Math and science::Algebra::Aluffi

Group Isomorphism

Group isomorphism iff bijection

Let $φ : G \to H$ be a group homomorphism. Then $φ$ is a group isomorphism iff it is a bijection.

Proof

Proof.

Forward. $φ$ is a set function, as group homomorphisms are defined as such. For $φ$ to be an isomorphism, it must be a bijective function.

Reverse. We must show that $φ$ has an inverse, and that its inverse is a group homomorphism from $H$ to $G$ . $φ$ has an inverse $φ^{- 1}$ because it is bijective. What remains is to show that $φ^{- 1}$ satisfies:

\forall a, b \in H, φ^{- 1} (a \cdot b) = φ^{- 1} (a) \cdot φ^{- 1} (b)

Let $a, b$ be elements of $H$ . Let $a^{'} = φ^{- 1} (a)$ and $b^{'} = φ^{- 1} (b)$ . We then have:

\begin{aligned} φ^{- 1} (a \cdot b) & = φ^{- 1} (φ (a^{'}) \cdot φ (b^{'})) \\ = φ^{- 1} (φ (a^{'} \cdot b^{'})) \\ = a^{'} \cdot b^{'} \\ = φ^{- 1} (a) \cdot φ^{- 1} (b) \end{aligned}

The proof is simple, but not trivial, and in that sense, it is noteworthy.

Source

Aluffi, p66