# Additive Group and Multiplicative Group of Field are not Isomorphic

## Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $\struct {F, +}$ denote the additive group of $F$.

Let $\struct {F_{\ne 0_F}, \times}$ denote the multiplicative group of $F$.

Then $\struct {F, +}$ and $\struct {F_{\ne 0_F}, \times}$ are not isomorphic to each other.

## Proof

Aiming for a contradiction, suppose $\phi: \struct {F_{\ne 0_F}, \times} \to \struct {F, +}$ is an isomorphism.

By definition:

$0_F$ is the identity of $\struct {F, +}$

and

$1_F$ is the identity of $\struct {F_{\ne 0_F}, \times}$.

We have that:

 $\ds 0_F$ $=$ $\ds \map \phi {1_F}$ Epimorphism Preserves Identity $\ds$ $=$ $\ds \map \phi {\paren {-1_F} \times \paren {-1_F} }$ $\ds$ $=$ $\ds \map \phi {-1_F} + \map \phi {-1_F}$ Definition of Group Isomorphism

and so by definition $F$ has characteristic $2$.

Let $x \in \struct {F_{\ne 0_F}, \times}$.

Then:

 $\ds \map \phi {x^2}$ $=$ $\ds \map \phi x + \map \phi x$ Definition of Group Isomorphism $\ds$ $=$ $\ds 0_F$ as $F$ has characteristic $2$

As $\phi$ is an isomorphism, it is also a monomorphism.

$\map \ker \phi = \set {1_F}$

and so $x^2 = 1$.

Thus $x = 1$ and so $\order F = 2$.

Thus $\order {F_{\ne 0_F} } = 1$.

So $\phi$ is a bijection from a set of cardinality $1$ to a set of cardinality $2$.

So $\phi$ cannot be a bijection and so cannot be an isomorphism.

Hence the result by Proof by Contradiction.

$\blacksquare$