomorphic to Group of Units of Integers Modulo n/Lemma}}{{qed|lemma}}
{{AimForCont}} $U$ 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.
From the {{Lemma|Cyclic Group of Order 8 is not isomorphic to Group of Units of Integers Modulo n}}:
:there are $5$ numbers $n$ with the property that $\map \phi n = 8$, and they are $15$, $16$, $20$, $24$ and $30$.
Hence:
{{begin-eqn}}
{{eqn | l = \paren {\Z / 15 \Z}^\times
      | o = \simeq
      | r = \paren {\Z / 3 \Z}^\times \times \paren {\Z / 5 \Z}^\times
      | c = {{Corollary|Chinese Remainder Theorem}}
}}
{{eqn | o = \simeq
      | r = C_2 \times C_4
      | c = Cyclicity Condition for Units of Ring of Integers Modulo $n$ and $\map \varphi 3 = 2, \map \varphi 5 = 4$
      | n = 1
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {\Z / 16 \Z}^\times
      | o = \simeq
      | r = \paren {\Z / 2^4 \Z}^\times
}}
{{eqn | o = \simeq
      | r = C_2 \times C_4
      | c = Isomorphism between Group of Units Ring of Integers Modulo $2^n$ and $C_2 \times C_{2^{n - 2} }$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {\Z / 20 \Z}^\times
      | o = \simeq
      | r = \paren {\Z / 2^2 \Z}^\times \times \paren {\Z / 5 \Z}^\times
      | c = {{Corollary|Chinese Remainder Theorem}}
}}
{{eqn | o = \simeq
      | r = C_2 \times \paren {\Z / 5 \Z}^\times
      | c = Isomorphism between Group of Units Ring of Integers Modulo $2^n$ and $C_2 \times C_{2^{n - 2} }$
}}
{{eqn | o = \simeq
      | r = C_2 \times C_4
      | c = Cyclicity Condition for Units of Ring of Integers Modulo $n$ and {{EulerPhiLink|5}} $\map \varphi 5 = 4$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {\Z / 24 \Z}^\times
      | o = \simeq
      | r = \paren {\Z / 2^3 \Z}^\times \times \paren {\Z / 3 \Z}^\times
      | c = {{Corollary|Chinese Remainder Theorem}}
}}
{{eqn | o = \simeq
      | r = C_2 \times C_2 \times \paren {\Z / 3 \Z}^\times
      | c = Isomorphism between Group of Units Ring of Integers Modulo $2^n$ and $C_2 \times C_{2^{n - 2} }$
}}
{{eqn | o = \simeq
      | r = C_2 \times C_2 \times C_2
      | c = Cyclicity Condition for Units of Ring of Integers Modulo $n$ and {{EulerPhiLink|3}}: $\map \varphi 3 = 2$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {\Z / 30 \Z}^\times
      | o = \simeq
      | r = \paren {\Z / 2 \Z}^\times \times \paren {\Z / 15 \Z}^\times
      | c = {{Corollary|Chinese Remainder Theorem}}
}}
{{eqn | o = \simeq
      | r = \set 1 \times \paren {\Z / 15 \Z}^\times
      | c = {{EulerPhiLink|2}}: $\map \phi 2 = 1$
}}
{{eqn | o = \simeq
      | r = C_2 \times C_4
      | c = from $(1)$
}}
{{end-eqn}}
Therefore, no multiplicative group of integers mod $n$ is isomorphic to $C_8$.
{{qed}}