Additive Group and Multiplicative Group of Field are not Isomorphic

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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.

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.

$\blacksquare$


Sources

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