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
$\phi: \struct {F_{\ne 0_F}, \times} \to \struct {F, +}$ 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:

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

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

Then:

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

From Kernel is Trivial iff 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.