Additive Group of Complex Numbers is Direct Product of Reals with Reals

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Theorem

Let $\struct {\C, +}$ be the additive group of complex numbers.

Let $\struct {\R, +}$ be the additive group of real numbers.


Then the direct product $\struct {\R, +} \times \struct {\R, +}$ is isomorphic with $\struct {\C, +}$.


Proof

We need to show that $\phi: \tuple {x, y} \mapsto x + y i$ is a group isomorphism.


$\phi$ is a group homomorphism

\(\displaystyle \forall a, b, c, d \in \R: \ \ \) \(\displaystyle \map \phi {a, b} + \map \phi {c, d}\) \(=\) \(\displaystyle \paren {a + b i} + \paren {c + d i}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + c} + \paren {b + d} i\) Complex Addition is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {a + c, b + d}\)

$\Box$


$\phi$ is bijective

We show that $\phi^{-1}: z \mapsto \tuple {\map \Re z, \map \Im z}$ is the inverse of $\phi$.

We have:

$\map \phi {\map {\phi^{-1} } z} = \map \phi {\map \Re z, \map \Im z} = \map \Re z + i \map \Im z = z$
$\map {\phi^{-1} } {\map \phi {a, b} } = \map {\phi^{-1} } {a + b i} = \tuple {a, b}$

hence $\phi^{-1}$ is the inverse of $\phi$.

$\Box$


Hence $\phi$ is a group isomorphism, and thus the direct product $\struct {\R, +} \times \struct {\R, +}$ is isomorphic with $\struct {\C, +}$.

$\blacksquare$


Sources