Theorem
Prove that the mapping \(f: U_{16} \to U_{16}\) sending \(x \to x^3\) is an automorphism.
Proof
Step 1
Let us define the mapping
Define \(f: U_{16} \to U_{16}\) by
\[
f(x) = x^3.
\]
Since \(U_{16}\) consists of the units modulo 16, for any \(x \in U_{16}\), \(x^3\) is also a unit modulo 16.
Step 2
To prove \(f\) is well-defined
We must verify that if \(x \in U_{16}\), then \(x^3 \in U_{16}\). Since \(x\) is a unit modulo 16, we have \(\gcd(x, 16) = 1\). Raising \(x\) to the third power preserves this property, so
\[
\gcd(x^3, 16) = 1.
\]
Thus, \(x^3 \in U_{16}\) and the mapping \(f\) is well-defined.
Step 3
To prove \(f\) is a homomorphism
To prove that \(f\) is a homomorphism, let \(x,y \in U_{16}\). Then:
\begin{align}
f(xy) &= (xy)^3 \nonumber\\
&= x^3 y^3 \quad \text{(since \(U_{16}\) is abelian)} \nonumber\\
&= f(x) f(y)\nonumber
\end{align}
Step 4
To prove \(f\) is a injective
Suppose \(f(x) = f(y)\) for \(x, y \in U_{16}\). Then,
\begin{align}
x^3 &= y^3 \nonumber\\
\Rightarrow (xy^{-1})^3 &= 1\nonumber
\end{align}
Thus, the kernel of \(f\) is given by:
\begin{align}
\ker f &= \{ x \in U_{16} \mid x^3 = 1 \}\nonumber
\end{align}
By verifying the elements of \(U_{16} = \{1, 3, 5, 7, 9, 11, 13, 15\}\), one finds that the only solution to \(x^3 = 1 \mod 16\) is \(x = 1\). Hence, \(\ker f = \{1\}\) and \(f\) is injective.
Step 5
To prove \(f\) is a surjective
Here, \(U_{16}\) is a finite group. We have already shown that the mapping
\[
f: U_{16} \to U_{16}, \quad f(x) = x^3,
\]
is an injective homomorphism. Injectivity means that no two distinct elements in \(U_{16}\) are mapped to the same element.
Since the domain and codomain of \(f\) are the same finite set, the injectivity of \(f\) guarantees that every element in \(U_{16}\) must be hit by \(f\). In other words, the function \(f\) is also surjective.
This is a standard result in finite set theory: an injective function on a finite set is a bijection.
Therefore, \(f\) is surjective.
