Math and science::Algebra::Aluffi
Group Isomorphism
Group isomorphism iff bijection
Let \( \varphi : G \to H \) be a group homomorphism. Then \( \varphi \) is a group isomorphism iff it is a bijection.
Proof
Proof.
Forward. \( \varphi \) is a set function, as group homomorphisms are defined as such. For \( \varphi \) to be an isomorphism, it must be a bijective function.
Reverse. We must show that \( \varphi \) has an inverse, and that its inverse is a group homomorphism from \( H \) to \( G \). \( \varphi \) has an inverse \( \varphi^{-1} \) because it is bijective. What remains is to show that \( \varphi^{-1} \) satisfies:
\[
\forall a,b \in H, \quad
\varphi^{-1}(a \cdot b) = \varphi^{-1}(a) \cdot \varphi^{-1}(b)
\]
Let \( a, b \) be elements of \( H \). Let \( a' = \varphi^{-1}(a) \) and \( b' = \varphi^{-1}(b) \). We then have:
\[
\begin{align}
\varphi^{-1}(a \cdot b) &= \varphi^{-1}(\varphi(a') \cdot \varphi(b')) \\
&= \varphi^{-1}(\varphi(a' \cdot b') \\
&= a' \cdot b' \\
&= \varphi^{-1}(a) \cdot \varphi^{-1}(b) \\
\end{align}
\]
The proof is simple, but not trivial, and in that sense, it is noteworthy.